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

@SuppressWarnings("unused")
public class Main {
    FastScanner in;
    PrintWriter out;
    int MOD = (int)1e9+7;
    long ceil(long a, long b){return (a + b - 1) / b;}
    long gcd(long a, long b){return b == 0 ? a : gcd(b, a % b);}
    long lcm(long a, long b){return a / gcd(a, b) * b;}

    void solve(){
        int N = in.nextInt(), R = in.nextInt();
        if(N < 10) R += 100 * (10 - N);
        out.println(R);
    }

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

    private void m() {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out);
        /*
        try {
            String path = "input.txt";
            out = new PrintWriter(new BufferedWriter(new FileWriter(new File(path))));
        }catch (IOException e){
            e.printStackTrace();
        }
        */
        solve();
        out.flush();
        in.close();
        out.close();
    }

    static class FastScanner {
        private Reader input;

        public FastScanner() {this(System.in);}
        public FastScanner(InputStream stream) {this.input = new BufferedReader(new InputStreamReader(stream));}
        public void close() {
            try {
                this.input.close();
            } catch (IOException e) {
                e.printStackTrace();
            }
        }

        public int nextInt() {return (int) nextLong();}

        public long nextLong() {
            try {
                int sign = 1;
                int b = input.read();
                while ((b < '0' || '9' < b) && b != '-' && b != '+') {
                    b = input.read();
                }
                if (b == '-') {
                    sign = -1;
                    b = input.read();
                } else if (b == '+') {
                    b = input.read();
                }
                long ret = b - '0';
                while (true) {
                    b = input.read();
                    if (b < '0' || '9' < b) return ret * sign;
                    ret *= 10;
                    ret += b - '0';
                }
            } catch (IOException e) {
                e.printStackTrace();
                return -1;
            }
        }

        public double nextDouble() {
            try {
                double sign = 1;
                int b = input.read();
                while ((b < '0' || '9' < b) && b != '-' && b != '+') {
                    b = input.read();
                }
                if (b == '-') {
                    sign = -1;
                    b = input.read();
                } else if (b == '+') {
                    b = input.read();
                }
                double ret = b - '0';
                while (true) {
                    b = input.read();
                    if (b < '0' || '9' < b) break;
                    ret *= 10;
                    ret += b - '0';
                }
                if (b != '.') return sign * ret;
                double div = 1;
                b = input.read();
                while ('0' <= b && b <= '9') {
                    ret *= 10;
                    ret += b - '0';
                    div *= 10;
                    b = input.read();
                }
                return sign * ret / div;
            } catch (IOException e) {
                e.printStackTrace();
                return Double.NaN;
            }
        }

        public char nextChar() {
            try {
                int b = input.read();
                while (Character.isWhitespace(b)) {
                    b = input.read();
                }
                return (char) b;
            } catch (IOException e) {
                e.printStackTrace();
                return 0;
            }
        }

        public String nextStr() {
            try {
                StringBuilder sb = new StringBuilder();
                int b = input.read();
                while (Character.isWhitespace(b)) {
                    b = input.read();
                }
                while (b != -1 && !Character.isWhitespace(b)) {
                    sb.append((char) b);
                    b = input.read();
                }
                return sb.toString();
            } catch (IOException e) {
                e.printStackTrace();
                return "";
            }
        }

        public String nextLine() {
            try {
                StringBuilder sb = new StringBuilder();
                int b = input.read();
                while (b != -1 && b != '\n') {
                    sb.append((char) b);
                    b = input.read();
                }
                return sb.toString();
            } catch (IOException e) {
                e.printStackTrace();
                return "";
            }
        }

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

        public int[] nextIntArrayDec(int n) {
            int[] res = new int[n];
            for (int i = 0; i < n; i++) {
                res[i] = nextInt() - 1;
            }
            return res;
        }

        public int[] nextIntArray1Index(int n) {
            int[] res = new int[n + 1];
            for (int i = 0; i < n; i++) {
                res[i + 1] = nextInt();
            }
            return res;
        }

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

        public long[] nextLongArrayDec(int n) {
            long[] res = new long[n];
            for (int i = 0; i < n; i++) {
                res[i] = nextLong() - 1;
            }
            return res;
        }

        public long[] nextLongArray1Index(int n) {
            long[] res = new long[n + 1];
            for (int i = 0; i < n; i++) {
                res[i + 1] = nextLong();
            }
            return res;
        }

        public double[] nextDoubleArray(int n) {
            double[] res = new double[n];
            for (int i = 0; i < n; i++) {
                res[i] = nextDouble();
            }
            return res;
        }
    }
}


import java.util.Scanner;
import java.util.Arrays;
class Main{
  public static void main(String[] args){
    Scanner scan = new Scanner(System.in);
    int N = scan.nextInt();
    int K = scan.nextInt();
    int count = 0;
    while(N >= K){
      N = N / K;
      count += 1;
    }
    System.out.println(count + 1);
  }
}

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

Score : 200 points

Problem Statement

Given is an integer N. Find the number of digits that N has in base K.

Notes

For information on base-K representation, see Positional notation - Wikipedia.

Constraints

All values in input are integers.

1 \leq N \leq 10^9

2 \leq K \leq 10

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of digits that N has in base K.

Sample Input 1

11 2

Sample Output 1

4

In binary, 11 is represented as 1011.

Sample Input 2

1010101 10

Sample Output 2

7

Sample Input 3

314159265 3

Sample Output 3

18

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.*;
public class Main{
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    long n = sc.nextInt();
    int k = sc.nextInt();
    long a = 0;
    int count = 0;
    for(int i=0 ; i<n ; i++){
      a = (long)Math.pow(k,i);
      if(n/a>0){
        count++;
      }else{
        count++;
        break;
      }
    }
    if(count==1){
      System.out.println(count);
    }else{
      System.out.println(count-1);
    }

  }
}


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

public class Main {

	public static void main(String[] args) {

		int n = 0;
		int k = 0;
		ArrayList<Integer> anser = new ArrayList<>();

		Scanner scan = new Scanner(System.in);

		n = scan.nextInt();
		k = scan.nextInt();

		while(n>=k) {
			anser.add(n%k);
			n=n/k;
		}

		if(n<k) {
			anser.add(n);
		}
		System.out.println(anser.size());
	}

}


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

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int a = sc.nextInt();
		int b = sc.nextInt();
		int c = sc.nextInt();
		if ((a == b && a != c) || (a == c && a != b) || (b == c && b != a)) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
	}
}

Score: 100 points

Problem Statement

A triple of numbers is said to be poor when two of those numbers are equal but the other number is different from those two numbers.

You will be given three integers A, B, and C. If this triple is poor, print Yes; otherwise, print No.

Constraints

A, B, and C are all integers between 1 and 9 (inclusive).

Input

Input is given from Standard Input in the following format:

A B C

Output

If the given triple is poor, print Yes; otherwise, print No.

Sample Input 1

5 7 5

Sample Output 1

Yes

A and C are equal, but B is different from those two numbers, so this triple is poor.

Sample Input 2

4 4 4

Sample Output 2

No

A, B, and C are all equal, so this triple is not poor.

Sample Input 3

4 9 6

Sample Output 3

No

Sample Input 4

3 3 4

Sample Output 4

Yes


import java.util.Scanner;
class Main{
 public static void main(String args[]){
   Scanner sc = new Scanner(System.in);
   int n = sc.nextInt();
   boolean flag = true;

   for(int i=0;i<n;i++) {
	   int a = sc.nextInt();
	   if(a%2==0&&((a%3)!=0)&&((a%5)!=0)) {
		   flag=false;
	   }
   }
   	System.out.println(flag?"APPROVED":"DENIED");
 }
}

Score: 200 points

Problem Statement

You are an immigration officer in the Kingdom of AtCoder. The document carried by an immigrant has some number of integers written on it, and you need to check whether they meet certain criteria.

According to the regulation, the immigrant should be allowed entry to the kingdom if and only if the following condition is satisfied:

All even numbers written on the document are divisible by 3 or 5.

