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

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);
        Solver solver = new Solver();
        solver.solve(in,out);
        out.close();

    }

    static class Solver{
        boolean[] visited;
        ArrayList<Integer>[] adjList;

        public void solve(InputReader in, OutputWriter out){
            char[] s = in.next().toCharArray();
            int a =0;
            int b=0;
            int c=0;
            for(char i:s){
                switch (i){
                    case 'a':a++; break;
                    case 'b':b++; break;
                    case 'c': c++; break;
                }
            }
            if(a==b && a==c && a==1){
                out.printf("Yes");
            }
            else{
                out.printf("No");
            }

        }

        private void dfs(int u){
            visited[u] = true;
            for(int v: adjList[u]){
                if(!visited[v]){
                    dfs(v);
                }
            }
        }

        private void bfs(int s) {
            Queue<Integer> q = new LinkedList<Integer>();
            q.add(s);
            visited[s] = true;
            while(!q.isEmpty()) {
                int u = q.remove();
                for(int v: adjList[u])
                    if(!visited[v]) {
                        visited[v] = true;
                        q.add(v);
                    }
            }
        }

        private  int[] nextAInt(int n, InputReader in){
            int arr[] = new int[n];
            for(int i=0;i<n;i++){
                arr[i]= in.nextInt();
            }
            return arr;
        }

    }

    static class InputReader {
        private InputStream stream;
        private byte[] buf = new byte[1 << 16];
        private int curChar;
        private int numChars;

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

        public int read() {
            if (this.numChars == -1) {
                throw new InputMismatchException();
            } else {
                if (this.curChar >= this.numChars) {
                    this.curChar = 0;

                    try {
                        this.numChars = this.stream.read(this.buf);
                    } catch (IOException var2) {
                        throw new InputMismatchException();
                    }

                    if (this.numChars <= 0) {
                        return -1;
                    }
                }

                return this.buf[this.curChar++];
            }
        }

        public int nextInt() {
            int c;
            for (c = this.read(); isSpaceChar(c); c = this.read()) {
                ;
            }

            byte sgn = 1;
            if (c == 45) {
                sgn = -1;
                c = this.read();
            }

            int res = 0;

            while (c >= 48 && c <= 57) {
                res *= 10;
                res += c - 48;
                c = this.read();
                if (isSpaceChar(c)) {
                    return res * sgn;
                }
            }

            throw new InputMismatchException();
        }

        public String next() {
            int c;
            while (isSpaceChar(c = this.read())) {
                ;
            }

            StringBuilder result = new StringBuilder();
            result.appendCodePoint(c);

            while (!isSpaceChar(c = this.read())) {
                result.appendCodePoint(c);
            }

            return result.toString();
        }

        public static boolean isSpaceChar(int c) {
            return c == 32 || c == 10 || c == 13 || c == 9 || c == -1;
        }

    }

    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 printf(String format, Object... objects) {
            writer.printf(format, objects);
        }

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

    }

}

Score : 100 points

Problem Statement

You are given a string S of length 3 consisting of a, b and c. Determine if S can be obtained by permuting abc.

Constraints

|S|=3

S consists of a, b and c.

Input

Input is given from Standard Input in the following format:

S

Output

If S can be obtained by permuting abc, print Yes; otherwise, print No.

Sample Input 1

bac

Sample Output 1

Yes

Swapping the first and second characters in bac results in abc.

Sample Input 2

bab

Sample Output 2

No

Sample Input 3

abc

Sample Output 3

Yes

Sample Input 4

aaa

Sample Output 4

No


import java.util.*;
public class Main {
    public static void main (String[] args) {
      Scanner sc = new Scanner(System.in);
      String str = sc.next();
      char[] array = str.toCharArray();
      Arrays.sort(array);
      String sortedStr = new String(array);
      String ans = (sortedStr.equals("abc"))? "Yes" : "No";

      System.out.println(ans);
    }
}

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

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.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		if(s.matches("^(?=.*a)(?=.*b)(?=.*c).*")) {
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}

	}

}


import java.io.*;
import java.util.Arrays;
import java.util.stream.IntStream;

public class Main {
	public static void main(String[] args) throws IOException {
		BufferedReader stdin = new BufferedReader(new InputStreamReader(System.in));
		int[] input = Arrays.stream(stdin.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
		int[] nums = IntStream.rangeClosed(input[0], input[1]).toArray();
		IntStream.range(0, nums.length).filter( i -> (i < input[2] || nums.length - 1 -input[2] < i) ).sorted().forEach(i -> System.out.println(nums[i]));
	}
}

Score : 200 points

Problem Statement

Print all the integers that satisfies the following in ascending order:

Among the integers between A and B (inclusive), it is either within the K smallest integers or within the K largest integers.

