import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.NoSuchElementException;

public class Main {
	public static void main(String args[]) {
		FastScanner scanner = new FastScanner();
		int n=scanner.nextInt();
		List<Integer>p=new ArrayList<>();
		List<Integer>m=new ArrayList<>();
		long ans=0;
		for(int i=0;i<n;i++){
			int a=scanner.nextInt();
			ans+=Math.abs(a);
			if(a>=0){
				p.add(a);
			}else{
				m.add(-a);
			}
		}
		if(m.size()%2==0){
			System.out.println(ans);
			return;
		}
		Collections.sort(p);
		Collections.sort(m);
		if(p.get(0)<m.get(0)){
			p.set(0,-p.get(0));
		}else{
			m.set(0, -m.get(0));
		}
		ans=0;
		for(int i=0;i<p.size();i++){
			ans+=p.get(i);
		}
		for(int i=0;i<m.size();i++){
			ans+=m.get(i);
		}
		System.out.println(ans);
	}
}

class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() { return Double.parseDouble(next());}
}


import java.io.*;
import java.util.Map;

/**
 * Copyright © 2018 Chris. All rights reserved.
 *
 * @author Chris
 * 2018/7/9 15:33
 * @see format
 */
public class Main {

    private static BufferedReader br;
    private static StreamTokenizer st;
    private static PrintWriter pw;

    static final int INF = 1000000007;
    static int MOD = 1000000007;

    static int[] getBit(long num) {
        int bit[] = new int[40];
        int cur = 0;
        while (num > 0) {
            bit[cur++] = (int) (num & 1);
            num >>= 1;
        }
        return bit;
    }

    private static long[][] multiMatrix(long a[][], long b[][]) {
        long c[][] = new long[2][2];
        for (int i = 0; i < 2; i++) {
            for (int j = 0; j < 2; j++) {
                for (int k = 0; k < 2; k++) {
                    c[i][j] = (c[i][j] + a[i][k] * b[k][j]) % MOD;
                }
            }
        }
        return c;
    }

    private static long[][] powMatrix(long a[][], int k) {
        if (k == 1) {
            return a;
        }
        int r = k % 2;
        long c[][] = powMatrix(a, k / 2);
        c = multiMatrix(c, c);
        if (r == 1) {
            c = multiMatrix(a, c);
        }
        return c;
    }

    private static void solve() throws IOException {
        int n = nextInt();
        int a[] = new int[n];
        int cnt = 0;
        long sum = 0;
        int min = Integer.MAX_VALUE;
        for (int i = 0; i < n; i++) {
            a[i] = nextInt();
            sum += Math.abs(a[i]);
            min = Math.min(min, Math.abs(a[i]));
            if (a[i] < 0) {
                cnt++;
            }
        }
        if (cnt % 2 == 1) {
            sum -= min << 1;
        }
        pw.println(sum);
    }

    static void swap(long a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void swap(int a[], int i, int j) {
        a[i] ^= a[j];
        a[j] ^= a[i];
        a[i] ^= a[j];
    }

    static void getDiv(Map<Integer, Integer> map, int n) {
        int sqrt = (int) Math.sqrt(n);
        for (int i = sqrt; i >= 2; i--) {
            if (n % i == 0) {
                getDiv(map, i);
                getDiv(map, n / i);
                return;
            }
        }
        map.put(n, map.getOrDefault(n, 0) + 1);
    }

    public static boolean[] generatePrime(int n) {
        boolean p[] = new boolean[n + 1];
        p[2] = true;

        for (int i = 3; i <= n; i += 2) {
            p[i] = true;
        }

        for (int i = 3; i <= Math.sqrt(n); i += 2) {
            if (!p[i]) {
                continue;
            }
            for (int j = i * i; j <= n; j += i << 1) {
                p[j] = false;
            }
        }
        return p;
    }

    static long llMod(long a, long b, long mod) {
        return (a * b - (long) ((double) a / mod * b + 0.5) * mod + mod) % mod;
    }

    static int pow(long a, long n) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = (ans * a) % MOD;
            }
            a = (a * a) % MOD;
            n >>= 1;
        }
        return (int) ans;
    }

    static int pow(long a, long n, long mod) {
        long ans = 1;
        while (n > 0) {
            if ((n & 1) == 1) {
                ans = llMod(ans, a, mod);
            }
            a = llMod(a, a, mod);
            n >>= 1;
        }
        return (int) ans;
    }

    private static long[][] initC(int n) {
        long c[][] = new long[n][n];

        for (int i = 0; i < n; i++) {
            c[i][0] = 1;
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j <= i; j++) {
                c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
            }
        }
        return c;
    }

    /**
     * ps: n >= m, choose m from n;
     */
    private static int c(long n, long m) {
        if (m > n) {
            n ^= m;
            m ^= n;
            n ^= m;
        }
        m = Math.min(m, n - m);

        long top = 1;
        long bot = 1;
        for (long i = n - m + 1; i <= n; i++) {
            top = (top * i) % MOD;
        }
        for (int i = 1; i <= m; i++) {
            bot = (bot * i) % MOD;
        }

        return (int) ((top * pow(bot, MOD - 2)) % MOD);
    }

    static int gcd(int a, int b) {
        if (a < b) {
            a ^= b;
            b ^= a;
            a ^= b;
        }
        while (b != 0) {
            int tmp = a % b;
            a = b;
            b = tmp;
        }
        return a;
    }

    static boolean even(long n) {
        return (n & 1) == 0;
    }

    public static void main(String args[]) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        st = new StreamTokenizer(br);
        pw = new PrintWriter(new OutputStreamWriter(System.out));
        st.ordinaryChar('\''); //指定单引号、双引号和注释符号是普通字符
        st.ordinaryChar('\"');
        st.ordinaryChar('/');

        solve();
        pw.flush();
    }

    static String next(int len) throws IOException {
        int b = br.read();
        while (b == '\n' || b == ' ' || b == '\r' || b == '\t') {
            b = br.read();
        }
        char ch[] = new char[len];
        ch[0] = (char) b;
        int idx = 1;
        while (idx < len && (b = br.read()) != ' ' && b != '\n' && b != '\r' && b != '\t') {
            ch[idx++] = (char) b;
        }
        return String.valueOf(ch).trim();
    }

    private static int nextInt() throws IOException {
        st.nextToken();
        return (int) st.nval;
    }

    private static double nextDouble() throws IOException {
        st.nextToken();
        return st.nval;
    }

    private static String[] nextSS(String reg) throws IOException {
        return br.readLine().split(reg);
    }

    private static String nextLine() throws IOException {
        return br.readLine();
    }
}

Global OOD problem IDs are disjoint from train, validation, and in-distribution test problem IDs.

Score : 400 points

Problem Statement

There are N integers, A_1, A_2, ..., A_N, arranged in a row in this order.

You can perform the following operation on this integer sequence any number of times:

Operation: Choose an integer i satisfying 1 \leq i \leq N-1. Multiply both A_i and A_{i+1} by -1.

Let B_1, B_2, ..., B_N be the integer sequence after your operations.

Find the maximum possible value of B_1 + B_2 + ... + B_N.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

-10^9 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the maximum possible value of B_1 + B_2 + ... + B_N.

Sample Input 1

3
-10 5 -4

Sample Output 1

19

If we perform the operation as follows:

Choose 1 as i, which changes the sequence to 10, -5, -4.

Choose 2 as i, which changes the sequence to 10, 5, 4.

we have B_1 = 10, B_2 = 5, B_3 = 4. The sum here, B_1 + B_2 + B_3 = 10 + 5 + 4 = 19, is the maximum possible result.

Sample Input 2

5
10 -4 -8 -11 3

Sample Output 2

30

Sample Input 3

11
-1000000000 1000000000 -1000000000 1000000000 -1000000000 0 1000000000 -1000000000 1000000000 -1000000000 1000000000

Sample Output 3

10000000000

The output may not fit into a 32-bit integer type.

Submitted source-code file size, in bytes.

Execution CPU time reported by official Project CodeNet metadata, in milliseconds.

Peak memory usage reported by official Project CodeNet metadata, in kilobytes.


import java.util.*;
import static java.lang.System.*;

public class Main {
	static Scanner sc = new Scanner(System.in);
	public static void main(String[] args) {
		int n = sc.nextInt();
        long[] a = new long[n];
        long sum=0;
        int cnt=0;
        int zero=0;
        for(int i=0;i<n;i++){
            a[i] = sc.nextInt();
            if(a[i]<0) cnt++;
            if(a[i]==0) zero++;
            sum+=Math.abs(a[i]);
        }
        long ans=0;
        for(int i=0;i<n;i++){
            // out.println(ans);
            ans=Math.max(ans,sum-2*Math.abs(a[i]));
        }
        if(cnt%2==0) ans=sum;
        if(cnt%2==1&&zero>0) ans=sum;
        out.println(ans);
	}
}

// int n = sc.nextInt();
// int[] array = new int[n];
// String s = sc.next();
// s.charAt(i)
// out.println();
// Math.min()


import java.util.*;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		int[] a = new int[N];
		int sum = 0;
		for(int i=0; i<N; i++) {
			a[i] = sc.nextInt();
			sum += a[i];
		}

		final long MOD = 998244353;

		long[][] dp = new long[2][sum+1];
		dp[0][0] = 1;
		int flg=0;
		for(int i=0; i<N; i++) {
			Arrays.fill(dp[flg^1], 0);
			for(int j=0; j<=sum; j++) {
				dp[flg^1][j] = (dp[flg^1][j] + dp[flg][j]*2) %MOD;
				if(j+a[i]<=sum)
					dp[flg^1][j+a[i]] = (dp[flg^1][j+a[i]] + dp[flg][j]) %MOD;
			}
			flg ^= 1;
		}
		long over = 0;
		for(int i=(sum+1)/2; i<=sum; i++)
			over = (over + dp[flg][i])%MOD;

		long ans = 1;
		for(int i=0; i<N; i++)
			ans = (ans * 3) % MOD;

		ans = (ans - over*3 + MOD*3) % MOD;
		if(sum%2==0) {
			long[][] dp2 = new long[2][sum/2+1];
			dp2[0][0] = 1;
			flg=0;
			for(int i=0; i<N; i++) {
				Arrays.fill(dp2[flg^1], 0);
				for(int j=0; j<=sum/2; j++) {
					dp2[flg^1][j] = (dp2[flg^1][j] + dp2[flg][j]) %MOD;
					if(j+a[i]<=sum/2)
						dp2[flg^1][j+a[i]] = (dp2[flg^1][j+a[i]] + dp2[flg][j]) %MOD;
				}
				flg ^= 1;
			}
			ans = (ans + dp2[flg][sum/2]*3 + MOD*3) % MOD;

		}
		System.out.println(ans);
		sc.close();
	}
}

Score : 600 points

Problem Statement

You are given N integers. The i-th integer is a_i.
Find the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the following condition is satisfied:

Let R, G and B be the sums of the integers painted red, green and blue, respectively. There exists a triangle with positive area whose sides have lengths R, G and B.