If the immigrant should be allowed entry according to the regulation, output APPROVED; otherwise, print DENIED.

Notes

The condition in the statement can be rephrased as "If x is an even number written on the document, x is divisible by 3 or 5".
Here "if" and "or" are logical terms.

Constraints

All values in input are integers.

1 \leq N \leq 100

1 \leq A_i \leq 1000

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \dots A_N

Output

If the immigrant should be allowed entry according to the regulation, print APPROVED; otherwise, print DENIED.

Sample Input 1

5
6 7 9 10 31

Sample Output 1

APPROVED

The even numbers written on the document are 6 and 10.

All of them are divisible by 3 or 5, so the immigrant should be allowed entry.

Sample Input 2

3
28 27 24

Sample Output 2

DENIED

28 violates the condition, so the immigrant should not be allowed entry.


import java.util.Arrays;
import java.util.List;
import java.util.Scanner;
import java.util.concurrent.atomic.AtomicBoolean;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] arg) {
        Scanner scan = new Scanner(System.in);
        int n = Integer.parseInt(scan.next());
        scan.nextLine();
        List<Integer> aList = Arrays.stream(scan.nextLine().split(" "))
                .map(Integer::parseInt).collect(Collectors.toList());

        AtomicBoolean bool = new AtomicBoolean(false);
        aList.forEach(a -> {
            if(a % 2 == 0) {
                if(a % 3 != 0 && a % 5 != 0) {
                    System.out.println("DENIED");
                    System.exit(0);
                }
            }
        });
        System.out.println("APPROVED");
    }
}


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		boolean approved = true;
		for(int i = 0; i < n ; i++) {
			int val = sc.nextInt();
			if(val % 2 == 0) {
				if(val % 3 == 0 || val % 5 == 0) {
					continue;
				}else {
					approved = false;
					break;
				}
			}
		}
		sc.close();
		if(approved) {
			System.out.println("APPROVED");
		}else {
			System.out.println("DENIED");
		}
	}

}


import java.util.Scanner;

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

    private void run() {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int[] A = new int[N];
        boolean isOk = true;
        for(int i = 0; i < N; i++) {
            A[i] = scanner.nextInt();
            if(A[i]  % 2 == 0) {
                if(!(A[i] % 6 == 0 || A[i] % 10 == 0)) {
                    isOk = false;
                }
            }
        }
        if(isOk){
            System.out.println("APPROVED");
        }else{
            System.out.println("DENIED");
        }
        return;
    }
}


import java.util.*;
import java.io.PrintWriter;
public class Main{
	public static void main(String[] args){
		Scanner sc=new Scanner(System.in);
		String s=sc.next();
		long[][] dp=new long[s.length()+1][2];
		dp[s.length()][1]=11;
		for(int i=s.length()-1; i>=0; i--){

				dp[i][0]=Math.min(dp[i+1][0]+(s.charAt(i)-'0'),dp[i+1][1]+(s.charAt(i)-'0'));
				dp[i][1]=Math.min(dp[i+1][0]+11-(s.charAt(i)-'0'),dp[i+1][1]+9-(s.charAt(i)-'0'));

		}
		System.out.println(Math.min(dp[0][0],dp[0][1]));
	}
}

Score: 500 points

Problem Statement

In the Kingdom of AtCoder, only banknotes are used as currency. There are 10^{100}+1 kinds of banknotes, with the values of 1, 10, 10^2, 10^3, \dots, 10^{(10^{100})}. You have come shopping at a mall and are now buying a takoyaki machine with a value of N. (Takoyaki is the name of a Japanese snack.)

To make the payment, you will choose some amount of money which is at least N and give it to the clerk. Then, the clerk gives you back the change, which is the amount of money you give minus N.

What will be the minimum possible number of total banknotes used by you and the clerk, when both choose the combination of banknotes to minimize this count?

Assume that you have sufficient numbers of banknotes, and so does the clerk.

Constraints

N is an integer between 1 and 10^{1,000,000} (inclusive).

Input

Input is given from Standard Input in the following format:

N

Output

Print the minimum possible number of total banknotes used by you and the clerk.

Sample Input 1

36

Sample Output 1

8

If you give four banknotes of value 10 each, and the clerk gives you back four banknotes of value 1 each, a total of eight banknotes are used.

The payment cannot be made with less than eight banknotes in total, so the answer is 8.

Sample Input 2

91

Sample Output 2

3

If you give two banknotes of value 100, 1, and the clerk gives you back one banknote of value 10, a total of three banknotes are used.

Sample Input 3

314159265358979323846264338327950288419716939937551058209749445923078164062862089986280348253421170

Sample Output 3

243


import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        try (Scanner sc = new Scanner(System.in)) {
            String S = sc.next();
            String T = sc.next();
            int A = sc.nextInt();
            int B = sc.nextInt();
            String U = sc.next();

            if (U.equals(S)) {
                A--;
            } else {
                B--;
            }

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

Score : 100 points

Problem Statement

We have A balls with the string S written on each of them and B balls with the string T written on each of them.

From these balls, Takahashi chooses one with the string U written on it and throws it away.

Find the number of balls with the string S and balls with the string T that we have now.

Constraints

S, T, and U are strings consisting of lowercase English letters.

The lengths of S and T are each between 1 and 10 (inclusive).

S \not= T

S=U or T=U.

1 \leq A,B \leq 10

A and B are integers.

Input

Input is given from Standard Input in the following format:

S T
A B
U

Output

Print the answer, with space in between.

Sample Input 1

red blue
3 4
red

Sample Output 1

2 4

Takahashi chose a ball with red written on it and threw it away.
Now we have two balls with the string S and four balls with the string T.

Sample Input 2

red blue
5 5
blue

Sample Output 2

5 4

Takahashi chose a ball with blue written on it and threw it away.
Now we have five balls with the string S and four balls with the string T.


import java.util.*;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.List;
import java.util.Collections;

public class Main {
	public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        Random rand = new Random();
        //ArrayList<Integer> list;
        List<Integer> list = new ArrayList<>(200000);

        int number = sc.nextInt();
        for(int i = 0; i < number ; i++){
            list.add(sc.nextInt());
        }

        Collections.sort(list);
        int count = 0;
        for(int i = 0; i < number - 1 ; i++){
            if(list.get(i).equals(list.get(i+1))){
                count++;
            }else{}
        }
        System.out.println(list);

        if(count == 0){
            System.out.println("YES");
        }else{
            System.out.println("NO");
        }

	}
}


import java.util.Scanner;

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

   for(int i=0;i<s.length();i++){
  System.out.print("x");
   }

 }
}

Score : 200 points

Problem Statement

Given is a string S. Replace every character in S with x and print the result.

Constraints

S is a string consisting of lowercase English letters.

The length of S is between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Replace every character in S with x and print the result.

Sample Input 1

sardine

Sample Output 1

xxxxxxx

Replacing every character in S with x results in xxxxxxx.