Constraints

1 \leq A \leq B \leq 10^9

1 \leq K \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B K

Output

Print all the integers that satisfies the condition above in ascending order.

Sample Input 1

3 8 2

Sample Output 1

3
4
7
8

3 is the first smallest integer among the integers between 3 and 8.

4 is the second smallest integer among the integers between 3 and 8.

7 is the second largest integer among the integers between 3 and 8.

8 is the first largest integer among the integers between 3 and 8.

Sample Input 2

4 8 3

Sample Output 2

4
5
6
7
8

Sample Input 3

2 9 100

Sample Output 3

2
3
4
5
6
7
8
9


import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int[] a = new int[3];
        a[0] = sc.nextInt();
        a[1] = sc.nextInt();
        a[2] = sc.nextInt();
        Arrays.sort(a);

        if (a[0] == a[1] && a[1] == a[2]) {
            System.out.println(0);
            return;
        }

        int ans = 0;
        int d1 = a[2] - a[0];
        int d2 = a[2] - a[1];
        ans += d1 / 2 + d2 / 2;

        if (d1 % 2 == 0) {
            if (d2 % 2 == 0) {

            } else {
                ans += 2;
            }
        } else {
            if (d2 % 2 == 0) {
                ans += 2;
            } else {
                ans += 1;
            }
        }

        System.out.println(ans);
    }
}

Score : 300 points

Problem Statement

You are given three integers A, B and C. Find the minimum number of operations required to make A, B and C all equal by repeatedly performing the following two kinds of operations in any order:

Choose two among A, B and C, then increase both by 1.

Choose one among A, B and C, then increase it by 2.

It can be proved that we can always make A, B and C all equal by repeatedly performing these operations.

Constraints

0 \leq A,B,C \leq 50

All values in input are integers.

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the minimum number of operations required to make A, B and C all equal.

Sample Input 1

2 5 4

Sample Output 1

2

We can make A, B and C all equal by the following operations:

Increase A and C by 1. Now, A, B, C are 3, 5, 5, respectively.

Increase A by 2. Now, A, B, C are 5, 5, 5, respectively.

Sample Input 2

2 6 3

Sample Output 2

5

Sample Input 3

31 41 5

Sample Output 3

23


import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.PrintStream;
import java.util.InputMismatchException;
import java.io.IOException;
import java.io.InputStream;

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

    static class TaskB {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int three = 2;
            int two = -1;
            if (n == 3) {
                out.print(2 + " " + 5 + " " + 63);
                return;
            }
            while (true) {
                //int y=1;
                int val = n - three;
                if (val > 0 && val <= 15000 && (val % 3 == 0 || val % 3 == 2)) {
                    two = val;
                    break;
                }
                three += 2;
            }
            System.out.println(two + " " + three);
            int ct = 2;
            if (two != -1) {
                while (two > 0) {

                    out.print(ct + " ");
                    ct += 2;
                    two--;
                }
                ct = 3;
                while (three > 0) {
                    out.print(ct + " ");
                    ct += 6;
                    three--;
                }
            }

        }

    }

    static class InputReader {
        private final InputStream stream;
        private final byte[] buf = new byte[8192];
        private int curChar;
        private int snumChars;

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

        public int read() {
            //*-*------clare------
//remeber while comparing 2 non primitive data type not to use ==
//remember Arrays.sort for primitive data has worst time case complexity of 0(n^2) bcoz it uses quick sort
//again silly mistakes ,yr kb tk krta rhega ye mistakes
//try to write simple codes ,break it into simple things
//knowledge>rating
        /*
        public class Main
        implements Runnable{
    public static void main(String[] args) {
        new Thread(null,new Main(),"Main",1<<26).start();

    }
    public void run() {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskC solver = new TaskC();//chenge the name of task
        solver.solve(1, in, out);
        out.close();
    }
         */

            if (snumChars == -1)
                throw new InputMismatchException();
            if (curChar >= snumChars) {
                curChar = 0;
                try {
                    snumChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (snumChars <= 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 {
                res *= 10;
                res += c - '0';
                c = read();
            } while (!isSpaceChar(c));
            return res * sgn;
        }

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

    }
}

Score : 600 points

Problem Statement

Nagase is a top student in high school. One day, she's analyzing some properties of special sets of positive integers.