Constraints

3 \leq N \leq 300

1 \leq a_i \leq 300(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
a_1
:
a_N

Output

Print the number, modulo 998244353, of ways to paint each of the integers red, green or blue so that the condition is satisfied.

Sample Input 1

4
1
1
1
2

Sample Output 1

18

We can only paint the integers so that the lengths of the sides of the triangle will be 1, 2 and 2, and there are 18 such ways.

Sample Input 2

6
1
3
2
3
5
2

Sample Output 2

150

Sample Input 3

20
3
1
4
1
5
9
2
6
5
3
5
8
9
7
9
3
2
3
8
4

Sample Output 3

563038556


import java.util.Scanner;
public class Main {
	public static void main(String[] args) {
		Scanner stdIn=new Scanner(System.in);
		int A=stdIn.nextInt();
		int B=stdIn.nextInt();
		int C=stdIn.nextInt();
		if(A<=C&&C<=B)
			System.out.println("Yes");
		else
			if(B<=C&&C<=A)
				System.out.println("Yes");
			else
				System.out.println("No");
	}
}

Score : 100 points

Problem Statement

There are three houses on a number line: House 1, 2 and 3, with coordinates A, B and C, respectively.
Print Yes if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print No otherwise.

Constraints

0\leq A,B,C\leq 100

A, B and C are distinct integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print Yes if we pass the coordinate of House 3 on the straight way from House 1 to House 2 without making a detour, and print No otherwise.

Sample Input 1

3 8 5

Sample Output 1

Yes

We pass the coordinate 5 on the straight way from the house at coordinate 3 to the house at coordinate 8.

Sample Input 2

7 3 1

Sample Output 2

No

Sample Input 3

10 2 4

Sample Output 3

Yes

Sample Input 4

31 41 59

Sample Output 4

No


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);

		int A = scan.nextInt();
		int B = scan.nextInt();
		int X = 0;
		int Y = 0;

			if (A >= B) {
				X = A;
				if (A-1 >= B) {
					Y = A-1;
				}else {
					Y = B;
				}
			} else if (A < B) {
				X = B;
				if (B-1 >= A) {
					Y = B-1;
				}else {
					Y = A;
				}
			}
		System.out.println(X+Y);
	}
}

Score : 100 points

Problem Statement

There are two buttons, one of size A and one of size B.

When you press a button of size X, you get X coins and the size of that button decreases by 1.

You will press a button twice. Here, you can press the same button twice, or press both buttons once.

At most how many coins can you get?

Constraints

All values in input are integers.

3 \leq A, B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

Print the maximum number of coins you can get.

Sample Input 1

5 3

Sample Output 1

9

You can get 5 + 4 = 9 coins by pressing the button of size 5 twice, and this is the maximum result.

Sample Input 2

3 4

Sample Output 2

7

Sample Input 3

6 6

Sample Output 3

12


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner scan = new Scanner(System.in);

		String InStr = scan.nextLine();
		String[] AryStr = InStr.split(" ");

		int A;
		int B;

		A = Integer.parseInt(AryStr[0]);
		B = Integer.parseInt(AryStr[1]);

		if (A < B) {
			A = B;
		} else {
            B = A;
        }
		B--;

        System.out.println(A + B);
	}
}


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Main {
  static class FastScanner {
    private BufferedReader reader = null;
    private StringTokenizer tokenizer = null;

    public FastScanner(InputStream in) {
      reader = new BufferedReader(new InputStreamReader(in));
      tokenizer = null;
    }

    public String next() {
      if (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

    public String nextLine() {
      if (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          return reader.readLine();
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }

      return tokenizer.nextToken("\n");
    }

    public long nextLong() {
      return Long.parseLong(next());
    }

    public int nextInt() {
      return Integer.parseInt(next());
    }

    public double nextDouble() {
      return Double.parseDouble(next());
    }

    public int[] nextIntArray(int n) {
      int[] a = new int[n];
      for (int i = 0; i < n; i++)
        a[i] = nextInt();
      return a;
    }

    public long[] nextLongArray(int n) {
      long[] a = new long[n];
      for (int i = 0; i < n; i++)
        a[i] = nextLong();
      return a;
    }
  }

  public static void main(String[] args) {
    FastScanner sc = new FastScanner(System.in);
    int n = sc.nextInt();
    int[] h = sc.nextIntArray(n);
    int count = 1;
    int max = h[0];
    for (int i = 1; i < n; i++) {
      if (h[i] >= max) {
        count++;
        max = h[i];
      }
    }
    System.out.println(count);
  }
}

Score : 200 points

Problem Statement

There are N mountains ranging from east to west, and an ocean to the west.

At the top of each mountain, there is an inn. You have decided to choose where to stay from these inns.

The height of the i-th mountain from the west is H_i.

You can certainly see the ocean from the inn at the top of the westmost mountain.

For the inn at the top of the i-th mountain from the west (i = 2, 3, ..., N), you can see the ocean if and only if H_1 \leq H_i, H_2 \leq H_i, ..., and H_{i-1} \leq H_i.

From how many of these N inns can you see the ocean?

Constraints

All values in input are integers.

1 \leq N \leq 20

1 \leq H_i \leq 100

Input

Input is given from Standard Input in the following format:

N
H_1 H_2 ... H_N

Output

Print the number of inns from which you can see the ocean.

Sample Input 1

4
6 5 6 8

Sample Output 1

3

You can see the ocean from the first, third and fourth inns from the west.

Sample Input 2

5
4 5 3 5 4

Sample Output 2

3

Sample Input 3

5
9 5 6 8 4

Sample Output 3

1


import java.util.Scanner;

class Main{

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);			//文字の入力
		int A = sc.nextInt();
		int B = sc.nextInt();
		int C = sc.nextInt();
		int D = sc.nextInt();
		int E = sc.nextInt();
		int K = sc.nextInt();

		if(E-A <= K){
			System.out.println("Yay!");
		}else{
			System.out.println(":(");

		}

	}

}

class Pair implements Comparable{
	int from;
	int end;
	@Override
	public int compareTo(Object other) {
		Pair otherpair = (Pair)other;

		return from - otherpair.from;
	}
}

Score: 100 points

Problem Statement

In AtCoder city, there are five antennas standing in a straight line. They are called Antenna A, B, C, D and E from west to east, and their coordinates are a, b, c, d and e, respectively.

Two antennas can communicate directly if the distance between them is k or less, and they cannot if the distance is greater than k.

Determine if there exists a pair of antennas that cannot communicate directly.

Here, assume that the distance between two antennas at coordinates p and q (p < q) is q - p.

Constraints

a, b, c, d, e and k are integers between 0 and 123 (inclusive).

a < b < c < d < e

Input

Input is given from Standard Input in the following format:

a
b
c
d
e
k

Output

Print :( if there exists a pair of antennas that cannot communicate directly, and print Yay! if there is no such pair.

Sample Input 1

1
2
4
8
9
15

Sample Output 1

Yay!

In this case, there is no pair of antennas that cannot communicate directly, because:

the distance between A and B is 2 - 1 = 1

the distance between A and C is 4 - 1 = 3

the distance between A and D is 8 - 1 = 7

the distance between A and E is 9 - 1 = 8

the distance between B and C is 4 - 2 = 2

the distance between B and D is 8 - 2 = 6

the distance between B and E is 9 - 2 = 7

the distance between C and D is 8 - 4 = 4

the distance between C and E is 9 - 4 = 5

the distance between D and E is 9 - 8 = 1

and none of them is greater than 15. Thus, the correct output is Yay!.

Sample Input 2

15
18
26
35
36
18

Sample Output 2

:(

In this case, the distance between antennas A and D is 35 - 15 = 20 and exceeds 18, so they cannot communicate directly.
Thus, the correct output is :(.


import java.util.*;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int count=0;
		int max=0;
		int a=0;
		int[] A = new int[5];
		int c=0;

		for(int i=0;i<5;i++) {
			A[i]=sc.nextInt();
			if(A[i]%10!=0) {
				max=Math.max((10-A[i]%10), max);
				c=i;
				}
		}

		for(int i=0;i<5;i++) {
			count+=A[i];
			for(int j=0;j<9;j++) {
				if(count%10!=0) {
					count++;
				}
			}
		}

		System.out.println(count-max);

		sc.close();
	}
}

Score: 200 points

Problem Statement

The restaurant AtCoder serves the following five dishes:

ABC Don (rice bowl): takes A minutes to serve.

ARC Curry: takes B minutes to serve.

AGC Pasta: takes C minutes to serve.

APC Ramen: takes D minutes to serve.

ATC Hanbagu (hamburger patty): takes E minutes to serve.

Here, the time to serve a dish is the time between when an order is placed and when the dish is delivered.

This restaurant has the following rules on orders:

An order can only be placed at a time that is a multiple of 10 (time 0, 10, 20, ...).

Only one dish can be ordered at a time.

No new order can be placed when an order is already placed and the dish is still not delivered, but a new order can be placed at the exact time when the dish is delivered.

E869120 arrives at this restaurant at time 0. He will order all five dishes. Find the earliest possible time for the last dish to be delivered.

Here, he can order the dishes in any order he likes, and he can place an order already at time 0.

Constraints

A, B, C, D and E are integers between 1 and 123 (inclusive).

Input

Input is given from Standard Input in the following format:

A
B
C
D
E

Output

Print the earliest possible time for the last dish to be delivered, as an integer.

Sample Input 1

29
20
7
35
120

Sample Output 1

215

If we decide to order the dishes in the order ABC Don, ARC Curry, AGC Pasta, ATC Hanbagu, APC Ramen, the earliest possible time for each order is as follows:

Order ABC Don at time 0, which will be delivered at time 29.

Order ARC Curry at time 30, which will be delivered at time 50.

Order AGC Pasta at time 50, which will be delivered at time 57.

Order ATC Hanbagu at time 60, which will be delivered at time 180.

Order APC Ramen at time 180, which will be delivered at time 215.

There is no way to order the dishes in which the last dish will be delivered earlier than this.