Sample Input 2

xxxx

Sample Output 2

xxxx

Sample Input 3

gone

Sample Output 3

xxxx


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

//System.out.println();
public class Main implements Runnable { //クラス名はMain

    public static void main(String[] args) {
    	Thread.setDefaultUncaughtExceptionHandler((t,e)->System.exit(1));
        new Thread(null, new Main(), "", 16 * 1024 * 1024).start(); //16MBスタックを確保して実行
    }

    public void run() {
    	FastScanner sc = new FastScanner(System.in);
        String s = sc.next();
        String rep = s.replaceAll(".", "x");
        System.out.println(rep);
    }
}

//高速なScanner
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;
	}
}


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

public class Main {
    static class pair implements Comparable<pair>{
        int x;
        int idx;
        pair(int x,int idx){
            this.x=x;
            this.idx=idx;
        }

        @Override
        public int compareTo(pair p) {
            if(p.x==x){
                return idx-p.idx;
            }
            return x-p.x;
        }
        }

    public static long ciel(long x,long y){
        long res=x/y;
        if(x%y!=0)res++;
        return res;
    }

    public static long gcd(long u,long v){
        if(v==0)return u;
        return gcd(v, u%v);
    }
    static int  mod= (int) (1e9+7);
    static long memo[][];
    static long inf= (long) 2e9;
    static int[]cost,weight;
    static ArrayList<Integer> primes;
    static ArrayList<Integer> primeFactors(int N)		// O(sqrt(N) / ln sqrt(N))
    {
        ArrayList<Integer> factors = new ArrayList<Integer>();		//take abs(N) in case of -ve integers
        int idx = 0, p = primes.get(idx);

        while(p * p <= N)
        {
            while(N % p == 0) { factors.add(p); N /= p; }
            p = primes.get(++idx);
        }

        if(N != 1)						// last prime factor may be > sqrt(N)
            factors.add(N);				// for integers whose largest prime factor has a power of 1
        return factors;
    }
    static int[] isComposite;
    static void sieve(int N)	// O(N log log N)
    {
        isComposite = new int[N+1];
        isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
        primes = new ArrayList<Integer>();

        for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N))
            if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
            {
                primes.add(i);
                if(1l * i * i <= N)
                    for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
                        isComposite[j] = 1;
            }
    }

    /*
     * 2. Primality Test
     *
     *  Preprocessing: call sieve with sqrt(N), O(sqrt(N) log log sqrt(N))
     *  Query: best case O(1), worst case O(sqrt(N) / log sqrt(N))
     */
    static boolean isPrime(int N)
    {
        if(N < isComposite.length)
            return isComposite[N] == 0;
        for(int p: primes)					//may stop if p * p > N
            if(N%p==0)
                return false;
        return true;
    }
    static long modPow(long a, long e, long mod)	// O(log e)
    {
        a %= mod;
        long res = 1;
        while(e > 0)
        {
            if((e & 1) == 1)
                res = (res * a) % mod;
            a = (a * a) % mod;
            e >>= 1;
        }
        return res;
    }
    public static void main(String[] args) throws Exception {
        Scanner sc=new Scanner(System.in);
        PrintWriter pw=new PrintWriter(System.out);
        String s=sc.next();
       int n=s.length();
       while (n-->0)pw.print("x");
        pw.close();
    }

    public static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

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

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

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

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public boolean ready() throws IOException, IOException {
            return br.ready();
        }

    }
}


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

public class Main {
    public static void main(String[] args) throws IOException {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        Set<Integer> set = new HashSet<>();
        for (int i = 0; i < n; i ++){
            int t = scan.nextInt();
            if (set.contains(t)){
                System.out.println("NO");
                return;
            }else
                set.add(t);
        }
        System.out.println("YES");
    }
}

Score : 300 points

Problem Statement

Given is a sequence of integers A_1, A_2, ..., A_N.
If its elements are pairwise distinct, print YES; otherwise, print NO.

Constraints

2 ≤ N ≤ 200000

1 ≤ A_i ≤ 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 ... A_N

Output

If the elements of the sequence are pairwise distinct, print YES; otherwise, print NO.

Sample Input 1

5
2 6 1 4 5

Sample Output 1

YES

The elements are pairwise distinct.

Sample Input 2

6
4 1 3 1 6 2

Sample Output 2

NO

The second and fourth elements are identical.

Sample Input 3

2
10000000 10000000

Sample Output 3

NO


import java.util.HashMap;
import java.util.Scanner;

public class Main {
	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);
		int N = scan.nextInt();
		String anser = "YES";

		HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();

		for (int i = 0; i < N; i++) {
			map.put(scan.nextInt(), i);
		}

		for (int i = 0; i < N; i++) {
			int size = map.size();
			map.remove(map.get(i));

			if (size - 1 > map.size()) {
				anser = "NO";
				break;
			}
		}
		System.out.println(anser);
	}

}


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		byte n = sc.nextByte(),c=0,i=0,j=0;
		int a[] = new int [n];

		while(c<n) {
			a[c] = sc.nextInt();
			c++;
		}
		sc.close();

		for(;i<n;i++) {
			for(;j<n;j++) {
				if(i==j) continue;
				if(a[i]==a[j]) {
					System.out.println("NO");
					return ;
				}
			}
		}

		System.out.println("YES");

	}

}


import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.LinkedList;
import java.util.NoSuchElementException;
import java.util.Queue;

class Main {

  static class Solver {
    final FastScanner sc;
    final PrintWriter writer;

    Solver(final FastScanner sc, final PrintWriter writer) {
      this.sc = sc;
      this.writer = writer;
    }

    void run() {
      int n = sc.nextInt();
      int k = sc.nextInt();
      int[] p = sc.nextIntArray(n);
      Queue<Integer> list = new LinkedList<>();
      int currentSum = 0;
      for (int i = 0; i < k; i++) {
        int tmp = p[i];
        currentSum += tmp;
        list.add(p[i]);
      }
      int max = currentSum;
      int pointer = 0;
      for (int i = k; i < n; i++) {
        int tmp = p[i];
        int first = list.poll();
        list.add(tmp);
        currentSum = currentSum + tmp - first;
        max = Math.max(max, currentSum);
        if (max == currentSum) {
          pointer = i - (k - 1);
        }
      }
      double predicts = 0;
      for (int i = pointer; i < pointer + k; i++) {
        int tmp = p[i];
        int tmpSum = 0;
        for (int j = 1; j <= tmp; j++) {
          tmpSum += j;
        }
        predicts += (double) tmpSum / tmp;
      }
      writer.println(predicts);
    }
  }

  public static void main(final String[] args) {
    final FastScanner sc = new FastScanner();
    try (final PrintWriter w = new PrintWriter(System.out)) {
      new Solver(sc, w).run();
      w.flush();
    }
  }

  // CRT stuff
  static class MOD {
    private static final int MOD = (int) 1e9 + 7;

    static long plus(long x, long y) {
      x %= MOD;
      y %= MOD;
      return (x + y) % MOD;
    }

    static long sub(long x, long y) {
      x %= MOD;
      y %= MOD;
      return (x - y + MOD) % MOD;
    }

    static long multi(long x, long y) {
      x %= MOD;
      y %= MOD;
      return x * y % MOD;
    }

    static long div(long x, long y) {
      x %= MOD;
      y %= MOD;
      return x * pow(y, MOD - 2) % MOD;
    }

    static long pow(final long a, final long p) {
      if (p == 0)
        return 1;
      if (p % 2 == 0) {
        final long halfP = p / 2;
        final long root = pow(a, halfP);
        return root * root % MOD;
      } else
        return a * pow(a, p - 1) % MOD;
    }
  }

  // FastScanner
  static class FastScanner {
    private final InputStream in = System.in;
    private final byte[] buffer = new byte[1024];
    private int ptr = 0;
    private int buflen = 0;

    public FastScanner() {
    }

    private boolean hasNextByte() {
      if (ptr < buflen) {
        return true;
      } else {
        ptr = 0;
        try {
          buflen = in.read(buffer);
        } catch (final IOException e) {
          e.printStackTrace();
        }
        if (buflen <= 0) {
          return false;
        }
      }
      return true;
    }

    private int readByte() {
      if (hasNextByte())
        return buffer[ptr++];
      else
        return -1;
    }

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

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

Score : 400 points

Problem Statement

We have N dice arranged in a line from left to right. The i-th die from the left shows p_i numbers from 1 to p_i with equal probability when thrown.

We will choose K adjacent dice, throw each of them independently, and compute the sum of the numbers shown. Find the maximum possible value of the expected value of this sum.

Constraints

1 ≤ K ≤ N ≤ 200000

1 ≤ p_i ≤ 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
p_1 ... p_N

Output

Print the maximum possible value of the expected value of the sum of the numbers shown.

Your output will be considered correct when its absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

5 3
1 2 2 4 5

Sample Output 1

7.000000000000

When we throw the third, fourth, and fifth dice from the left, the expected value of the sum of the numbers shown is 7. This is the maximum value we can achieve.

Sample Input 2

4 1
6 6 6 6

Sample Output 2

3.500000000000

Regardless of which die we choose, the expected value of the number shown is 3.5.