She thinks that a set S = \{a_{1}, a_{2}, ..., a_{N}\} of distinct positive integers is called special if for all 1 \leq i \leq N, the gcd (greatest common divisor) of a_{i} and the sum of the remaining elements of S is not 1.

Nagase wants to find a special set of size N. However, this task is too easy, so she decided to ramp up the difficulty. Nagase challenges you to find a special set of size N such that the gcd of all elements are 1 and the elements of the set does not exceed 30000.

Constraints

3 \leq N \leq 20000

Input

Input is given from Standard Input in the following format:

N

Output

Output N space-separated integers, denoting the elements of the set S. S must satisfy the following conditions :

The elements must be distinct positive integers not exceeding 30000.

The gcd of all elements of S is 1, i.e. there does not exist an integer d > 1 that divides all elements of S.

S is a special set.

If there are multiple solutions, you may output any of them. The elements of S may be printed in any order. It is guaranteed that at least one solution exist under the given contraints.

Sample Input 1

3

Sample Output 1

2 5 63

\{2, 5, 63\} is special because gcd(2, 5 + 63) = 2, gcd(5, 2 + 63) = 5, gcd(63, 2 + 5) = 7. Also, gcd(2, 5, 63) = 1. Thus, this set satisfies all the criteria.

Note that \{2, 4, 6\} is not a valid solution because gcd(2, 4, 6) = 2 > 1.

Sample Input 2

4

Sample Output 2

2 5 20 63

\{2, 5, 20, 63\} is special because gcd(2, 5 + 20 + 63) = 2, gcd(5, 2 + 20 + 63) = 5, gcd(20, 2 + 5 + 63) = 10, gcd(63, 2 + 5 + 20) = 9. Also, gcd(2, 5, 20, 63) = 1. Thus, this set satisfies all the criteria.


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

public class Main {
    static int[] w = new int[100010];
    static int[] v = new int[100010];
    static int[][] dp = new int[110][100010];
    static int k;

    static int chmin(int a, int b) {
        if (a > b) {
            a = b;
            return a;
        }
        return a;
    }

    static int chmax(int a, int b) {
        if (a < b) {
            a = b;
            return a;
        }
        return a;
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int train = sc.nextInt();
        int fTrain = sc.nextInt();
        int bus = sc.nextInt();
        int fBus = sc.nextInt();
        System.out.println(Math.min(train, fTrain) + Math.min(bus, fBus));

    }
}

Score : 100 points

Problem Statement

You planned a trip using trains and buses.
The train fare will be A yen (the currency of Japan) if you buy ordinary tickets along the way, and B yen if you buy an unlimited ticket.
Similarly, the bus fare will be C yen if you buy ordinary tickets along the way, and D yen if you buy an unlimited ticket.

Find the minimum total fare when the optimal choices are made for trains and buses.

Constraints

1 \leq A \leq 1 000

1 \leq B \leq 1 000

1 \leq C \leq 1 000

1 \leq D \leq 1 000

All input values are integers.

Input

Input is given from Standard Input in the following format:

A
B
C
D

Output

Print the minimum total fare.

Sample Input 1

600
300
220
420

Sample Output 1

520

The train fare will be 600 yen if you buy ordinary tickets, and 300 yen if you buy an unlimited ticket.
Thus, the optimal choice for trains is to buy an unlimited ticket for 300 yen.
On the other hand, the optimal choice for buses is to buy ordinary tickets for 220 yen.

Therefore, the minimum total fare is 300 + 220 = 520 yen.