Sample Input 2

101
86
119
108
57

Sample Output 2

481

If we decide to order the dishes in the order AGC Pasta, ARC Curry, ATC Hanbagu, APC Ramen, ABC Don, the earliest possible time for each order is as follows:

Order AGC Pasta at time 0, which will be delivered at time 119.

Order ARC Curry at time 120, which will be delivered at time 206.

Order ATC Hanbagu at time 210, which will be delivered at time 267.

Order APC Ramen at time 270, which will be delivered at time 378.

Order ABC Don at time 380, which will be delivered at time 481.

There is no way to order the dishes in which the last dish will be delivered earlier than this.

Sample Input 3

123
123
123
123
123

Sample Output 3

643

This is the largest valid case.


import java.util.Scanner;

public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);

        long n = sc.nextLong();
        long a = sc.nextLong();
        long b = sc.nextLong();
        long c = sc.nextLong();
        long d = sc.nextLong();
        long e = sc.nextLong();

        long min = Math.min(a,b);
        min = Math.min(min, c);
        min = Math.min(min, d);
        min = Math.min(min, e);

//        long minute = n/min+1;
        long minute=0;

        if (n%min==0){
            minute = n/min;
        }else{
            minute = n/min+1;
        }

        System.out.println(minute+4);

    }

}

Score: 300 points

Problem Statement

In 2028 and after a continuous growth, AtCoder Inc. finally built an empire with six cities (City 1, 2, 3, 4, 5, 6)!

There are five means of transport in this empire:

Train: travels from City 1 to 2 in one minute. A train can occupy at most A people.

Bus: travels from City 2 to 3 in one minute. A bus can occupy at most B people.

Taxi: travels from City 3 to 4 in one minute. A taxi can occupy at most C people.

Airplane: travels from City 4 to 5 in one minute. An airplane can occupy at most D people.

Ship: travels from City 5 to 6 in one minute. A ship can occupy at most E people.

For each of them, one vehicle leaves the city at each integer time (time 0, 1, 2, ...).

There is a group of N people at City 1, and they all want to go to City 6.

At least how long does it take for all of them to reach there?
You can ignore the time needed to transfer.

Constraints

1 \leq N, A, B, C, D, E \leq 10^{15}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A
B
C
D
E

Output

Print the minimum time required for all of the people to reach City 6, in minutes.

Sample Input 1

5
3
2
4
3
5

Sample Output 1

7

One possible way to travel is as follows.
First, there are N = 5 people at City 1, as shown in the following image:

In the first minute, three people travels from City 1 to City 2 by train. Note that a train can only occupy at most three people.

In the second minute, the remaining two people travels from City 1 to City 2 by train, and two of the three people who were already at City 2 travels to City 3 by bus. Note that a bus can only occupy at most two people.

In the third minute, two people travels from City 2 to City 3 by train, and another two people travels from City 3 to City 4 by taxi.

From then on, if they continue traveling without stopping until they reach City 6, all of them can reach there in seven minutes.

There is no way for them to reach City 6 in 6 minutes or less.

Sample Input 2

10
123
123
123
123
123

Sample Output 2

5

All kinds of vehicles can occupy N = 10 people at a time.
Thus, if they continue traveling without stopping until they reach City 6, all of them can reach there in five minutes.

Sample Input 3

10000000007
2
3
5
7
11

Sample Output 3

5000000008

Note that the input or output may not fit into a 32-bit integer type.


import java.util.Scanner;

public class Main {
	static long[]m = new long[5];
	static long Mod = 1000_000_007;

	public static void main(String[] args) {
		Main main = new Main();
		main.solve();
	}

	void solve() {
		Scanner sc = new Scanner(System.in);
		int n = Integer.parseInt(sc.next());
		long min = Mod;
		for(int i=0;i<5;i++) {
			m[i]=Long.parseLong(sc.next());
			if(m[i]<min)min=m[i];
		}
		sc.close();

		long res = 0;
		if(min<n) {
			res = n/min;
		}
		res+=5;
		if(res > Mod) {
			while(res>Mod*100)res-=Mod*100;
			while(res>Mod)res-=Mod;
		}
		System.out.println(res);
	}

}


import java.io.*;
import java.util.*;
import java.util.Map.Entry;

public class Main {

	final boolean isDebug = false;
	String fileName = "input.txt";
	FastScanner sc;
	PrintWriter out;
	final int MOD = (int)1e9+7;
	final int INF = Integer.MAX_VALUE / 2;

	void solve() throws Exception{
		int a = sc.nextInt(), b = sc.nextInt(), c = sc.nextInt();
		if(a == b && b == c) System.out.println("Yes");
		else System.out.println("No");

	}

	/* end solve */

	/* main */
	public static void main(String[] args) throws Exception {
		new Main().m();
	}

	void m() throws Exception {
		long S = System.currentTimeMillis();
		sc = (isDebug) ? new FastScanner(new FileInputStream(fileName)) : new FastScanner(System.in);
		out = new PrintWriter(System.out);
		solve();
		out.flush();
		long G = System.currentTimeMillis();
		if(isDebug){
			System.out.println("---Debug---");
			System.out.printf("%8d ms", (G-S));
		}
	}
	/* end main */
}
/* end Main */

class FastScanner {
    private InputStream in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;
    public FastScanner(InputStream in) {
		this.in = in;
	}
    private boolean hasNextByte() {
        if (ptr < buflen) {
            return true;
        }else{
            ptr = 0;
            try {
                buflen = in.read(buffer);
            } catch (IOException e) {
                e.printStackTrace();
            }
            if (buflen <= 0) {
                return false;
            }
        }
        return true;
    }
    private int readByte() {
    	if (hasNextByte()) return buffer[ptr++];
    	else return -1;
    }
    private static boolean isPrintableChar(int c){
    	return 33 <= c && c <= 126;
    }
    public boolean hasNext() {
    	while(hasNextByte() && !isPrintableChar(buffer[ptr]))
    		ptr++; return hasNextByte();
    }
    public String next() {
        if (!hasNext()) throw new NoSuchElementException();
        StringBuilder sb = new StringBuilder();
        int b = readByte();
        while(isPrintableChar(b)) {
            sb.appendCodePoint(b);
            b = readByte();
        }
        return sb.toString();
    }
    public long nextLong() {
        if (!hasNext()) throw new NoSuchElementException();
        long n = 0;
        boolean minus = false;
        int b = readByte();
        if (b == '-') {
            minus = true;
            b = readByte();
        }
        if (b < '0' || '9' < b) {
            throw new NumberFormatException();
        }
        while(true){
            if ('0' <= b && b <= '9') {
                n *= 10;
                n += b - '0';
            }else if(b == -1 || !isPrintableChar(b)){
                return minus ? -n : n;
            }else{
                throw new NumberFormatException();
            }
            b = readByte();
        }
    }
    public int nextInt() {
        long nl = nextLong();
        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
        return (int) nl;
    }
    public double nextDouble() {
    	return Double.parseDouble(next());
    }
}

Score : 100 points

Problem Statement

You are given three integers A, B and C.

Determine if there exists an equilateral triangle whose sides have lengths A, B and C.

Constraints

All values in input are integers.

1 \leq A,B,C \leq 100

Input

Input is given from Standard Input in the following format:

A B C

Output

If there exists an equilateral triangle whose sides have lengths A, B and C, print Yes; otherwise, print No.

Sample Input 1

2 2 2

Sample Output 1

Yes

There exists an equilateral triangle whose sides have lengths 2, 2 and 2.

Sample Input 2

3 4 5

Sample Output 2

No

There is no equilateral triangle whose sides have lengths 3, 4 and 5.


import java.io.*;
import java.util.Arrays;
import java.util.StringJoiner;
import java.util.StringTokenizer;
import java.util.function.Function;

public class Main {

    static int N;
    static String S;

    public static void main(String[] args) {
        FastScanner sc = new FastScanner(System.in);
        N = sc.nextInt();
        S = sc.next();

        System.out.println(solve() ? "Yes" : "No");
    }

    static boolean solve() {
        int r = 0;
        int b = 0;
        for (int i = 0; i < N; i++) {
            if( S.charAt(i) == 'R' ) {
                r++;
            } else {
                b++;
            }
        }
        return r > b;
    }

    @SuppressWarnings("unused")
    static class FastScanner {
        private BufferedReader reader;
        private StringTokenizer tokenizer;

        FastScanner(InputStream in) {
            reader = new BufferedReader(new InputStreamReader(in));
            tokenizer = null;
        }

        String next() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

        String nextLine() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    return reader.readLine();
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken("\n");
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        int[] nextIntArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        int[] nextIntArray(int n, int delta) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt() + delta;
            return a;
        }

        long[] nextLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }
    }

    static <A> void writeLines(A[] as, Function<A, String> f) {
        PrintWriter pw = new PrintWriter(System.out);
        for (A a : as) {
            pw.println(f.apply(a));
        }
        pw.flush();
    }

    static void writeLines(int[] as) {
        PrintWriter pw = new PrintWriter(System.out);
        for (int a : as) pw.println(a);
        pw.flush();
    }

    static void writeLines(long[] as) {
        PrintWriter pw = new PrintWriter(System.out);
        for (long a : as) pw.println(a);
        pw.flush();
    }

    static int max(int... as) {
        int max = Integer.MIN_VALUE;
        for (int a : as) max = Math.max(a, max);
        return max;
    }

    static int min(int... as) {
        int min = Integer.MAX_VALUE;
        for (int a : as) min = Math.min(a, min);
        return min;
    }

    static void debug(Object... args) {
        StringJoiner j = new StringJoiner(" ");
        for (Object arg : args) {
            if (arg instanceof int[]) j.add(Arrays.toString((int[]) arg));
            else if (arg instanceof long[]) j.add(Arrays.toString((long[]) arg));
            else if (arg instanceof double[]) j.add(Arrays.toString((double[]) arg));
            else if (arg instanceof Object[]) j.add(Arrays.toString((Object[]) arg));
            else j.add(arg.toString());
        }
        System.err.println(j.toString());
    }
}

Score : 200 points

Problem Statement

There are N people numbered 1 to N. Each person wears a red hat or a blue hat.

You are given a string s representing the colors of the people. Person i wears a red hat if s_i is R, and a blue hat if s_i is B.

Determine if there are more people wearing a red hat than people wearing a blue hat.

Constraints

1 \leq N \leq 100

|s| = N

s_i is R or B.

Input

Input is given from Standard Input in the following format:

N
s

Output

If there are more people wearing a red hat than there are people wearing a blue hat, print Yes; otherwise, print No.