Sample Input 3

10 4
17 13 13 12 15 20 10 13 17 11

Sample Output 3

32.000000000000


import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.StringTokenizer;
import java.util.TreeSet;

import java.lang.Integer;

public class Main {
	static  ArrayList<Pair>adj[];
	static PrintWriter out=new PrintWriter(System.out);
	 public static int mod = 1000000007;

	static int notmemo[][][][];
	static int k;
	static int a[];
	static int b[];
	static int m;
	static char c[];

	static int trace[];
	static int h;
	static int x;
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
	s=sc.nextLine();
	c=s.toCharArray();
	k=sc.nextInt();
	n=c.length;
	memo=new long[c.length+1][k+1][2];
	for (int i = 0; i < c.length+1; i++) {
		for (int j = 0; j <k+1; j++) {
			Arrays.fill(memo[i][j], -1);
		}
	}
	int count=0;

	System.out.println(dp(0,0,0));
	}
	static String s;
		static boolean isOn(int S, int j) { return (S & 1 << j) != 0; }
	static int y, d;
	static void extendedEuclid(int a, int b)
	{
		if(b == 0) { x = 1; y = 0; d = a; return; }
		extendedEuclid(b, a % b);
		int x1 = y;
		int y1 = x - a / b * y;
		x = x1; y = y1;
	}
	static long dp(int i,int count,int state) {
		if(count>k) {
			return 0;
		}
	if(i==n) {
		if(count==k) {
			return 1;
		}
		else
		return 0;
	}

	if(memo[i][count][state]!=-1) {
		return memo[i][count][state];
	}

	long total=0;
	for (int j = 0; j <10; j++) {
		if(state==0&&j>Integer.parseInt(""+c[i])) {
			continue;
		}

		else if(state==0&&j==Integer.parseInt(""+c[i])) {
		if(j==0)
			total+=dp(i+1,count,0);
		else
			total+=dp(i+1,count+1,0);

		}
		else if(j==0) {
			total+=dp(i+1,count,1);
		}
		else {
			total+=dp(i+1,count+1,1);
		}
	}

	return memo[i][count][state]=total;

	}
	static boolean f=true;
	static class SegmentTree {		// 1-based DS, OOP
		int N; 			//the number of elements in the array as a power of 2 (i.e. after padding)
		int[] array, sTree, lazy;

		SegmentTree(int[] in)
		{
			array = in; N = in.length - 1;
			sTree = new int[N*4];		//no. of nodes = 2*N - 1, we add one to cross out index zero
			lazy = new int[N*4];

		//	build(1,1,N);
		}

		void build(int node, int b, int e)	// O(n)
		{
			if(b == e)
				sTree[node] = array[b];
			else
			{
				int mid = b + e >> 1;
				build(node<<1,b,mid);
				build(node<<1|1,mid+1,e);
				sTree[node] = sTree[node<<1]&sTree[node<<1|1];
			}
		}

		void update_point(int index, int val)			// O(log n)
		{
			index += N - 1;
			sTree[index] += val;
			while(index>1)
			{
				index >>= 1;
				sTree[index] = sTree[index<<1] + sTree[index<<1|1];
			}
		}

		void update_range(int i, int j,int val)		// O(log n)
		{
			update_range(1,1,N,i,j,val);
		}

		void update_range(int node, int b, int e, int i, int j,int val)
		{
			if(i > e || j < b)
				return;
			if(b >= i && e <= j)
			{
				sTree[node]|= val;
				lazy[node]|=val;

			//LAZY LAZY
			}
			else
			{
				int mid = b + e >> 1;
				propagate(node, b, mid, e);
				update_range(node<<1,b,mid,i,j,val);
				update_range(node<<1|1,mid+1,e,i,j,val);
				sTree[node] = sTree[node<<1]&sTree[node<<1|1];
			}

		}
		void propagate(int node, int b, int mid, int e)
		{

			lazy[node<<1] |= lazy[node];
			lazy[(node<<1)|1] |= lazy[node];
			sTree[node<<1] |= lazy[node];
			sTree[(node<<1)|1] |= lazy[node];
			lazy[node] = 0;

		}

		int query(int i, int j)
		{
			return query(1,1,N,i,j);
		}

		int query(int node, int b, int e, int i, int j)	// O(log n)
		{
			if(i>e || j <b)
				return -1;
			if(b>= i && e <= j)
				return sTree[node];
			int mid = b + e >> 1;
			propagate(node, b, mid, e);
			int q1 = query(node<<1,b,mid,i,j);
			int q2 = query(node<<1|1,mid+1,e,i,j);
			return q1 &q2;

		}
	}
static long memo[][][];

		static long nCr1(int n , int k)
	{
		if(n < k)
			return 0;
		if(k == 0 || k == n)
			return 1;
		if(comb[n][k] != -1)
			return comb[n][k];
		if(n - k < k)
			return comb[n][k] = nCr1(n, n - k);
		return comb[n][k] = ((nCr1(n - 1, k - 1)%mod) +( nCr1(n - 1, k)%mod))%mod;
	}

static class UnionFind {
	int[] p, rank, setSize;
	int numSets;
	int max[];
	public UnionFind(int N)
	{
		max=new int[N];
		p = new int[numSets = N];
		rank = new int[N];
		setSize = new int[N];
		for (int i = 0; i < N; i++) {  p[i] = i; setSize[i] = 1;
		max[i]=i;
		}
	}

	public int findSet(int i) { return p[i] == i ? i : (p[i] = findSet(p[i])); }

	public boolean isSameSet(int i, int j) { return findSet(i) == findSet(j); }

	public void unionSet(int i, int j)
	{
		if (isSameSet(i, j))
			return;
		numSets--;
		int x = findSet(i), y = findSet(j);
		if(rank[x] > rank[y]) { p[y] = x; setSize[x] += setSize[y];
		max[x]=Math.max(max[x], max[y]);

		}
		else
		{	p[x] = y; setSize[y] += setSize[x];
			if(rank[x] == rank[y]) rank[y]++;
			max[y]=Math.max(max[x],max[y]);

		}
	}

	private int getmax(int j) {
	return max[findSet(j)];
	}

	public int numDisjointSets() { return numSets; }

	public int sizeOfSet(int i) { return setSize[findSet(i)]; }
}

/**private static void trace(int i, int time) {
if(i==n)
	return;

long ans=dp(i,time);
if(time+a[i].t<a[i].burn&&(ans==dp(i+1,time+a[i].t)+a[i].cost)) {

	trace(i+1, time+a[i].t);

	l1.add(a[i].idx);
    return;
}
    trace(i+1,time);

}**/

	static class incpair implements Comparable<incpair>
{
	int a; long b;
	int idx;
	incpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i; }

	public int compareTo(incpair e){ return (int) (b - e.b);	}
}
	static class decpair implements Comparable<decpair>
{
	int a; long b;int idx;

	decpair(int a, long dirg,int i) {this.a = a;	b = dirg; idx=i;}

	public int compareTo(decpair e){ return (int) (e.b- b);	}
}

	static long allpowers[];
	static class Quad implements Comparable<Quad>{
		int u;
		int v;
		char state;
		int turns;
		public Quad(int i, int j, char c, int k) {
		u=i;
		v=j;
		state=c;
		turns=k;
		}
		public int compareTo(Quad e){ return (int) (turns - e.turns);	}

	}
	static long dirg[][];
	static Edge[] driver;

	static int n;
		static class Edge implements Comparable<Edge>
	{
		int node; long cost;

		Edge(int a, long dirg) { node = a;	cost = dirg; }

		public int compareTo(Edge e){ return (int) (cost - e.cost);	}
	}

	  static long manhatandistance(long x,long x2,long y,long y2) {
		  return  Math.abs(x-x2)+Math.abs(y-y2);
	  }
	  static long fib[];
	  static long fib(int n) {
		  if(n==1||n==0) {
			  return 1;
		  }
		  if(fib[n]!=-1) {
			  return fib[n];
		  }
		  else
			  return  fib[n]=  ((fib(n-2)%mod+fib(n-1)%mod)%mod);
	  }
	  static class Pair {
			int a, b;