Sample Input 2

555
555
400
200

Sample Output 2

755

Sample Input 3

549
817
715
603

Sample Output 3

1152


import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.HashSet;
import java.util.List;
import java.util.Map;
import java.util.Optional;
import java.util.Set;
import java.util.stream.IntStream;

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

	private void solve() {
		Scanner sc = new Scanner(System.in);
		int A = sc.nextInt();
		int B = sc.nextInt();
		int C = sc.nextInt();
		int D = sc.nextInt();
		System.out.println(Math.min(A, B) + Math.min(C, D));
	}

	class Scanner {
		private InputStream in;
		private byte[] buffer = new byte[1024];
		private int index;
		private int length;

		public Scanner(InputStream in) {
			this.in = in;
		}

		private boolean isPrintableChar(int c) {
			return '!' <= c && c <= '~';
		}

		private boolean isDigit(int c) {
			return '0' <= c && c <= '9';
		}

		private boolean hasNextByte() {
			if (index < length) {
				return true;
			} else {
				try {
					length = in.read(buffer);
					index = 0;
				} catch (IOException e) {
					e.printStackTrace();
				}
				return length > 0;
			}
		}

		private boolean hasNext() {
			while (hasNextByte() && !isPrintableChar(buffer[index])) {
				index++;
			}
			return hasNextByte();
		}

		private int readByte() {
			return hasNextByte() ? buffer[index++] : -1;
		}

		public String next() {
			if (!hasNext()) {
				throw new RuntimeException("no input");
			}
			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 RuntimeException("no input");
			}
			long value = 0L;
			boolean minus = false;
			int b = readByte();
			if (b == '-') {
				minus = true;
				b = readByte();
			}
			while (isPrintableChar(b)) {
				if (isDigit(b)) {
					value = value * 10 + (b - '0');
				}
				b = readByte();
			}
			return minus ? -value : value;
		}

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

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

	class DfsResolver {
		Graph g;
		Set<Integer> used;

		public DfsResolver(Graph g) {
			this.g = g;
			used = new HashSet<>();
		}

		void search(int start) {
			used.add(start);
			//for (;false;) {}
		}
	}

	class MapSetGraph implements Graph {
		Map<Integer, Set<Edge>> map = new HashMap<>();
		int vertexNum = 0;

		@Override
		public void link(int from, int to, long cost) {
			if (!map.containsKey(from)) {
				map.put(from, new HashSet<>());
			}
			map.get(from).add(new Edge(to, cost));
			if (vertexNum < from + 1) {
				vertexNum = from + 1;
			}
			if (vertexNum < to + 1) {
				vertexNum = to + 1;
			}
		}

		@Override
		public Optional<Long> getCost(int from, int to) {
			if (map.containsKey(from)) {
				for (Edge e : map.get(from)) {
					if (e.to == to) {
						return Optional.of(e.cost);
					}
				}
			}
			return Optional.ofNullable(null);
		}

		@Override
		public int getVertexNum() {
			// linkメソッドで渡される頂点IDが0-indexの前提に立った実装
			return vertexNum;
		}

		@Override
		public Set<Edge> getAdjs(int from) {
			if (!map.containsKey(from)) {
				return new HashSet<>();
			}
			return map.get(from);
		}
	}

	interface CombCalculator {
		long comb(int n, int m);
	}

	interface MobiusFunction {
		int get(int n);
	}

	/**
	 * メビウス関数をエラトステネスの篩っぽく前計算するクラスです。
	 * 計算量はO(1)で、前計算でO(N logN)です。
	 */
	class SieveMobiusFunction implements MobiusFunction {
		int size;
		int[] mobiusFunctionValues;

		public SieveMobiusFunction(int size) {
			this.size = size;
			mobiusFunctionValues = new int[size];

			mobiusFunctionValues[0] = 0;
			mobiusFunctionValues[1] = 1;
			for (int i = 2; i < size; i++) {
				mobiusFunctionValues[i] = 1;
			}
			for (int i = 2; i * i < size; i++) {
				for (int k = 1; i * i * k < size; k++) {
					mobiusFunctionValues[i * i * k] *= 0;
				}
			}

			for (int i = 2; i < size; i++) {
				if (mobiusFunctionValues[i] == 1) {
					for (int k = 1; i * k < size; k++) {
						mobiusFunctionValues[i * k] *= -2;
					}
				}
				if (mobiusFunctionValues[i] > 1) {
					mobiusFunctionValues[i] = 1;
				}
				if (mobiusFunctionValues[i] < -1) {
					mobiusFunctionValues[i] = -1;
				}
			}
		}

		@Override
		public int get(int n) {
			if (n > size) {
				throw new RuntimeException("n is greater than size.");
			}
			if (n < 0) {
				return 0;
			}
			return mobiusFunctionValues[n];
		}
	}