Sample Input 1

4
RRBR

Sample Output 1

Yes

There are three people wearing a red hat, and one person wearing a blue hat.

Since there are more people wearing a red hat than people wearing a blue hat, the answer is Yes.

Sample Input 2

4
BRBR

Sample Output 2

No

There are two people wearing a red hat, and two people wearing a blue hat.

Since there are as many people wearing a red hat as people wearing a blue hat, the answer is No.


import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        // int n = Integer.parseInt(sc.next());
        String s = sc.next();

        int ans = 0;
        int tmp = 0;
        for (int i = 0; i < s.length(); i++) {
            if (s.substring(i, i + 1).equals("A") || s.substring(i, i + 1).equals("T")
                    || s.substring(i, i + 1).equals("G") || s.substring(i, i + 1).equals("C")) {
                tmp++;
            } else {
                ans = Math.max(tmp, ans);
                tmp=0;
            }
        }
        System.out.println(ans);
        sc.close();
    }
}

Score : 200 points

Problem Statement

You are given a string S consisting of uppercase English letters. Find the length of the longest ACGT string that is a substring (see Notes) of S.

Here, a ACGT string is a string that contains no characters other than A, C, G and T.

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

S is a string of length between 1 and 10 (inclusive).

Each character in S is an uppercase English letter.

Input

Input is given from Standard Input in the following format:

S

Output

Print the length of the longest ACGT string that is a substring of S.

Sample Input 1

ATCODER

Sample Output 1

3

Among the ACGT strings that are substrings of ATCODER, the longest one is ATC.

Sample Input 2

HATAGAYA

Sample Output 2

5

Among the ACGT strings that are substrings of HATAGAYA, the longest one is ATAGA.

Sample Input 3

SHINJUKU

Sample Output 3

0

Among the ACGT strings that are substrings of SHINJUKU, the longest one is (the empty string).


import java.util.ArrayList;
import java.util.Scanner;

public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        // TODO code application logic here
        Scanner sc = new Scanner(System.in);
        String AtCoder = sc.next();
        String[] text = AtCoder.split("");
        int count = 0;
        int answer = 0;
        for (int i = 0; i < AtCoder.length(); i++) {
            if (text[i].equals("A") || text[i].equals("C") || text[i].equals("G") || text[i].equals("T")) {
                count++;
            } else {
                if (answer < count) {
                    answer = count;
                }
                count = 0;
            }
        }

        System.out.print(answer);
    }

}


import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {

	private static final String KEYWORD = "AC";

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int q = scanner.nextInt();
			scanner.nextLine();
			String s = scanner.nextLine();
			List<Integer> result = new ArrayList<>();
			for (int i = 0; i < q; i++) {
				int l = scanner.nextInt();
				int r = scanner.nextInt();
				int count = 0;
				scanner.nextLine();
				String substring = s.substring(l - 1, r);
				char previous = ' ';
				for (char current : substring.toCharArray()) {
					if (('A' == previous) && ('C' == current)) {
						count++;
					}
					previous = current;
				}
				result.add(count);
			}
			result.stream().forEach(System.out::println);
		}
	}
}

Score : 300 points

Problem Statement

You are given a string S of length N consisting of A, C, G and T. Answer the following Q queries:

Query i (1 \leq i \leq Q): You will be given integers l_i and r_i (1 \leq l_i < r_i \leq N). Consider the substring of S starting at index l_i and ending at index r_i (both inclusive). In this string, how many times does AC occurs as a substring?

Notes

A substring of a string T is a string obtained by removing zero or more characters from the beginning and the end of T.

For example, the substrings of ATCODER include TCO, AT, CODER, ATCODER and (the empty string), but not AC.

Constraints

2 \leq N \leq 10^5

1 \leq Q \leq 10^5

S is a string of length N.

Each character in S is A, C, G or T.

1 \leq l_i < r_i \leq N

Input

Input is given from Standard Input in the following format:

N Q
S
l_1 r_1
:
l_Q r_Q

Output

Print Q lines. The i-th line should contain the answer to the i-th query.

Sample Input 1

8 3
ACACTACG
3 7
2 3
1 8

Sample Output 1

2
0
3

Query 1: the substring of S starting at index 3 and ending at index 7 is ACTAC. In this string, AC occurs twice as a substring.

Query 2: the substring of S starting at index 2 and ending at index 3 is CA. In this string, AC occurs zero times as a substring.

Query 3: the substring of S starting at index 1 and ending at index 8 is ACACTACG. In this string, AC occurs three times as a substring.


import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String[] chars={"a","b","c","d","e","f","g","h","i","j","h","u","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
        long output=1;
        long mod=1000000007;
        int n=sc.nextInt();
        String s=sc.next();
        for(int i=0;i<26;i++){
            output*=(long)(s.length()-s.replaceAll(chars[i],"").length()+1);
            output%=mod;
        }
        System.out.println(output==0?mod-1:output-1);
    }
}

Score : 200 points

Problem Statement

You are given a string S of length N.
Among its subsequences, count the ones such that all characters are different, modulo 10^9+7. Two subsequences are considered different if their characters come from different positions in the string, even if they are the same as strings.

Here, a subsequence of a string is a concatenation of one or more characters from the string without changing the order.

Constraints

1 \leq N \leq 100000

S consists of lowercase English letters.

|S|=N

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the number of the subsequences such that all characters are different, modulo 10^9+7.

Sample Input 1

4
abcd

Sample Output 1

15

Since all characters in S itself are different, all its subsequences satisfy the condition.

Sample Input 2

3
baa

Sample Output 2

5

The answer is five: b, two occurrences of a, two occurrences of ba. Note that we do not count baa, since it contains two as.

Sample Input 3

5
abcab

Sample Output 3

17


import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.nio.charset.Charset;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

public class Main {
    public static void main(String[] args) throws Exception {
        boolean local = false;
        boolean async = false;

        Charset charset = Charset.forName("ascii");

        FastIO io = local ? new FastIO(new FileInputStream("D:\\DATABASE\\TESTCASE\\Code.in"), System.out, charset) : new FastIO(System.in, System.out, charset);
        Task task = new Task(io, new Debug(local));

        if (async) {
            Thread t = new Thread(null, task, "dalt", 1 << 27);
            t.setPriority(Thread.MAX_PRIORITY);
            t.start();
            t.join();
        } else {
            task.run();
        }

        if (local) {
            io.cache.append("\n\n--memory -- \n" + ((Runtime.getRuntime().totalMemory() - Runtime.getRuntime().freeMemory()) >> 20) + "M");
        }

        io.flush();
    }

    public static class Task implements Runnable {
        final FastIO io;
        final Debug debug;
        int inf = (int) 1e8;
        int mod = (int) 1e9 + 7;

        public int mod(int val) {
            val %= mod;
            if (val < 0) {
                val += mod;
            }
            return val;
        }

        public int mod(long val) {
            val %= mod;
            if (val < 0) {
                val += mod;
            }
            return (int) val;
        }

        int bitAt(int x, int i) {
            return (x >> i) & 1;
        }

        int bitAt(long x, int i) {
            return (int) ((x >> i) & 1);
        }

        public Task(FastIO io, Debug debug) {
            this.io = io;
            this.debug = debug;
        }

        @Override
        public void run() {
            solve();
        }

        public void solve() {
            int n = io.readInt();
            int[] colors = new int[n + 1];
            int[] dp = new int[n + 1];
            Map<Integer, Integer> registries = new HashMap<>();
            for (int i = 1; i <= n; i++) {
                colors[i] = io.readInt();
                if (i == 1) {
                    dp[i] = 1;
                } else {
                    int last = registries.getOrDefault(colors[i], 0);
                    dp[i] = dp[i - 1];
                    if (last != i - 1) {
                        dp[i] = mod(dp[i] + dp[last]);
                    }
                }
                registries.put(colors[i], i);
            }

            io.cache.append(dp[n]);
        }
    }

    public static class FastIO {
        public final StringBuilder cache = new StringBuilder();
        private final InputStream is;
        private final OutputStream os;
        private final Charset charset;
        private StringBuilder defaultStringBuf = new StringBuilder(1 << 8);
        private byte[] buf = new byte[1 << 13];
        private int bufLen;
        private int bufOffset;
        private int next;

        public FastIO(InputStream is, OutputStream os, Charset charset) {
            this.is = is;
            this.os = os;
            this.charset = charset;
        }

        public FastIO(InputStream is, OutputStream os) {
            this(is, os, Charset.forName("ascii"));
        }

        private int read() {
            while (bufLen == bufOffset) {
                bufOffset = 0;
                try {
                    bufLen = is.read(buf);
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
                if (bufLen == -1) {
                    return -1;
                }
            }
            return buf[bufOffset++];
        }

        public void skipBlank() {
            while (next >= 0 && next <= 32) {
                next = read();
            }
        }

        public int readInt() {
            int sign = 1;

            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+' ? 1 : -1;
                next = read();
            }

            int val = 0;
            if (sign == 1) {
                while (next >= '0' && next <= '9') {
                    val = val * 10 + next - '0';
                    next = read();
                }
            } else {
                while (next >= '0' && next <= '9') {
                    val = val * 10 - next + '0';
                    next = read();
                }
            }

            return val;
        }

        public long readLong() {
            int sign = 1;