			Pair(int x, int y) {
				a = x;
				b = y;
			}

		}

	static long[][] comb;

static  class Triple implements Comparable<Triple>{

int l;
int r;
long cost;
int idx;
		public Triple(int a,int b,long l1,int l2) {
			l=a;
			r=b;
			cost=l1;
			idx=l2;
		}
		public int compareTo(Triple x)
		{
			if(l!=x.l || idx==x.idx)
				return l-x.l;
			return -idx;
		}

	}

	static TreeSet<Long> primeFactors(long N)		// O(sqrt(N) / ln sqrt(N))
	{
		TreeSet<Long> factors = new TreeSet<Long>();		//take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while(p * p <= N)
		{
			while(N % p == 0) { factors.add((long) p); N /= p; }
			if(primes.size()>idx+1)
			p = primes.get(++idx);
			else
				break;
		}

		if(N != 1)						// last prime factor may be > sqrt(N)
			factors.add(N);				// for integers whose largest prime factor has a power of 1
		return factors;
	}

	static boolean visited[];
	/**static int bfs(int s)
	{
		Queue<Integer> q = new LinkedList<Integer>();
		q.add(s);
		int count=0;
		int maxcost=0;
		int dist[]=new int[n];
		dist[s]=0;
		while(!q.isEmpty())
		{

			int u = q.remove();
			if(dist[u]==k) {
				break;
			}
			for(Pair v: adj[u])
			{
				maxcost=Math.max(maxcost, v.cost);

			if(!visited[v.v])	{

				visited[v.v]=true;
				q.add(v.v);
				dist[v.v]=dist[u]+1;
				maxcost=Math.max(maxcost, v.cost);
			}
			}

		}
		return maxcost;
	}**/
	public static boolean FindAllElements(int n, int k) {
	 int sum = k;
     int[] A = new int[k];
      Arrays.fill(A, 0, k, 1);

      for (int i = k - 1; i >= 0; --i) {

           while (sum + A[i] <= n) {

              sum += A[i];
              A[i] *= 2;
          }
      }
      if(sum==n) {
      	return true;
      }
      else
    	return false;
	}

	static boolean vis2[][];

	static boolean f2=false;
	static long[][] matMul(long[][] a2, long[][] b, int p, int q, int r)	//C(p x r) = A(p x q) x (q x r) -- O(p x q x r)
	{
		long[][] C = new long[p][r];
		for(int i = 0; i < p; ++i) {
			for(int j = 0; j < r; ++j) {
				for(int k = 0; k < q; ++k) {
					C[i][j] = (C[i][j]+(a2[i][k]%mod * b[k][j]%mod))%mod;
		 C[i][j]%=mod;
				}

			}
		}
		return C;
	}

	static ArrayList<Pair> a1[];
	static int memo1[];

	static boolean vis[];
	static TreeSet<Integer> set=new TreeSet<Integer>();

	static long modPow(long ways, long count, long mod)	// O(log e)
	{
		ways %= mod;
		long res = 1;
		while(count > 0)
		{
			if((count & 1) == 1)
				res = (res * ways) % mod;
			ways = (ways * ways) % mod;
			count >>= 1;
		}
		return res%mod;
	}

	static long gcd(long ans,long b) {
		if(b==0) {
			return ans;
		}
		return gcd(b,ans%b);
	}
	static int[] isComposite;
	static int[] valid;

	static ArrayList<Integer> primes;
	static ArrayList<Integer> l;
	static void sieve(int N)	// O(N log log N)
	{
		isComposite = new int[N+1];
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();
		l=new ArrayList<>();
		valid=new int[N+1];
		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N))
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				l.add(i);
				valid[i]=1;

				for (int j = i*2; j <=N; j +=i) {	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;

					}
			}

		for(int i=0 ; i<primes.size() ; i++) {
			for(int j:primes) {
				if(primes.get(i)*j>N) {
					break;
				}
				valid[primes.get(i)*j]=1;
			}
		}
	}

	public static int[] schuffle(int[] a2) {
		for (int i = 0; i < a2.length; i++) {
			int x=(int)(Math.random()*a2.length);
		int temp=a2[x];
		a2[x]=a2[i];
		a2[i]=temp;
		}
		return a2;
	}
		static int V;
	static long INF=(long) 1E16;

		static class Edge2
		{
			int node;
		long cost;
			long next;
			Edge2(int a, int c,Long long1) { node = a;	cost = long1; next=c;}

		}
static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

		public String next() throws IOException {
			while (st == null || !st.hasMoreTokens())
				st = new StringTokenizer(br.readLine());
			return st.nextToken();
		}

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

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

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

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

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}

Score : 500 points

Problem Statement

Find the number of integers between 1 and N (inclusive) that contains exactly K non-zero digits when written in base ten.

Constraints

1 \leq N < 10^{100}

1 \leq K \leq 3

Input

Input is given from Standard Input in the following format:

N
K

Output

Print the count.

Sample Input 1

100
1

Sample Output 1

19

The following 19 integers satisfy the condition:

1,2,3,4,5,6,7,8,9,10,20,30,40,50,60,70,80,90,100

Sample Input 2

25
2

Sample Output 2

14

The following 14 integers satisfy the condition:

11,12,13,14,15,16,17,18,19,21,22,23,24,25

Sample Input 3

314159
2

Sample Output 3

937

Sample Input 4

9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999
3

Sample Output 4

117879300


import java.util.*;

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

    int h = sc.nextInt();
    int a = sc.nextInt();

    int result = h / a;
    if (h%a>0) {
      result ++;
    }

    System.out.println(result);

  }
}

Score : 100 points

Problem Statement

Serval is fighting with a monster.

The health of the monster is H.

In one attack, Serval can decrease the monster's health by A.
There is no other way to decrease the monster's health.

Serval wins when the monster's health becomes 0 or below.

Find the number of attacks Serval needs to make before winning.

Constraints

1 \leq H \leq 10^4

1 \leq A \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H A

Output

Print the number of attacks Serval needs to make before winning.

Sample Input 1

10 4

Sample Output 1

3

After one attack, the monster's health will be 6.

After two attacks, the monster's health will be 2.

After three attacks, the monster's health will be -2.

Thus, Serval needs to make three attacks to win.

Sample Input 2

1 10000

Sample Output 2

1

Sample Input 3

10000 1

Sample Output 3

10000


import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int num1 = scan.nextInt();
		int num2 = scan.nextInt();
		int i;
		int sum = 0;

		for(i = 0 ; i < num2 ; i++) {
			int attack_point = scan.nextInt();
			num1 -= attack_point;
		}
		scan.close();
		if(num1 > 0) {
			System.out.println("NO");
		}
		else System.out.println("YES");
	}
}

Score : 200 points

Problem Statement

Raccoon is fighting with a monster.

The health of the monster is H.

Raccoon can use N kinds of special moves. Using the i-th move decreases the monster's health by A_i.
There is no other way to decrease the monster's health.

Raccoon wins when the monster's health becomes 0 or below.

If Raccoon can win without using the same move twice or more, print Yes; otherwise, print No.

Constraints

1 \leq H \leq 10^9

1 \leq N \leq 10^5

1 \leq A_i \leq 10^4

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H N
A_1 A_2 ... A_N

Output

If Raccoon can win without using the same move twice or more, print Yes; otherwise, print No.

Sample Input 1

10 3
4 5 6

Sample Output 1

Yes

The monster's health will become 0 or below after, for example, using the second and third moves.