	/**
	 * メビウス関数を定義通り計算するクラスです。
	 * 計算量はO(logN)です。
	 */
	class PrimeFactorizationMobiusFunction implements MobiusFunction {
		@Override
		public int get(int n) {
			if (n <= 0) {
				return 0;
			}
			if (n == 1) {
				return 1;
			}
			int num = 0;
			for (int i = 2; i < n; i++) {
				if (n % i == 0) {
					n /= i;
					num++;
					if (n % i == 0) {
						return 0;
					}
				}
			}
			return num % 2 == 0 ? -1 : 1;
		}
	}

	/**
	 * 組み合わせ計算を階乗の値で行うクラスです(MOD対応)
	 * 階乗とその逆元は前計算してテーブルに格納します。
	 * C(N, N) % M の計算量は O(1)、 前計算でO(max(N, logM))です。
	 * sizeを1e8より大きい値で実行するとMLEの危険性があります。
	 * また素数以外のMODには対応していません(逆元の計算に素数の剰余環の性質を利用しているため)。
	 */
	class FactorialTableCombCalculator implements CombCalculator {
		int size;
		long[] factorialTable;
		long[] inverseFactorialTable;
		long mod;

		public FactorialTableCombCalculator(int size, long mod) {
			this.size = size;
			factorialTable = new long[size + 1];
			inverseFactorialTable = new long[size + 1];
			this.mod = mod;

			factorialTable[0] = 1L;
			for (int i = 1; i <= size; i++) {
				factorialTable[i] = (factorialTable[i - 1] * i) % mod;
			}
			inverseFactorialTable[size] = inverse(factorialTable[size], mod);
			for (int i = size - 1; i >= 0; i--) {
				inverseFactorialTable[i] = (inverseFactorialTable[i + 1] * (i + 1)) % mod;
			}
		}

		private long inverse(long n, long mod) {
			return pow(n, mod - 2, mod);
		}

		private long pow(long n, long p, long mod) {
			if (p == 0) {
				return 1L;
			}
			long half = pow(n, p / 2, mod);
			long ret = (half * half) % mod;
			if (p % 2 == 1) {
				ret = (ret * n) % mod;
			}
			return ret;
		}

		@Override
		public long comb(int n, int m) {
			if (n > size) {
				throw new RuntimeException("n is greater than size.");
			}
			if (n < 0 || m < 0 || n < m) {
				return 0L;
			}
			return (((factorialTable[n] * inverseFactorialTable[m]) % mod) * inverseFactorialTable[n - m]) % mod;
		}
	}

	/**
	 * 組み合わせ計算をテーブルで実装したクラスです(MOD対応)
	 * 前計算でO(N^2), combはO(1)で実行できます
	 * sizeを2 * 1e4より大きい値で実行するとMLEの危険性があります
	 */
	class TableCombCalculator implements CombCalculator {
		long[][] table;
		int size;

		public TableCombCalculator(int size, long mod) {
			this.size = size;
			table = new long[size + 1][];

			table[0] = new long[1];
			table[0][0] = 1L;
			for (int n = 1; n <= size; n++) {
				table[n] = new long[n + 1];
				table[n][0] = 1L;
				for (int m = 1; m < n; m++) {
					table[n][m] = (table[n - 1][m - 1] + table[n - 1][m]) % mod;
				}
				table[n][n] = 1L;
			}
		}

		@Override
		public long comb(int n, int m) {
			if (n > size) {
				throw new RuntimeException("n is greater than size.");
			}
			if (n < 0 || m < 0 || n < m) {
				return 0L;
			}
			return table[n][m];
		}
	}

	interface Graph {
		void link(int from, int to, long cost);
		Optional<Long> getCost(int from, int to);
		int getVertexNum();
		Set<Edge> getAdjs(int from);
		class Edge {
			int to;
			long cost;
			public Edge(int to, long cost) {
				this.to = to;
				this.cost = cost;
			}

			public int getTo() {return to;}
			public long getCost() {return cost;}

			@Override
			public boolean equals(Object obj) {
				if (obj == null || !(obj instanceof Edge)) {
					return false;
				}
				Edge e = (Edge)obj;
				return this.to == e.to && this.cost == e.cost;
			}

			@Override
			public int hashCode() {
				return this.to;
			}
		}
	}

	interface FlowResolver {
		long maxFlow(int from, int to);
	}

	/**
	 * グラフの行列による実装
	 * 接点数の大きいグラフで使うとMLEで死にそう
	 */
	class ArrayGraph implements Graph {
		private Long[][] costArray;
		private int vertexNum;

		public ArrayGraph(int n) {
			costArray = new Long[n][];
			for (int i = 0; i < n; i++) {
				costArray[i] = new Long[n];
			}
			vertexNum = n;
		}