            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+' ? 1 : -1;
                next = read();
            }

            long val = 0;
            if (sign == 1) {
                while (next >= '0' && next <= '9') {
                    val = val * 10 + next - '0';
                    next = read();
                }
            } else {
                while (next >= '0' && next <= '9') {
                    val = val * 10 - next + '0';
                    next = read();
                }
            }

            return val;
        }

        public double readDouble() {
            boolean sign = true;
            skipBlank();
            if (next == '+' || next == '-') {
                sign = next == '+';
                next = read();
            }

            long val = 0;
            while (next >= '0' && next <= '9') {
                val = val * 10 + next - '0';
                next = read();
            }
            if (next != '.') {
                return sign ? val : -val;
            }
            next = read();
            long radix = 1;
            long point = 0;
            while (next >= '0' && next <= '9') {
                point = point * 10 + next - '0';
                radix = radix * 10;
                next = read();
            }
            double result = val + (double) point / radix;
            return sign ? result : -result;
        }

        public String readString(StringBuilder builder) {
            skipBlank();

            while (next > 32) {
                builder.append((char) next);
                next = read();
            }

            return builder.toString();
        }

        public String readString() {
            defaultStringBuf.setLength(0);
            return readString(defaultStringBuf);
        }

        public int readLine(char[] data, int offset) {
            int originalOffset = offset;
            while (next != -1 && next != '\n') {
                data[offset++] = (char) next;
                next = read();
            }
            return offset - originalOffset;
        }

        public int readString(char[] data, int offset) {
            skipBlank();

            int originalOffset = offset;
            while (next > 32) {
                data[offset++] = (char) next;
                next = read();
            }

            return offset - originalOffset;
        }

        public int readString(byte[] data, int offset) {
            skipBlank();

            int originalOffset = offset;
            while (next > 32) {
                data[offset++] = (byte) next;
                next = read();
            }

            return offset - originalOffset;
        }

        public char readChar() {
            skipBlank();
            char c = (char) next;
            next = read();
            return c;
        }

        public void flush() {
            try {
                os.write(cache.toString().getBytes(charset));
                os.flush();
                cache.setLength(0);
            } catch (IOException e) {
                throw new RuntimeException(e);
            }
        }

        public boolean hasMore() {
            skipBlank();
            return next != -1;
        }
    }

    public static class Debug {
        private boolean allowDebug;

        public Debug(boolean allowDebug) {
            this.allowDebug = allowDebug;
        }

        public void assertTrue(boolean flag) {
            if (!allowDebug) {
                return;
            }
            if (!flag) {
                fail();
            }
        }

        public void fail() {
            throw new RuntimeException();
        }

        public void assertFalse(boolean flag) {
            if (!allowDebug) {
                return;
            }
            if (flag) {
                fail();
            }
        }

        private void outputName(String name) {
            System.out.print(name + " = ");
        }

        public void debug(String name, int x) {
            if (!allowDebug) {
                return;
            }

            outputName(name);
            System.out.println("" + x);
        }

        public void debug(String name, long x) {
            if (!allowDebug) {
                return;
            }
            outputName(name);
            System.out.println("" + x);
        }

        public void debug(String name, double x) {
            if (!allowDebug) {
                return;
            }
            outputName(name);
            System.out.println("" + x);
        }

        public void debug(String name, int[] x) {
            if (!allowDebug) {
                return;
            }
            outputName(name);
            System.out.println(Arrays.toString(x));
        }

        public void debug(String name, long[] x) {
            if (!allowDebug) {
                return;
            }
            outputName(name);
            System.out.println(Arrays.toString(x));
        }

        public void debug(String name, double[] x) {
            if (!allowDebug) {
                return;
            }
            outputName(name);
            System.out.println(Arrays.toString(x));
        }

        public void debug(String name, Object x) {
            if (!allowDebug) {
                return;
            }
            outputName(name);
            System.out.println("" + x);
        }

        public void debug(String name, Object... x) {
            if (!allowDebug) {
                return;
            }
            outputName(name);
            System.out.println(Arrays.deepToString(x));
        }
    }
}

Score : 700 points

Problem Statement

There are N stones arranged in a row. The i-th stone from the left is painted in the color C_i.

Snuke will perform the following operation zero or more times:

Choose two stones painted in the same color. Repaint all the stones between them, with the color of the chosen stones.

Find the number of possible final sequences of colors of the stones, modulo 10^9+7.

Constraints

1 \leq N \leq 2\times 10^5

1 \leq C_i \leq 2\times 10^5(1\leq i\leq N)

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
C_1
:
C_N

Output

Print the number of possible final sequences of colors of the stones, modulo 10^9+7.

Sample Input 1

5
1
2
1
2
2

Sample Output 1

3

We can make three sequences of colors of stones, as follows:

(1,2,1,2,2), by doing nothing.

(1,1,1,2,2), by choosing the first and third stones to perform the operation.

(1,2,2,2,2), by choosing the second and fourth stones to perform the operation.

Sample Input 2

6
4
2
5
4
2
4

Sample Output 2

5

Sample Input 3

7
1
3
1
2
3
3
2

Sample Output 3

5


import java.io.IOException;
import java.io.InputStream;
import java.lang.reflect.Array;
import java.util.*;

public class Main {
    long MOD = 1000000007;
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        int H = sc.nextInt();
        int W = sc.nextInt();
        int h = sc.nextInt();
        int w = sc.nextInt();
        System.out.println((H-h)*(W-w));
    }

    public static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
        private boolean hasNextByte() {
            if (ptr < buflen) {
                return true;
            }else{
                ptr = 0;
                try {
                    buflen = in.read(buffer);
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (buflen <= 0) {
                    return false;
                }
            }
            return true;
        }
        private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
        private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
        public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while(isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) {
                throw new NumberFormatException();
            }
            while(true){
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                }else if(b == -1 || !isPrintableChar(b)){
                    return minus ? -n : n;
                }else{
                    throw new NumberFormatException();
                }
                b = readByte();
            }
        }
        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
            return (int) nl;
        }
        public double nextDouble() { return Double.parseDouble(next());}
    }
    public static class BasicMath {
        private final int MAX = 510000;
        private final int MOD = 1000000007;
        private final long[] fac = new long[MAX];
        private final long[] finv= new long[MAX];
        private final long[] inv = new long[MAX];
        public long factorial(long num){
            if (num<2)return 1;
            else return num*factorial(num-1);
        }
        public int arraySum(int[] array){
            int ans = 0;
            for (int value : array) ans += value;
            return ans;
        }
        public double log(double base,double antilogarithm){
            return Math.log(antilogarithm)/Math.log(base);
        }
        public int gcd(int x,int y){
            if (y==0)return x;
            else return gcd(y,x%y);
        }
        public int lcm(int x,int y){
            return x/gcd(x,y)*y;
        }
        public HashMap<Long,Long> factorization(long num){
            HashMap<Long,Long> hash = new HashMap<>();
            long n = num;
            long count = 2;
            while (n>1){
                while (n%count==0){
                    n/=count;
                    if (hash.containsKey(count))hash.put(count,hash.get(count)+1);
                    else hash.put(count,1L);
                }
                count++;
            }
            return hash;
        }
        public int sum(int[] num){
            int ans = 0;
            for (int i = 0; i < num.length;i++){
                ans += num[i];
            }
            return ans;
        }
        public int[] reduce(int small, int big){
            int ins = 2;
            while (ins<=small){
                if (small%ins==0&&big%ins==0){
                    small/=ins;
                    big/=ins;
                    ins = 2;
                }else {
                    ins++;
                }
            }
            return new int[]{small, big};
        }
        public int reduceCount(int small, int big){
            int ins = 2;
            int ans = 0;
            while (ins<=small){
                if (small%ins==0&&big%ins==0){
                    small/=ins;
                    big/=ins;
                    ins = 2;
                    ans++;
                }else {
                    ins++;
                }
            }
            return ans;
        }
        public int binarySearch(int[] array, int target){
            int pos = -1;
            int left = 0;
            int right = array.length-1;
            int middle;
            while (pos==-1&&left<=right){
                middle = (left+right)/2;
                if (array[middle]==target)pos=middle;
                else if (array[middle]>target)right=middle-1;
                else left=middle+1;
            }
            return pos;
        }
        public long modPow(long a, long n, long mod){
            long res = 1;
            while (n>0){
                if ((n&1)!=0)res = res*a%mod;
                a=a*a%mod;
                n >>= 1;
            }
            return res;
        }
        public long modinv(long a, long mod){return modPow(a,mod-2,mod);}
        public void COMinit(){
            fac[0]=1;
            fac[1]=1;
            finv[0]=1;
            finv[1]=1;
            inv[1]=1;
            for (int i=2;i<MAX;i++){
                fac[i]=fac[i-1]*i%MOD;
                inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
                finv[i]=finv[i-1]*inv[i]%MOD;
            }
        }
        public long COM(int n, int k){
            if (n<k)return 0;
            if (n<0||k<0)return 0;
            return fac[n]*(finv[k]*finv[n-k]%MOD)%MOD;
        }
        public long HCOM(int n, int k){
            return COM(n+k-1,k);
        }
    }
    public static class Queue<T>{
        private final int max = 1000000;
        private final T[] queue = (T[]) new Object[max];
        private int tail = 0; private int head = 0;
        public void init(){this.head = 0; tail = 0;}
        public boolean isEmpty(){return (head==tail);}
        public boolean isFull(){return (head == (tail+1)%max);}
        public void enqueue(T v){
            if (isFull()){
                System.out.println("error: queue is full.");
                return;
            }
            queue[tail++] = v;
            if (tail == max) tail = 0;
        }
        public T dequeue(){
            if (isEmpty()){
                System.out.println("error: queue is empty");
                return null;
            }
            T res = queue[head];
            ++head;
            if (head == max) head = 0;
            return res;
        }
    }
    public static class Stack<T>{
        private final int max = 100000;
        private final T[] stack =(T[]) new Object[max];
        private int top = 0;
        public boolean isEmpty(){return (top==0);}
        public boolean isFull(){return (top==max);}
        public void init(){top=0;}
        public void push(T v){
            if (isFull()){
                System.out.println("error: stack is full.");
                return;
            }
            stack[top++] = v;
        }
        public T pop(){
            if (isEmpty()){
                System.out.println("error: stack is empty");
                return null;
            }
            return stack[--top];
        }
    }
    public static class Pair<T>{
        private T A;
        private T B;
        Pair(T a,T b){
            A=a;
            B=b;
        }
        public void changeA(T a){A=a;}
        public void changeB(T b){B=b;}
        public void changeAB(T a,T b){A=a;B=b;}
        public static Comparator numberSortWithA = new Comparator() {
            @Override
            public int compare(Object o1, Object o2) {
                Pair p = (Pair)o1;
                Pair q = (Pair)o2;
                if ((int)p.A<(int)q.A)return -1;
                else if ((int)p.A==(int)q.A)return 0;
                else return 1;
            }
        };
        public static Comparator numberSortWithB = new Comparator() {
            @Override
            public int compare(Object o1, Object o2) {
                Pair p = (Pair)o1;
                Pair q = (Pair)o2;
                if ((int)p.B<(int)q.B)return -1;
                else if ((int)p.B==(int)q.B)return 0;
                else return 1;
            }
        };
    }
    public static class Graph{
        private final HashMap<Integer,HashSet<Integer>> graph = new HashMap<>();
        private final int Size;
        private int[] Depth;
        Graph(int size){
            for (int i=0;i<size;i++)graph.put(i,new HashSet<>());
            Size = size;
        }