Sample Input 2

20 3
4 5 6

Sample Output 2

No

Sample Input 3

210 5
31 41 59 26 53

Sample Output 3

Yes

Sample Input 4

211 5
31 41 59 26 53

Sample Output 4

No


import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Scanner;
import java.util.function.Supplier;

public class Main {

    static void solve() {
        int h = nextInt();
        int n = nextInt();

        int sum = 0;
        for (int i = 0; i < n; i++) {
            int a = nextInt();
            sum += a;
        }
        if (sum >= h)
            out.println("Yes");
        else
            out.println("No");

    }

    static final int MOD = 1_000_000_007;
    static long[] fac, finv, inv;

    // 階乗(n!)
    static long factorial(long n) {
        long ans = 1;
        for (long i = n; i > 0; i--) {
            ans = ans * i % MOD;
        }
        return ans;
    }

    // nCkの初期化
    static void comInit(int max) {
        fac = new long[max];
        finv = new long[max];
        inv = new long[max];
        fac[0] = fac[1] = 1;
        finv[0] = 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;
        }
    }

    // nCkの計算
    static 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;
    }

    static Map<Integer, List<Integer>> primeFactors = new HashMap<>();

    // 素因数分解
    static List<Integer> primeFactorize(int n) {
        if (primeFactors.containsKey(n))
            return primeFactors.get(n);

        List<Integer> list = new ArrayList<>();
        if (n % 2 == 0) {
            list.addAll(primeFactorize(n / 2));
            list.add(2);
        } else {
            boolean ok = true;
            for (int j = 3; j * j <= n; j += 2) {
                if (n % j == 0) {
                    list.addAll(primeFactorize(n / j));
                    list.addAll(primeFactorize(j));
                    ok = false;
                    break;
                }
            }
            if (ok) {
                if (n != 1)
                    list.add(n);
            }
        }

        primeFactors.put(n, Collections.unmodifiableList(list));
        return list;
    }

    // 2〜nまでの数値を素因数分解したmapを返す.
    // map: 数値 -> (素因数)
    // ex)
    // 2 -> (2)
    // 4 -> (2,2)
    // 12 -> (2,2,3)
    @Deprecated // primeFactorizeを使うべき
    static Map<Integer, List<Integer>> primeFactorizeTo(int n) {
        Map<Integer, List<Integer>> map = new HashMap<>();
        for (int i = 2; i <= n; i++) {
            List<Integer> list = new ArrayList<>();
            int tmp = i;
            for (int j = i - 1; j > 1; j--) {
                if (tmp % j == 0) {
                    if (map.containsKey(j)) {
                        list.addAll(map.get(j));
                    } else {
                        list.add(j);
                    }
                    tmp /= j;
                }
            }
            if (tmp != 1)
                list.add(tmp);
            map.put(i, list);
        }
        return map;
    }

    static PrintWriter out;
    static Scanner sc;
    static boolean debugEnabled;
    static long start;
    static long end;

    static int[][] newIntArray(int h, int w, int value) {
        int[][] ret = new int[h][w];
        for (int i = 0; i < h; i++) {
            for (int j = 0; j < w; j++) {
                ret[i][j] = value;
            }
        }
        return ret;
    }

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

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

    static String nextString() {
        return sc.next();
    }

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

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

    static List<Integer> nextIntList(int n) {
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextInt());
        }
        return list;
    }

    static List<Double> nextDoubleList(int n) {
        List<Double> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add((double) nextInt());
        }
        return list;
    }

    static List<Long> nextLongList(int n) {
        List<Long> list = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            list.add(nextLong());
        }
        return list;
    }

    static char[][] nextCharArray(int h, int w) {
        char[][] c = new char[h][w];
        for (int i = 0; i < h; i++) {
            String str = nextString();
            for (int j = 0; j < w; j++) {
                c[i][j] = str.charAt(j);
            }
        }
        return c;
    }

    // 昇順
    static <T extends Comparable<? super T>> List<T> sort(List<T> list) {
        return sortAsc(list);
    }

    // 昇順
    static <T extends Comparable<? super T>> List<T> sortAsc(List<T> list) {
        Collections.sort(list);
        return list;
    }

    // 降順
    static <T extends Comparable<? super T>> List<T> sortDesc(List<T> list) {
        Collections.sort(list, (e1, e2) -> e2.compareTo(e1));
        return list;
    }

    // greatest common divisor
    // 最大公約数
    static long gcd(long a, long b) {
        if (b == 0) {
            return a;
        }
        return gcd(b, a % b);
    }

    // least common multiple
    // 最小公倍数
    static long lcm(long a, long b) {
        if (a > b) {
            return (a / gcd(a, b)) * b;
        } else {
            return (b / gcd(a, b)) * a;
        }
    }

    // least common multiple
    // 最小公倍数
    // 素因数分解されたListを引数とした場合
    static List<Integer> lcm(List<Integer> a, List<Integer> b) {
        List<Integer> ret = new ArrayList<>(a);
        List<Integer> tmp = new ArrayList<>(b);
        for (Integer i : a) {
            if (tmp.contains(i))
                tmp.remove(i);
        }
        ret.addAll(tmp);
        return ret;
    }

    // baseのn乗を計算を返す
    static int pow(int base, int n) {
        int ret = 1;
        for (int i = 0; i < n; i++) {
            ret *= base;
        }
        return ret;
    }

    // return n^k mod m
    static long powMod(long n, long k, long m) {
        if (k == 0) {
            return 1;
        } else if (k % 2 == 1) {
            return powMod(n, k - 1, m) * n % m;
        } else {
            long tmp = powMod(n, k / 2, m);
            return tmp * tmp % m;
        }
    }

    // intをlength桁のbit文字列に変換
    static String toBitString(int length, int n) {
        StringBuilder sb = new StringBuilder();
        for (int i = length - 1; i >= 0; i--) {
            if ((n >> i) % 2 == 1) {
                sb.append("1");
            } else {
                sb.append("0");
            }
        }
        return sb.toString();
    }

    static void setDebugEnabled() {
        String className = new Object() {
        }.getClass().getEnclosingClass().getSimpleName();

        if (!className.equals("Main"))
            debugEnabled = true;
    }

    static void debug(String msg) {
        if (debugEnabled)
            System.err.println(msg);
    }

    static void debug(Supplier<String> msg) {
        if (debugEnabled)
            System.err.println(msg.get());
    }

    static class UnionFind {
        int[] parent;

        UnionFind(int n) {
            parent = new int[n];
            init();
        }

        void init() {
            for (int i = 0; i < parent.length; i++) {
                parent[i] = i;
            }
        }

        int getRoot(int i) {
            if (parent[i] == i) {
                return i;
            } else {
                return parent[i] = getRoot(parent[i]);
            }
        }

        boolean isSame(int x, int y) {
            return getRoot(x) == getRoot(y);
        }

        void unit(int x, int y) {
            x = getRoot(x);
            y = getRoot(y);
            if (x != y) {
                parent[x] = y;
            }

        }
    }

    public static void main(String[] args) {
        preProcess();
        // System.exit(0)で終了した場合も、解答を出力するためにはshutdown hookの設定が必要
        // しかし、これを有効化すると100[msec]程度処理時間が長くなる
        // Runtime.getRuntime().addShutdownHook(new Thread(() -> postProcess()));
        solve();
        postProcess();
    }

    static void preProcess() {
        start = System.nanoTime();
        setDebugEnabled();
        out = new PrintWriter(System.out);
        sc = new Scanner(System.in);
    }

    static void postProcess() {
        out.flush();
        sc.close();
        end = System.nanoTime();
        debug(() -> "elapsed time: " + (end - start) / 1_000_000 + "[msec]");
    }

}


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

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int h = sc.nextInt();
        int n = sc.nextInt();
        int atk = 0;
        for(int i = 0; i < n; i++){
            atk += sc.nextInt();
        }
        if(atk >= h){
            System.out.println("Yes");
        }else{
            System.out.println("No");
        }
    }
}


import java.util.*;

class Main {
    static long[] memo;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long H = sc.nextLong();
        int N = (int)H;
        int count = 0;
        memo = new long[N*2];
        memo[0] = 0;
        memo[1] = 1;
        while(H != 0) {
            H = H/2;
            count++;
        }
        long ans = calc(count);
        int sum = 0;
        for(int i=0; i<=count; i++) {
            sum += memo[i];
        }
        System.out.println(sum);
    }
    public static long calc(int n){
        //memo[n]が0じゃなければreturn
        if (memo[n] != 0) return memo[n];
        //memo[n]が0だった場合は以下の処理
        if (n == 0) return 0;
        if (n == 1) return 1;
        long ans = 0;
        ans += calc(n-1) + calc(n-1);
        //呼び直しを起こさないようmemo[n]に値を格納
        memo[n] = ans;
        return ans;
    }
}

Score : 400 points

Problem Statement

Caracal is fighting with a monster.