		@Override
		public void link(int from, int to, long cost) {
			costArray[from][to] = new Long(cost);
		}

		@Override
		public Optional<Long> getCost(int from, int to) {
			return Optional.ofNullable(costArray[from][to]);
		}

		@Override
		public int getVertexNum() {
			return vertexNum;
		}

		@Override
		public Set<Edge> getAdjs(int from) {
			Set<Edge> set = new HashSet<>();
			for (int to = 0; to < vertexNum; to++) {
				if (costArray[from][to] != null) {
					set.add(new Edge(to, costArray[from][to]));
				}
			}
			return set;
		}
	}

	/**
	 * DFS(深さ優先探索)による実装
	 * 計算量はO(E*MaxFlow)のはず (E:辺の数, MaxFlow:最大フロー)
	 */
	class DfsFlowResolver implements FlowResolver {
		private Graph graph;
		public DfsFlowResolver(Graph graph) {
			this.graph = graph;
		}

		/**
		 * 最大フロー(最小カット)を求める
		 * @param from 始点(source)のID
		 * @param to 終点(target)のID
		 * @return 最大フロー(最小カット)
		 */
		public long maxFlow(int from, int to) {
			long sum = 0L;
			long currentFlow;
			do {
				currentFlow = flow(from, to, Long.MAX_VALUE / 3, new boolean[graph.getVertexNum()]);
				sum += currentFlow;
			} while (currentFlow > 0);
			return sum;
		}

		/**
		 * フローの実行 グラフの更新も行う
		 * @param from 現在いる節点のID
		 * @param to 終点(target)のID
		 * @param current_flow ここまでの流量
		 * @param passed 既に通った節点か否かを格納した配列
		 * @return 終点(target)に流した流量/戻りのグラフの流量
		 */
		private long flow(int from, int to, long current_flow, boolean[] passed) {
			passed[from] = true;
			if (from == to) {
				return current_flow;
			}
			for (int id = 0; id < graph.getVertexNum(); id++) {
				if (passed[id]) {
					continue;
				}
				Optional<Long> cost = graph.getCost(from, id);
				if (cost.orElse(0L) > 0) {
					long nextFlow = current_flow < cost.get() ? current_flow : cost.get();
					long returnFlow = flow(id, to, nextFlow, passed);
					if (returnFlow > 0) {
						graph.link(from, id, cost.get() - returnFlow);
						graph.link(id, from, graph.getCost(id, from).orElse(0L) + returnFlow);
						return returnFlow;
					}
				}
			}
			return 0L;
		}
	}

	/**
	 * 1-indexedのBIT配列
	 */
	class BinaryIndexedTree {
		private long[] array;

		public BinaryIndexedTree(int size) {
			this.array = new long[size + 1];
		}

		/**
		 * 指定した要素に値を加算する
		 * 計算量はO(logN)
		 * @param index 加算する要素の添字
		 * @param value 加算する量
		 */
		public void add(int index, long value) {
			for (int i = index; i < array.length; i += (i & -i)) {
				array[i] += value;
			}
		}

		/**
		 * 1〜指定した要素までの和を取得する
		 * 計算量はO(logN)
		 * @param index 和の終端となる要素の添字
		 * @return 1〜indexまでの和
		 */
		public long getSum(int index) {
			long sum = 0L;
			for (int i = index; i > 0; i -= (i & -i)) {
				sum += array[i];
			}
			return sum;
		}
	}

	/**
	 * 1-indexedの2次元BIT配列
	 */
	class BinaryIndexedTree2D {
		private long[][] array;

		public BinaryIndexedTree2D(int size1, int size2) {
			this.array = new long[size1 + 1][];
			for (int i = 1; i <= size1; i++) {
				this.array[i] = new long[size2 + 1];
			}
		}

		/**
		 * 指定した要素に値を加算する
		 * 計算量はO(logN * logN)
		 * @param index1 加算する要素の1次元目の添字
		 * @param index2 加算する要素の2次元目の添字
		 * @param value 加算する量
		 */
		public void add(int index1, int index2, long value) {
			for (int i1 = index1; i1 < array.length; i1 += (i1 & -i1)) {
				for (int i2 = index2; i2 < array.length; i2 += (i2 & -i2)) {
					array[i1][i2] += value;
				}
			}
		}