        public void addEdge(int node1,int node2){
            if (graph.containsKey(node1)&&graph.containsKey(node2)){
                graph.get(node1).add(node2);
                graph.get(node2).add(node1);
            }
        }
        public void addDirectEdge(int from,int to){
            if (graph.containsKey(from)&&graph.containsKey(to)){
                graph.get(from).add(to);
            }
        }
        public int[] Depth(int root){
            Queue<Integer> q = new Queue<>();
            q.enqueue(root);
            boolean[] seen = new boolean[Size];
            Arrays.fill(seen, false);
            seen[root]=true;
            int[] depth = new int[Size];
            depth[root]=0;
            while (!q.isEmpty()){
                int v = q.dequeue();
                for (int u : graph.get(v)){
                    if (!seen[u]){
                        seen[u]=true;
                        q.enqueue(u);
                        depth[u]=depth[v]+1;
                    }
                }
            }
            Depth = depth.clone();
            return depth;
        }
    }
    public static class UnionFindTree{
        int[] parent;
        int[] rank;
        int[] Size;
        public UnionFindTree(int size){
            this.parent = new int[size];
            this.rank = new int[size];
            this.Size = new int[size];
            for (int i=0;i<size;i++){
                makeSet(i);
            }
        }
        public void makeSet(int i){
            parent[i]=i;
            rank[i]=0;
            Size[i]=1;
        }
        public void union(int x, int y){
            int xRoot = find(x);
            int yRoot = find(y);
            if (rank[xRoot]>rank[yRoot]){
                parent[yRoot] = xRoot;
                Size[xRoot]+=Size[yRoot];
            }else if (rank[xRoot]<rank[yRoot]){
                parent[xRoot] = yRoot;
                Size[yRoot]+=Size[xRoot];
            }else if (xRoot!=yRoot){
                parent[yRoot] = xRoot;
                rank[xRoot]++;
                Size[xRoot]+=Size[yRoot];
            }
        }
        public int find(int i){
            if (i!= parent[i]){
                parent[i]=find(parent[i]);
            }
            return parent[i];
        }
        public boolean same(int x, int y){
            return find(x) == find(y);
        }
        public int getSize(int i){return Size[find(i)];}
    }
}

Score : 100 points

Problem Statement

There are H rows and W columns of white square cells.

You will choose h of the rows and w of the columns, and paint all of the cells contained in those rows or columns.

How many white cells will remain?

It can be proved that this count does not depend on what rows and columns are chosen.

Constraints

All values in input are integers.

1 \leq H, W \leq 20

1 \leq h \leq H

1 \leq w \leq W

Input

Input is given from Standard Input in the following format:

H W
h w

Output

Print the number of white cells that will remain.

Sample Input 1

3 2
2 1

Sample Output 1

1

There are 3 rows and 2 columns of cells. When two rows and one column are chosen and painted in black, there is always one white cell that remains.

Sample Input 2

5 5
2 3

Sample Output 2

6

Sample Input 3

2 4
2 4

Sample Output 3

0


import java.util.*;

class Main
{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int a = Integer.parseInt(sc.next());
		int b = Integer.parseInt(sc.next());

		int h = Integer.parseInt(sc.next());
		int w = Integer.parseInt(sc.next());

		int black = h * b + w * a - w * h;

		System.out.println(a * b - black);
	}
}


import java.util.*;

public class Main{
	public static void main(String[] args){
    	Scanner scan = new Scanner(System.in);
      	int counter = 0;
      	int N = scan.nextInt();
      	int M = scan.nextInt();
      	int C = scan.nextInt();
      	int[] B = new int[M];
      	for(int i = 0; i < M; i++){
        		B[i] = scan.nextInt();
        }
      	for(int k = 0; k < N; k++){
      		int[] A = new int[M];
      		for(int i = 0; i < M; i++){
        		A[i] = scan.nextInt();
        	}
      		int sahen = 0;
      		for(int j = 0; j < M; j++){
        		sahen = sahen + A[j] * B[j];
        	}
      		if(sahen + C > 0){
        		counter++;
        	}
        }
      	System.out.print(counter);

    }

}


import java.util.HashMap;
	import java.util.Map;
	import java.util.Map.Entry;
	import java.util.Scanner;
	import java.util.TreeMap;

	public class Main{

		public static void main(String[] args) {

			Scanner scanner = new Scanner(System.in);

			Map<Integer,Integer> list = new HashMap<Integer,Integer>();

			String[] temp = scanner.nextLine().split(" ");
			int n = Integer.parseInt(temp[0]);
			int m = Integer.parseInt(temp[1]);
			int t1 = 0;
			int t2 = 0;
			for(int i = 0; i < n; i ++) {
				temp = scanner.nextLine().split(" ");
				t1 = Integer.parseInt(temp[0]);
				t2 = Integer.parseInt(temp[1]);
				if(list.containsKey(t1)) {

					t2 = list.get(t1);
					list.put(t1, t2);
				}else {
					list.put(t1, t2);
				}
			}
			TreeMap<Integer, Integer> sorted = new TreeMap<>(list);

			long ans = 0;
			int kakaku = 0;
			int kazu = 0;
			int flag = m;
			for(Entry<Integer,Integer> entry : sorted.entrySet()) {
				kakaku = entry.getKey();
				kazu = entry.getValue();

				for(int i = 0; i < kazu ; i++) {
					flag--;
					ans += kakaku;
					if(flag == 0) {
						break;
					}
				}
				if(flag == 0) {
					break;
				}
			}

			System.out.println(ans);

		}

	}

Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.


import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        String s;
        String k="2019/04/30";
        Scanner input = new Scanner(System.in);
        s = input.nextLine();
        if(s.contains("2019"))
        {
        if(s.equals(k))
        {
            System.out.println("Heisei");
        }
        else
        {
            System.out.println("TBD");
        }
        }
    }
}

Score : 100 points

Problem Statement

You are given a string S as input. This represents a valid date in the year 2019 in the yyyy/mm/dd format. (For example, April 30, 2019 is represented as 2019/04/30.)

Write a program that prints Heisei if the date represented by S is not later than April 30, 2019, and prints TBD otherwise.

Constraints

S is a string that represents a valid date in the year 2019 in the yyyy/mm/dd format.

Input

Input is given from Standard Input in the following format:

S

Output

Print Heisei if the date represented by S is not later than April 30, 2019, and print TBD otherwise.

Sample Input 1

2019/04/30

Sample Output 1

Heisei

Sample Input 2

2019/11/01

Sample Output 2

TBD


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.AbstractMap;
import java.util.NoSuchElementException;

/** テンプレート */
public class Main {
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		int N = sc.nextInt();

//		int M =  sc.nextInt();
//
		double sum = 0;
		for(int i = 0; i < N; i++) {
			double x = sc.nextDouble();
			String s = 	sc.next();

			if(s.equals("JPY")) {
				sum += x ;
			}else {
				sum += x *  380000.0;
			}
		}

		//************************************/
		// ここから出力処理
		//************************************/
		PrintWriter out = new PrintWriter(System.out);

		out.println(sum);
		// 最後に必ずFlush
		out.flush();
	}
}

///** テンプレート */
//public class Main {
//	public static void main(String[] args) {
//		FastScanner sc = new FastScanner();
//		int N = sc.nextInt();
//		int M =  sc.nextInt();
//
//		int[] ary = new int[N];
//		for(int i = 0; i < N; i++) {
//			ary[i] = sc.nextInt();
//		}
//
//		//************************************/
//		// ここから出力処理
//		//************************************/
//		PrintWriter out = new PrintWriter(System.out);
//
//
//		out.println("hoge");
//
//		// 最後に必ずFlush
//		out.flush();
//	}
//}

/** スキャン用 */
class FastScanner {
	private final InputStream in = System.in;
	private final byte[] buffer = new byte[1024];
	private int ptr = 0;
	private int buflen = 0;

	/** クラス内部用だよ */
	private boolean hasNextByte() {
		if (ptr < buflen) {
			return true;
		} else {
			ptr = 0;
			try {
				buflen = in.read(buffer);
			} catch (IOException e) {
				e.printStackTrace();
			}
			if (buflen <= 0) {
				return false;
			}
		}
		return true;
	}

	/** クラス内部用だよ */
	private int readByte() {
		if (hasNextByte()) {
			return buffer[ptr++];
		} else {
			return -1;
		}
	}

	/**
	 * クラス内部用だよ
	 * ASCII の文字の内、表示用の文字を返す関数
	 *
	 * @return 改行とか制御文字じゃない、表示用文字ならtrue
	 * */
	private static boolean isPrintableChar(int c) {
		return 33 <= c && c <= 126;
	}

	/**
	 * @return 改行文字とか空白以外を除いた、次の文字があればtrue
	 * */
	public boolean hasNext() {
		while (hasNextByte() && !isPrintableChar(buffer[ptr])) {
			ptr++;
		}
		return hasNextByte();
	}

	/**
	 *
	 * @return 次の文字列
	 */
	public String next() {
		if (!hasNext()) {
			throw new NoSuchElementException();
		}

		StringBuilder sb = new StringBuilder();
		int b = readByte();
		while (isPrintableChar(b)) {
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	/**
	 *
	 * @return 次のLong
	 */
	public long nextLong() {
		if (!hasNext())
			throw new NoSuchElementException();
		long n = 0;
		boolean minus = false;
		int b = readByte();
		if (b == '-') {
			minus = true;
			b = readByte();
		}
		if (b < '0' || '9' < b) {
			throw new NumberFormatException();
		}
		while (true) {
			if ('0' <= b && b <= '9') {
				n *= 10;
				n += b - '0';
			} else if (b == -1 || !isPrintableChar(b)) {
				return minus ? -n : n;
			} else {
				throw new NumberFormatException();
			}
			b = readByte();
		}
	}

	/**
	 *
	 * @return 次のInt
	 */
	public int nextInt() {
		long nl = nextLong();
		if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE)
			throw new NumberFormatException();
		return (int) nl;
	}

	/**
	 *
	 * @return 次のDouble
	 */
	public double nextDouble() {
		return Double.parseDouble(next());
	}
}

class Pair<K, V> extends AbstractMap.SimpleEntry<K, V> {
	/** serialVersionUID. */
	private static final long serialVersionUID = 6411527075103472113L;

	public Pair(final K key, final V value) {
		super(key, value);
	}
}

Score : 200 points

Problem Statement

Takahashi received otoshidama (New Year's money gifts) from N of his relatives.

You are given N values x_1, x_2, ..., x_N and N strings u_1, u_2, ..., u_N as input. Each string u_i is either JPY or BTC, and x_i and u_i represent the content of the otoshidama from the i-th relative.