The health of the monster is H.

Caracal can attack by choosing one monster. When a monster is attacked, depending on that monster's health, the following happens:

If the monster's health is 1, it drops to 0.

If the monster's health, X, is greater than 1, that monster disappears. Then, two new monsters appear, each with the health of \lfloor X/2 \rfloor.

(\lfloor r \rfloor denotes the greatest integer not exceeding r.)

Caracal wins when the healths of all existing monsters become 0 or below.

Find the minimum number of attacks Caracal needs to make before winning.

Constraints

1 \leq H \leq 10^{12}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H

Output

Find the minimum number of attacks Caracal needs to make before winning.

Sample Input 1

2

Sample Output 1

3

When Caracal attacks the initial monster, it disappears, and two monsters appear, each with the health of 1.

Then, Caracal can attack each of these new monsters once and win with a total of three attacks.

Sample Input 2

4

Sample Output 2

7

Sample Input 3

1000000000000

Sample Output 3

1099511627775


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

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

		if (n == m) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}

	}
}

Score : 100 points

Problem Statement

Takahashi is participating in a programming contest, AXC001. He has just submitted his code to Problem A.

The problem has N test cases, all of which must be passed to get an AC verdict.

Takahashi's submission has passed M cases out of the N test cases.

Determine whether Takahashi's submission gets an AC.

Constraints

1 \leq N \leq 100

0 \leq M \leq N

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

If Takahashi's submission gets an AC, print Yes; otherwise, print No.

Sample Input 1

3 3

Sample Output 1

Yes

All three test cases have been passed, so his submission gets an AC.

Sample Input 2

3 2

Sample Output 2

No

Only two out of the three test cases have been passed, so his submission does not get an AC.

Sample Input 3

1 1

Sample Output 3

Yes


import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Mufaddal Naya
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        BComparingStrings solver = new BComparingStrings();
        solver.solve(1, in, out);
        out.close();
    }

    static class BComparingStrings {
        public void solve(int testNumber, InputReader c, OutputWriter w) {
            int a = c.readInt(), b = c.readInt();
            String s1 = "", s2 = "";
            for (int i = 0; i < b; i++) {
                s1 += a;
            }
            for (int j = 0; j < a; j++) {
                s2 += b;
            }
//        w.printLine(s1,s2);
//        w.printLine(s1.compareTo(s2));
            if (s1.compareTo(s2) > 0) {
                w.printLine(s2);
            } else {
                w.printLine(s1);
            }
        }

    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void print(Object... objects) {
            for (int i = 0; i < objects.length; i++) {
                if (i != 0) {
                    writer.print(' ');
                }
                writer.print(objects[i]);
            }
        }

        public void printLine(Object... objects) {
            print(objects);
            writer.println();
        }

        public void close() {
            writer.close();
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private InputReader.SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public int readInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public boolean isSpaceChar(int c) {
            if (filter != null) {
                return filter.isSpaceChar(c);
            }
            return isWhitespace(c);
        }

        public static boolean isWhitespace(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }
}

Score : 200 points

Problem Statement

Given are 1-digit positive integers a and b. Consider these two strings: the concatenation of b copies of the digit a, and the concatenation of a copies of the digit b. Which of these is lexicographically smaller?

Constraints

1 \leq a \leq 9

1 \leq b \leq 9

a and b are integers.

Input

Input is given from Standard Input in the following format:

a b

Output

Print the lexicographically smaller of the two strings. (If the two strings are equal, print one of them.)

Sample Input 1

4 3

Sample Output 1

3333

We have two strings 444 and 3333. Between them, 3333 is the lexicographically smaller.

Sample Input 2

7 7

Sample Output 2

7777777


import java.io.*;
import java.math.*;
import java.util.*;
import java.lang.*;

public class Main implements Runnable {
    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;
        private BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();

            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }

                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        public String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        public int nextInt() {
            int c = read();

            while (isSpaceChar(c))
                c = read();

            int sgn = 1;

            if (c == '-') {
                sgn = -1;
                c = read();
            }

            int res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));

            return res * sgn;
        }

        public long nextLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;

            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));
            return res * sgn;
        }

        public double nextDouble() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            double res = 0;
            while (!isSpaceChar(c) && c != '.') {
                if (c == 'e' || c == 'E')
                    return res * Math.pow(10, nextInt());
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            if (c == '.') {
                c = read();
                double m = 1;
                while (!isSpaceChar(c)) {
                    if (c == 'e' || c == 'E')
                        return res * Math.pow(10, nextInt());
                    if (c < '0' || c > '9')
                        throw new InputMismatchException();
                    m /= 10;
                    res += (c - '0') * m;
                    c = read();
                }
            }
            return res * sgn;
        }

        public String readString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            }
            while (!isSpaceChar(c));

            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public String next() {
            return readString();
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);
        }
    }

    public static void main(String args[]) throws Exception {
        new Thread(null, new Main(), "Main", 1 << 26).start();
    }

    static long gcd(long a, long b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }

    static long modPow(long a, long p, long m) {
        if (a == 1) return 1;
        long ans = 1;
        while (p > 0) {
            if (p % 2 == 1) ans = (ans * a) % m;
            a = (a * a) % m;
            p >>= 1;
        }
        return ans;
    }

    static long modInv(long a, long m) {
        return modPow(a, m - 2, m);
    }

    static class Radix {

        static int getMax(int arr[], int n) {
            int mx = arr[0];
            for (int i = 1; i < n; i++)
                if (arr[i] > mx)
                    mx = arr[i];
            return mx;
        }

        static void countSort(int arr[], int n, int exp) {
            int output[] = new int[n];
            int i;
            int count[] = new int[10];
            Arrays.fill(count, 0);

            for (i = 0; i < n; i++)
                count[(arr[i] / exp) % 10]++;

            for (i = 1; i < 10; i++)
                count[i] += count[i - 1];

            for (i = n - 1; i >= 0; i--) {
                output[count[(arr[i] / exp) % 10] - 1] = arr[i];
                count[(arr[i] / exp) % 10]--;
            }

            for (i = 0; i < n; i++)
                arr[i] = output[i];
        }

        static void radixsort(int arr[], int n) {
            int m = getMax(arr, n);

            for (int exp = 1; m / exp > 0; exp *= 10)
                countSort(arr, n, exp);
        }

    }

    class DSU {
        int par[], size[], connectedComponent;

        DSU(int n) {
            par = new int[n];
            size = new int[n];
            for (int i = 0; i < n; i++) {
                size[i] = 1;
                par[i] = i;
            }
            connectedComponent = n;
        }

        int root(int n) {
            if (n == par[n]) return n;
            return par[n] = root(par[n]);
        }

        boolean union(int u, int v) {
            return f_union(root(u), root(v));
        }

        boolean f_union(int u, int v) {
            if (u == v) return false;
            if (size[u] > size[v]) {
                size[u] += size[v];
                par[v] = u;
            } else {
                size[v] += size[u];
                par[u] = v;
            }
            connectedComponent--;
            return true;
        }