		/**
		 * (1,1)〜指定した要素までの矩形和を取得する
		 * 計算量はO(logN * logN)
		 * @param index1 和の終端となる要素の1次元目の添字
		 * @param index2 和の終端となる要素の2次元目の添字
		 * @return (1,1)〜(index1,index2)までの矩形和
		 */
		public long getSum(int index1, int index2) {
			long sum = 0L;
			for (int i1 = index1; i1 > 0; i1 -= (i1 & -i1)) {
				for (int i2 = index2; i2 > 0; i2 -= (i2 & -i2)) {
					sum += array[i1][i2];
				}
			}
			return sum;
		}
	}

	interface UnionFind {
		void union(int A, int B);
		boolean judge(int A, int B);
		Set<Integer> getSet(int id);
	}

	/**
	 * ArrayUnionFindの拡張
	 * MapSetで根の添字から根にぶら下がる頂点の集合が取得できるようにした
	 * getSetメソッドをO(logN * logN)に落とせているはず
	 * ただしunionメソッドは2倍の計算量になっているので注意(オーダーは変わらないはず)
	 */
	class SetUnionFind extends ArrayUnionFind {
		Map<Integer, Set<Integer>> map;
		public SetUnionFind(int size) {
			super(size);
			map = new HashMap<>();
			for (int i = 0; i < size; i++) {
				map.put(i, new HashSet<>());
				map.get(i).add(i);
			}
		}

		@Override
		protected void unionTo(int source, int dest) {
			super.unionTo(source, dest);
			map.get(dest).addAll(map.get(source));
		}

		@Override
		public Set<Integer> getSet(int id) {
			return map.get(root(id));
		}
	}

	/**
	 * 配列によるUnionFindの実装
	 * getSetメソッドはO(NlogN)なのでTLEに注意
	 */
	class ArrayUnionFind implements UnionFind {
		int[] parent;
		int[] rank;
		int size;
		public ArrayUnionFind(int size) {
			parent = new int[size];
			for (int i = 0; i < size; i++) {
				parent[i] = i;
			}
			rank = new int[size];
			this.size = size;
		}

		@Override
		public void union(int A, int B) {
			int rootA = root(A);
			int rootB = root(B);
			if (rootA != rootB) {
				if (rank[rootA] < rank[rootB]) {
					unionTo(rootA, rootB);
				} else {
					unionTo(rootB, rootA);
					if (rank[rootA] == rank[rootB]) {
						rank[rootA]++;
					}
				}
			}
		}

		protected void unionTo(int source, int dest) {
			parent[source] = dest;
		}

		@Override
		public boolean judge(int A, int B) {
			return root(A) == root(B);
		}

		@Override
		public Set<Integer> getSet(int id) {
			Set<Integer> set = new HashSet<>();
			IntStream.range(0, size).filter(i -> judge(i, id)).forEach(set::add);
			return set;
		}

		protected int root(int id) {
			if (parent[id] == id) {
				return id;
			}
			parent[id] = root(parent[id]);
			return parent[id];
		}
	}

	/**
	 * 素数のユーティリティ
	 */
	class PrimeNumberUtils {
		boolean[] isPrimeArray;
		List<Integer> primes;

		/**
		 * 素数判定の上限となる値を指定してユーティリティを初期化
		 * @param limit 素数判定の上限(この値以上が素数であるか判定しない)
		 */
		public PrimeNumberUtils(int limit) {
			if (limit > 10000000) {
				System.err.println("上限の値が高すぎるため素数ユーティリティの初期化でTLEする可能性が大変高いです");
			}
			primes = new ArrayList<>();
			isPrimeArray = new boolean[limit];
			if (limit > 2) {
				primes.add(2);
				isPrimeArray[2] = true;
			}

			for (int i = 3; i < limit; i += 2) {
				if (isPrime(i, primes)) {
					primes.add(i);
					isPrimeArray[i] = true;
				}
			}
		}

		public List<Integer> getPrimeNumberList() {
			return primes;
		}

		public boolean isPrime(int n) {
			return isPrimeArray[n];
		}

		private boolean isPrime(int n, List<Integer> primes) {
			for (int prime : primes) {
				if (n % prime == 0) {
					return false;
				}
				if (prime > Math.sqrt(n)) {
					break;
				}
			}
			return true;
		}
	}

	interface BitSet {
		void set(int index, boolean bit);
		boolean get(int index);
		void shiftRight(int num);
		void shiftLeft(int num);
		void or(BitSet bitset);
		void and(BitSet bitset);
	}