For example, if x_1 = 10000 and u_1 = JPY, the otoshidama from the first relative is 10000 Japanese yen; if x_2 = 0.10000000 and u_2 = BTC, the otoshidama from the second relative is 0.1 bitcoins.

If we convert the bitcoins into yen at the rate of 380000.0 JPY per 1.0 BTC, how much are the gifts worth in total?

Constraints

2 \leq N \leq 10

u_i = JPY or BTC.

If u_i = JPY, x_i is an integer such that 1 \leq x_i \leq 10^8.

If u_i = BTC, x_i is a decimal with 8 decimal digits, such that 0.00000001 \leq x_i \leq 100.00000000.

Input

Input is given from Standard Input in the following format:

N
x_1 u_1
x_2 u_2
:
x_N u_N

Output

If the gifts are worth Y yen in total, print the value Y (not necessarily an integer).

Output will be judged correct when the absolute or relative error from the judge's output is at most 10^{-5}.

Sample Input 1

2
10000 JPY
0.10000000 BTC

Sample Output 1

48000.0

The otoshidama from the first relative is 10000 yen. The otoshidama from the second relative is 0.1 bitcoins, which is worth 38000.0 yen if converted at the rate of 380000.0 JPY per 1.0 BTC. The sum of these is 48000.0 yen.

Outputs such as 48000 and 48000.1 will also be judged correct.

Sample Input 2

3
100000000 JPY
100.00000000 BTC
0.00000001 BTC

Sample Output 2

138000000.0038

In this case, outputs such as 138001000 and 1.38e8 will also be judged correct.


import java.util.Scanner;

public class Main {
    public static void main(String [] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int a = scanner.nextInt();
        int b = scanner.nextInt();
        int c = scanner.nextInt();

        int[] length = new int[n];
        for (int i = 0; i < n; ++i) {
            length[i] = scanner.nextInt();
        }
        int res = dfs(length, 0, a, b, c, 0, 0, 0);
        System.out.println(res);

    }

    static int dfs(int [] length, int current, int a, int b, int c, int sumA, int sumB, int sumC) {
        if (current == length.length) {
            if (sumA != 0 && sumB != 0 && sumC != 0) {
                return Math.abs(a - sumA) + Math.abs(b - sumB) + Math.abs(c - sumC) - 30;
            }
            return Integer.MAX_VALUE;
        }
        int res1 = dfs(length, current + 1, a, b, c, sumA, sumB, sumC);
        int res2 = dfs(length, current + 1, a, b ,c,sumA + length[current], sumB, sumC) + 10;
        int res3 = dfs(length, current + 1, a, b, c, sumA,sumB + length[current], sumC) + 10;
        int res4 = dfs(length, current + 1, a, b, c, sumA, sumB, sumC + length[current]) + 10;
        return Math.min(Math.min(res1, res2), Math.min(res3, res4));
    }
}

Score : 300 points

Problem Statement

You have N bamboos. The lengths (in centimeters) of these are l_1, l_2, ..., l_N, respectively.

Your objective is to use some of these bamboos (possibly all) to obtain three bamboos of length A, B, C. For that, you can use the following three kinds of magics any number:

Extension Magic: Consumes 1 MP (magic point). Choose one bamboo and increase its length by 1.

Shortening Magic: Consumes 1 MP. Choose one bamboo of length at least 2 and decrease its length by 1.

Composition Magic: Consumes 10 MP. Choose two bamboos and combine them into one bamboo. The length of this new bamboo is equal to the sum of the lengths of the two bamboos combined. (Afterwards, further magics can be used on this bamboo.)

At least how much MP is needed to achieve the objective?

Constraints

3 \leq N \leq 8

1 \leq C < B < A \leq 1000

1 \leq l_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N A B C
l_1
l_2
:
l_N

Output

Print the minimum amount of MP needed to achieve the objective.

Sample Input 1

5 100 90 80
98
40
30
21
80

Sample Output 1

23

We are obtaining three bamboos of lengths 100, 90, 80 from five bamboos 98, 40, 30, 21, 80. We already have a bamboo of length 80, and we can obtain bamboos of lengths 100, 90 by using the magics as follows at the total cost of 23 MP, which is optimal.

Use Extension Magic twice on the bamboo of length 98 to obtain a bamboo of length 100. (MP consumed: 2)

Use Composition Magic on the bamboos of lengths 40, 30 to obtain a bamboo of length 70. (MP consumed: 10)

Use Shortening Magic once on the bamboo of length 21 to obtain a bamboo of length 20. (MP consumed: 1)

Use Composition Magic on the bamboo of length 70 obtained in step 2 and the bamboo of length 20 obtained in step 3 to obtain a bamboo of length 90. (MP consumed: 10)

Sample Input 2

8 100 90 80
100
100
90
90
90
80
80
80

Sample Output 2

0

If we already have all bamboos of the desired lengths, the amount of MP needed is 0. As seen here, we do not necessarily need to use all the bamboos.

Sample Input 3

8 1000 800 100
300
333
400
444
500
555
600
666

Sample Output 3

243


import java.util.Arrays;
import java.util.Scanner;

public class Main {

    void run() {
        Scanner sc = new Scanner(System.in);
        int a = sc.nextInt(), b = sc.nextInt();

        if (b % a == 0) {
            System.out.println(a + b);
        } else {
            System.out.println(b - a);
        }
    }

    void debug(Object...os) {
        System.err.println(Arrays.deepToString(os));
    }

    public static void main(String[] args) {
        new Main().run();
    }
}

Score : 100 points

Problem Statement

You are given positive integers A and B.

If A is a divisor of B, print A + B; otherwise, print B - A.

Constraints

All values in input are integers.

1 \leq A \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

If A is a divisor of B, print A + B; otherwise, print B - A.

Sample Input 1

4 12

Sample Output 1

16

As 4 is a divisor of 12, 4 + 12 = 16 should be printed.

Sample Input 2

8 20

Sample Output 2

12

Sample Input 3

1 1

Sample Output 3

2

1 is a divisor of 1.


import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		//		new Main().solveA();
		//		new Main().solveB();
		new Main().solveC();
		// new Main().solveD();
	}

	private void solveA() {
		Scanner scanner = null;
		int lineAB = 0;
		int lineBC = 0;

		try {
			scanner = new Scanner(System.in);
			lineAB = scanner.nextInt();
			lineBC = scanner.nextInt();

			if (lineBC % lineAB == 0) {
				System.out.println(lineAB + lineBC);
			} else {
				System.out.println(lineBC - lineAB);
			}

			System.out.println("");

		} finally {
			if (scanner != null) {
				scanner.close();
			}
		}
	}

	private void solveB() {
		Scanner scanner = null;
		int nNin = 0;
		int mSyurui = 0;

		try {
			scanner = new Scanner(System.in);
			String line1 = scanner.nextLine();
			String[] wkStr1 = line1.split(" ");
			nNin = Integer.parseInt(wkStr1[0]);
			mSyurui = Integer.parseInt(wkStr1[1]);

			List<List<Integer>> wkList = new ArrayList<List<Integer>>();

			for (int i = 0; i < nNin; i++) {
				String wk = scanner.nextLine();
				String[] wkStr = wk.split(" ");
				List<Integer> wkInnerList = new ArrayList<Integer>();
				for (int j = 1; j < wkStr.length; j++) {
					wkInnerList.add(Integer.parseInt(wkStr[j]));
				}
				wkList.add(wkInnerList);
			}

			int result = 0;
			List<Integer> alreadyFinishList = new ArrayList<Integer>();
			for (int i = 0; i < wkList.size(); i++) {
				List<Integer> wkInnerList = wkList.get(i);
				for (int i2 = 0; i2 < wkInnerList.size(); i2++) {
					int syokuNum = wkInnerList.get(i2);
					if (alreadyFinishList.contains(syokuNum)) {
						continue;
					}
					if (isContain(nNin, wkList, i, syokuNum)) {
						result++;
					}
					alreadyFinishList.add(syokuNum);
				}

			}
			System.out.println(result);

		} finally {
			if (scanner != null) {
				scanner.close();
			}
		}
	}

	private boolean isContain(int nNin, List<List<Integer>> wkList, int i, int syokuNum) {
		int result = 0;
		for (int j = 0; j < nNin; j++) {
			if (i == j || j >= wkList.size()) {
				continue;
			}

			List<Integer> wkSarchList = wkList.get(j);
			if (wkSarchList.contains(syokuNum)) {
				result++;
			}

		}
		return result == nNin - 1;
	}

	public long maximuKouyakuNArgs(long currentKouyaku, long[] nums, int start) {
		long res = currentKouyaku;
		if (start < nums.length && start != nums.length) {
			res = maximumKouyaku2Args(nums[start], currentKouyaku);
			return maximuKouyakuNArgs(res, nums, start + 1);
		} else {
			return res;
		}
	}

	public long maximumKouyaku2Args(long num1, long num2) {
		try {
			long wkVal1;
			long wkVal2;
			if (num1 < num2) {
				wkVal1 = num2;
				wkVal2 = num1;
			} else {
				wkVal1 = num1;
				wkVal2 = num2;
			}
			long res = wkVal1 % wkVal2;
			if (res != 0) {
				return maximumKouyaku2Args(wkVal2, res);
			} else {
				return wkVal2;
			}

		} catch (Exception e) {
			System.out.println("num1 : " + num1 + " / num2:" + num2);
			e.printStackTrace();
			return -1;
		}

	}

	private void solveC() {
		Scanner scanner = null;
		int monsterNum = 0;

		try {
			scanner = new Scanner(System.in);
			monsterNum = scanner.nextInt();
			long[] monsters = new long[monsterNum];
			for (int i = 0; i < monsterNum; i++) {
				monsters[i] = scanner.nextInt();
			}
			//
			//			Arrays.sort(monsters);
			//			int minHP = Integer.MAX_VALUE;
			//			if (monsters[0] == monsters[monsters.length - 1]) {
			//				System.out.println(monsters[0]);
			//				return;
			//			} else {
			//				minHP = getMinHP(monsters, minHP);
			//			}
			System.out.println(maximuKouyakuNArgs(monsters[0], monsters, 0));

		} finally {
			if (scanner != null) {
				scanner.close();
			}
		}
	}

	private Map<Integer, Integer> wkHP = new HashMap<Integer, Integer>();

	private int getMinHP(int[] monsters, int minHP) {
		int resultHP = minHP;
		try {
			if (wkHP.get(minHP) != null) {
				return wkHP.get(minHP);
			}
			for (int i = 0; i < monsters.length; i++) {
				if (i == 0) {
					continue;
				}
				if (monsters[i] % monsters[0] == 0) {
					resultHP = Math.min(monsters[0], resultHP);
				} else {
					resultHP = Math.min(monsters[i] % monsters[0], resultHP);
				}
			}
			if (resultHP > 2) {
				resultHP = getMinHP(monsters, resultHP);
			}
			wkHP.put(minHP, resultHP);
		} catch (Exception e) {
			e.printStackTrace();
		}
		return resultHP;

	}

	private void solveD() {
		Scanner scanner = null;
		int lineAB = 0;
		int lineBC = 0;
		int lineCA = 0;

		try {
			scanner = new Scanner(System.in);
			lineAB = scanner.nextInt();

			System.out.println("");

		} finally {
			if (scanner != null) {
				scanner.close();
			}
		}
	}
}

Score : 300 points

Problem Statement

There are N monsters, numbered 1, 2, ..., N.