        DSU clone_DSU() {
            DSU t = new DSU(par.length);
            t.connectedComponent = connectedComponent;
            for (int i = 0; i < par.length; i++) {
                t.par[i] = par[i];
                t.size[i] = size[i];
            }
            return t;
        }
    }

    class segmentTree {
        long query(int cur, int l, int r, int start, int end, long tree[], long a[], long[] lazy) {
            if (l > end || r < start) return 0;
            if (lazy[cur] != 0) {
                tree[cur] += lazy[cur];
                if (l != r) {
                    lazy[2 * cur + 1] += lazy[cur];
                    lazy[2 * cur + 2] += lazy[cur];
                }
                lazy[cur] = 0;
            }
            if (l >= start && r <= end) return tree[cur];
            int mid = (l + r) >> 1;
            long p1 = query(2 * cur + 1, l, mid, start, end, tree, a, lazy);
            long p2 = query(2 * cur + 2, mid + 1, r, start, end, tree, a, lazy);
            return p1 + p2;
        }

        void build(int cur, int l, int r, long tree[], long a[]) {
            if (l == r) {
                tree[cur] = a[l];
                return;
            }
            int mid = (l + r) >> 1;
            build(2 * cur + 1, l, mid, tree, a);
            build(2 * cur + 2, mid + 1, r, tree, a);
            tree[cur] = tree[2 * cur + 1] + tree[2 * cur + 2];
        }

        void update(int cur, int start, int end, long val, int l, int r, long tree[], long a[], long[] lazy) {
            if (l > end || r < start) return;
            if (lazy[cur] != 0) {
                tree[cur] += (r - l + 1) * lazy[cur];
                if (l != r) {
                    lazy[2 * cur + 1] += lazy[cur];
                    lazy[2 * cur + 2] += lazy[cur];
                }
                lazy[cur] = 0;
            }
            if (l >= start && r <= end) {
                tree[cur] += (r - l + 1) * val;
                if (l != r) {
                    lazy[cur * 2 + 1] += val;
                    lazy[cur * 2 + 2] += val;
                }
                return;
            }
            int mid = (l + r) >> 1;
            update(2 * cur + 1, start, end, val, l, mid, tree, a, lazy);
            update(2 * cur + 2, start, end, val, mid + 1, r, tree, a, lazy);
            tree[cur] = tree[2 * cur + 1] + tree[2 * cur + 2];
        }
    }

    public void run() {
        InputReader sc = new InputReader(System.in);
        // Scanner sc=new Scanner(System.in);
        //  Random sc=new Random();
        PrintWriter out = new PrintWriter(System.out);

        int n=sc.nextInt();
        int a[]=new int[n];
        for (int i = 0; i < n; i++) {
            a[i]=sc.nextInt();
        }
        ArrayList<Integer> l[]=new ArrayList[n];
        for (int i = 0; i <n ; i++) {
            l[i]=new ArrayList<>();
        }
        HashMap<Integer,Integer> h=new HashMap<>();
        long ans=0;
        for (int i = n-1; i >=0; i--) {
            int m=a[i];
            int d=2;
            while (d*d<=m){
                int c=0;
                while (m%d==0){
                    m/=d;
                    c++;
                }
                if(c>0){
                    l[i].add(d);
                    l[i].add(c);
                    h.put(d,Math.max(c,h.getOrDefault(d,0)));
                }
                if(d==2)d=3;
                else d+=2;
            }
            if(m!=1){
                l[i].add(m);
                l[i].add(1);
                h.put(m,Math.max(1,h.getOrDefault(m,0)));
            }

        }

        long lcm=1,mod=1000000007;
        for(int key:h.keySet()){
            lcm=(lcm*modPow(key,h.get(key),mod))%mod;
        }

        for (int i = 0; i <n ; i++) {
            ans=(ans+(lcm*modInv(a[i],mod))%mod)%mod;
        }

        out.println(ans);

        out.close();
    }

}

Score : 500 points

Problem Statement

Given are N positive integers A_1,...,A_N.

Consider positive integers B_1, ..., B_N that satisfy the following condition.

Condition: For any i, j such that 1 \leq i < j \leq N, A_i B_i = A_j B_j holds.

Find the minimum possible value of B_1 + ... + B_N for such B_1,...,B_N.

Since the answer can be enormous, print the sum modulo (10^9 +7).

Constraints

1 \leq N \leq 10^4

1 \leq A_i \leq 10^6

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 minimum possible value of B_1 + ... + B_N for B_1,...,B_N that satisfy the condition, modulo (10^9 +7).

Sample Input 1

3
2 3 4

Sample Output 1

13

Let B_1=6, B_2=4, and B_3=3, and the condition will be satisfied.

Sample Input 2

5
12 12 12 12 12

Sample Output 2

5

We can let all B_i be 1.

Sample Input 3

3
1000000 999999 999998

Sample Output 3

996989508

Print the sum modulo (10^9+7).


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

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

	public void solve() throws IOException {
		FastScanner fs = new FastScanner();
		int h = fs.nextInt();
		int w = fs.nextInt();
		int n = fs.nextInt();

		int max = Math.max(h, w);
		if(n % max == 0) {
			System.out.println(n/max);
		}else {
			System.out.println(n/max + 1);
		}
	}

	public class FastScanner {

		BufferedReader reader;
		private StringTokenizer st;

		public FastScanner() {
			st = null;
			reader = new BufferedReader(new InputStreamReader(System.in));
		}

		public String next() throws IOException {
			if (st == null || !st.hasMoreElements()) {
				st = new StringTokenizer(reader.readLine());
			}
			return st.nextToken();
		}

		public String nextLine() throws IOException {
			st = null;
			String readLine = null;
			readLine = reader.readLine();
			return readLine;
		}

		public int nextInt() throws NumberFormatException, IOException {
			return Integer.parseInt(next());
		}

		public long nextLong() throws NumberFormatException, IOException {
			return Long.parseLong(next());
		}

		public int[] nextIntArr(int n) throws NumberFormatException, IOException {
			int[] retArr = new int[n];
			for (int i = 0; i < n; i++) {
				retArr[i] = nextInt();
			}
			return retArr;
		}

		public long[] nextLongArr(int n) throws NumberFormatException, IOException {
			long[] retArr = new long[n];
			for (int i = 0; i < n; i++) {
				retArr[i] = nextLong();
			}
			return retArr;
		}

		public void close() throws IOException {
			reader.close();
		}
	}

}

Score : 100 points

Problem Statement

We have a grid with H rows and W columns, where all the squares are initially white.

You will perform some number of painting operations on the grid.
In one operation, you can do one of the following two actions:

Choose one row, then paint all the squares in that row black.

Choose one column, then paint all the squares in that column black.

At least how many operations do you need in order to have N or more black squares in the grid?
It is guaranteed that, under the conditions in Constraints, having N or more black squares is always possible by performing some number of operations.

Constraints

1 \leq H \leq 100

1 \leq W \leq 100

1 \leq N \leq H \times W

All values in input are integers.

Input

Input is given from Standard Input in the following format:

H
W
N

Output

Print the minimum number of operations needed.

Sample Input 1

3
7
10

Sample Output 1

2

You can have 14 black squares in the grid by performing the "row" operation twice, on different rows.

Sample Input 2

14
12
112

Sample Output 2

8

Sample Input 3

2
100
200

Sample Output 3

2


import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.io.BufferedWriter;
import java.io.Writer;
import java.io.OutputStreamWriter;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author prakharjain
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        OutputWriter out = new OutputWriter(outputStream);
        APainting solver = new APainting();
        solver.solve(1, in, out);
        out.close();
    }

    static class APainting {
        public void solve(int testNumber, InputReader in, OutputWriter out) {
            int r = in.nextInt();
            int c = in.nextInt();
            int n = in.nextInt();

            out.println(Math.min((n + r - 1) / r, (n + c - 1) / c));
        }

    }

    static class OutputWriter {
        private final PrintWriter writer;

        public OutputWriter(OutputStream outputStream) {
            writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
        }

        public OutputWriter(Writer writer) {
            this.writer = new PrintWriter(writer);
        }

        public void close() {
            writer.close();
        }

        public void println(int i) {
            writer.println(i);
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private InputReader.SpaceCharFilter filter;

        public InputReader(InputStream stream) {
            this.stream = stream;
        }

        public static boolean isWhitespace(int c) {
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public int read() {
            if (numChars == -1) {
                throw new InputMismatchException();
            }
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0) {
                    return -1;
                }
            }
            return buf[curChar++];
        }

        public int nextInt() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') {
                    throw new InputMismatchException();
                }
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

        public boolean isSpaceChar(int c) {
            if (filter != null) {
                return filter.isSpaceChar(c);
            }
            return isWhitespace(c);
        }

        public interface SpaceCharFilter {
            public boolean isSpaceChar(int ch);

        }

    }
}