Initially, the health of Monster i is A_i.

Below, a monster with at least 1 health is called alive.

Until there is only one alive monster, the following is repeated:

A random alive monster attacks another random alive monster.

As a result, the health of the monster attacked is reduced by the amount equal to the current health of the monster attacking.

Find the minimum possible final health of the last monster alive.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible final health of the last monster alive.

Sample Input 1

4
2 10 8 40

Sample Output 1

2

When only the first monster keeps on attacking, the final health of the last monster will be 2, which is minimum.

Sample Input 2

4
5 13 8 1000000000

Sample Output 2

1

Sample Input 3

3
1000000000 1000000000 1000000000

Sample Output 3

1000000000


import java.util.*;
import java.io.*;

public class Main{
    static int[] dp, a;
    static int amin=10;
    static final int num[]={0,2,5,5,4,5,6,3,7,6};
    static ArrayList<Integer> ans=new ArrayList<>();

    static int dp(int i){
        int max=0;
        if(i<amin) return Integer.MIN_VALUE;
        if(dp[i]>0) return dp[i];
        for(int k:a) if(dp(i-num[k])+1>max) max=dp(i-num[k])+1;
        dp[i]=max;
        return dp[i];
    }
    public static void main(String[] args) throws Exception {
        FastScanner fsc=new FastScanner();
        final int n=fsc.nextInt(), m=fsc.nextInt();
        a=new int[m];
        dp=new int[Math.max(n+1, 8)];
        for(int i=0;i<m;i++){
            a[i]=fsc.nextInt();
            dp[num[a[i]]]=1;
            amin=Math.min(amin, a[i]);
        }
        fsc.close();
        if(dp[2]>0){
            dp[4]=2; dp[6]=3;
            if(dp[3]>0){
                dp[5]=2; dp[7]=3;
            }
            else if(dp[5]>0) dp[7]=2;
        }
        else if(dp[3]>0){
            dp[6]=2;
            if(dp[4]>0) dp[7]=2;
        }
        int d=dp(n);
        //for(int i=0;i<=n;i++) System.out.println("dp["+i+"]:"+dp[i]);
        int now=n;
        Arrays.sort(a);
        for(int i=d;i>=1;i--){
            for(int j=a.length-1;j>=0;j--){
                if(now-num[a[j]]<0) continue;
                if(i==1){
                    if(now-num[a[j]]==0){
                        ans.add(a[j]);
                        now=now-num[a[j]];
                        break;
                    }
                }
                else{
                    if(dp[now-num[a[j]]]==i-1){
                        ans.add(a[j]);
                        now-=num[a[j]];
                        break;
                    }
                }
            }
        }
        if(now!=0) throw new Exception();
        if(ans.size()!=d) throw new Exception();
        StringBuilder sb=new StringBuilder("");
        for(int k:ans) sb.append(k);
        System.out.println(sb);
    }
    static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
        private boolean hasNextByte(){
            if(ptr<buflen) return true;
            else{
                ptr = 0;
                try{
                    buflen = in.read(buffer);
                }
                catch(IOException e){
                    e.printStackTrace();
                }
                if(buflen<=0) return false;
            }
        return true;
        }
        private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
        private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
        public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while(isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) throw new NumberFormatException();
            while(true){
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                }
                else if(b == -1 || !isPrintableChar(b)) return minus ? -n : n;
                else throw new NumberFormatException();
                b = readByte();
            }
        }
        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
            return (int) nl;
        }
        public double nextDouble() { return Double.parseDouble(next());}
        public void close(){
            try{in.close();}
            catch(IOException e){e.printStackTrace();}
        }
    }
}

Score : 400 points

Problem Statement

Find the largest integer that can be formed with exactly N matchsticks, under the following conditions:

Every digit in the integer must be one of the digits A_1, A_2, ..., A_M (1 \leq A_i \leq 9).

The number of matchsticks used to form digits 1, 2, 3, 4, 5, 6, 7, 8, 9 should be 2, 5, 5, 4, 5, 6, 3, 7, 6, respectively.

Constraints

All values in input are integers.

2 \leq N \leq 10^4

1 \leq M \leq 9

1 \leq A_i \leq 9

A_i are all different.

There exists an integer that can be formed by exactly N matchsticks under the conditions.

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_M

Output

Print the largest integer that can be formed with exactly N matchsticks under the conditions in the problem statement.

Sample Input 1

20 4
3 7 8 4

Sample Output 1

777773

The integer 777773 can be formed with 3 + 3 + 3 + 3 + 3 + 5 = 20 matchsticks, and this is the largest integer that can be formed by 20 matchsticks under the conditions.

Sample Input 2

101 9
9 8 7 6 5 4 3 2 1

Sample Output 2

71111111111111111111111111111111111111111111111111

The output may not fit into a 64-bit integer type.

Sample Input 3

15 3
5 4 6

Sample Output 3

654


import java.io.*;
import java.util.Scanner;
import java.util.TreeSet;

public class Main{
    /*public static void */
    public static void main(String []args) throws IOException
    {
        Scanner sc = new Scanner (System.in);

        int n = sc.nextInt();
        int k = sc.nextInt();

        if(n%k==1)
            System.out.println("YES");
        else
            System.out.println("NO");
    }
}

Score : 100 points

Problem Statement

Determine if we can choose K different integers between 1 and N (inclusive) so that no two of them differ by 1.

Constraints

1\leq N,K\leq 100

N and K are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

If we can choose K integers as above, print YES; otherwise, print NO.

Sample Input 1

3 2

Sample Output 1

YES

We can choose 1 and 3.

Sample Input 2

5 5

Sample Output 2

NO

Sample Input 3

31 10

Sample Output 3

YES

Sample Input 4

10 90

Sample Output 4

NO


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.UncheckedIOException;
import java.util.StringTokenizer;

class Main{
	static boolean[][] mati=new boolean[4][4];
	static int[] ns = {0,1,2,3};
	public static void main(String[] args) {
		SC sc=new SC(System.in);
		for(int i=0; i<3; i++) {
			int f=sc.nextInt()-1;
			int t=sc.nextInt()-1;
			mati[f][t]=true;
			mati[t][f]=true;
		}
		permute(ns, ns.length);
		pl("NO");
	}
	private static void swap(int[] v, int i, int j) {
		int t = v[i]; v[i] = v[j]; v[j] = t;
	}

	public static void permute(int[] v, int n) {
		if (n == 1) {
			if(mati[ns[0]][ns[1]] && mati[ns[1]][ns[2]] && mati[ns[2]][ns[3]]) {
				pl("YES");
				System.exit(0);
			}
			else {
				//
			}
		}
		else {
			for (int i = 0; i < n; i++) {
				permute(v, n-1);
				if (n % 2 == 1) {
					swap(v, 0, n-1);
				} else {
					swap(v, i, n-1);
				}
			}
		}
	}
	public static void pl(Object o) {
		System.out.println(o);
	}
	public static void pl() {
		System.out.println();
	}
	public static void p(Object o) {
		System.out.print(o);
	}
	static class SC {
		private BufferedReader reader = null;
		private StringTokenizer tokenizer = null;
		public SC(InputStream in) {
			reader = new BufferedReader(new InputStreamReader(in));
		}
		public String next() {
			if (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new UncheckedIOException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int nextInt() {
			return Integer.parseInt(next());
		}
		public long nextLong() {
			return Long.parseLong(next());
		}
		public double nextDouble() {
			return Double.parseDouble(next());
		}
		public String nextLine() {
			try {
				return reader.readLine();
			} catch (IOException e) {
				throw new UncheckedIOException(e);
			}
		}
	}
}

Score : 200 points

Problem Statement

There are four towns, numbered 1,2,3 and 4.
Also, there are three roads. The i-th road connects different towns a_i and b_i bidirectionally.
No two roads connect the same pair of towns. Other than these roads, there is no way to travel between these towns, but any town can be reached from any other town using these roads.

Determine if we can visit all the towns by traversing each of the roads exactly once.

Constraints

1 \leq a_i,b_i \leq 4(1\leq i\leq 3)

a_i and b_i are different. (1\leq i\leq 3)

No two roads connect the same pair of towns.

Any town can be reached from any other town using the roads.

Input

Input is given from Standard Input in the following format:

a_1 b_1
a_2 b_2
a_3 b_3

Output

If we can visit all the towns by traversing each of the roads exactly once, print YES; otherwise, print NO.

Sample Input 1

4 2
1 3
2 3

Sample Output 1

YES

We can visit all the towns in the order 1,3,2,4.

Sample Input 2

3 2
2 4
1 2

Sample Output 2

NO

Sample Input 3

2 1
3 2
4 3

Sample Output 3

YES


import java.util.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		double t = scn.nextDouble();
		int x = scn.nextInt();
		System.out.println(t/x);
	}
}

Score : 100 points

Problem Statement

In order to pass the entrance examination tomorrow, Taro has to study for T more hours.

Fortunately, he can leap to World B where time passes X times as fast as it does in our world (World A).

While (X \times t) hours pass in World B, t hours pass in World A.

How many hours will pass in World A while Taro studies for T hours in World B?

Constraints

All values in input are integers.

1 \leq T \leq 100

1 \leq X \leq 100

Input

Input is given from Standard Input in the following format:

T X

Output

Print the number of hours that will pass in World A.

The output will be regarded as correct when its absolute or relative error from the judge's output is at most 10^{-3}.

Sample Input 1

8 3

Sample Output 1

2.6666666667

While Taro studies for eight hours in World B where time passes three times as fast, 2.6666... hours will pass in World A.

Note that an absolute or relative error of at most 10^{-3} is allowed.

Sample Input 2

99 1

Sample Output 2

99.0000000000

Sample Input 3

1 100

Sample Output 3

0.0100000000