	/**
	 * Longの配列によるBitSetの実装
	 * get/setはO(1)
	 * shift/or/andはO(size / 64)
	 */
	class LongBit implements BitSet {
		long[] bitArray;

		public LongBit(int size) {
			bitArray = new long[((size + 63) / 64)];
		}

		@Override
		public void set(int index, boolean bit) {
			int segment = index / 64;
			int inIndex = index % 64;
			if (bit) {
				bitArray[segment] |= 1L << inIndex;
			} else {
				bitArray[segment] &= ~(1L << inIndex);
			}
		}

		@Override
		public boolean get(int index) {
			int segment = index / 64;
			int inIndex = index % 64;
			return (bitArray[segment] & (1L << inIndex)) != 0L;
		}

		@Override
		public void shiftRight(int num) {
			int shiftSeg = num / 64;
			int shiftInI = num % 64;
			for (int segment = 0; segment < bitArray.length; segment++) {
				int sourceSeg = segment + shiftSeg;
				if (sourceSeg < bitArray.length) {
					bitArray[segment] = bitArray[sourceSeg] >>> shiftInI;
					if (shiftInI > 0 && sourceSeg + 1 < bitArray.length) {
						bitArray[segment] |= bitArray[sourceSeg + 1] << (64 - shiftInI);
					}
				} else {
					bitArray[segment] = 0L;
				}
			}
		}

		@Override
		public void shiftLeft(int num) {
			int shiftSeg = num / 64;
			int shiftInI = num % 64;
			for (int segment = bitArray.length - 1; segment >= 0; segment--) {
				int sourceSeg = segment - shiftSeg;
				if (sourceSeg >= 0) {
					bitArray[segment] = bitArray[sourceSeg] << shiftInI;
					if (shiftInI > 0 && sourceSeg > 0) {
						bitArray[segment] |= bitArray[sourceSeg - 1] >>> (64 - shiftInI);
					}
				} else {
					bitArray[segment] = 0L;
				}
			}
		}

		public long getLong(int segment) {
			return bitArray[segment];
		}

		@Override
		public void or(BitSet bitset) {
			if (!(bitset instanceof LongBit)) {
				return;
			}
			for (int segment = 0; segment < bitArray.length; segment++) {
				bitArray[segment] |= ((LongBit)bitset).getLong(segment);
			}
		}

		@Override
		public void and(BitSet bitset) {
			if (!(bitset instanceof LongBit)) {
				return;
			}
			for (int segment = 0; segment < bitArray.length; segment++) {
				bitArray[segment] &= ((LongBit)bitset).getLong(segment);
			}
		}
	}

}


import java.util.Scanner;

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 >= c) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }

}

Score : 100 points

Problem Statement

An elementary school student Takahashi has come to a variety store.

He has two coins, A-yen and B-yen coins (yen is the currency of Japan), and wants to buy a toy that costs C yen. Can he buy it?

Note that he lives in Takahashi Kingdom, and may have coins that do not exist in Japan.

Constraints

All input values are integers.

1 \leq A, B \leq 500

1 \leq C \leq 1000

Input

Input is given from Standard Input in the following format:

A B C

Output

If Takahashi can buy the toy, print Yes; if he cannot, print No.

Sample Input 1

50 100 120

Sample Output 1

Yes

He has 50 + 100 = 150 yen, so he can buy the 120-yen toy.

Sample Input 2

500 100 1000

Sample Output 2

No

He has 500 + 100 = 600 yen, but he cannot buy the 1000-yen toy.

Sample Input 3

19 123 143

Sample Output 3

No

There are 19-yen and 123-yen coins in Takahashi Kingdom, which are rather hard to use.

Sample Input 4

19 123 142

Sample Output 4

Yes


import java.util.*;

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

        int N = sc.nextInt();

        Map<String, Integer> cp = new HashMap<>();
        for(int i = 0; i < N; i++) {
            String str = sc.next();
            if( cp.containsKey( str ) ) {
                int x = cp.get(str);
                cp.put(str, ++x);
            } else {
                cp.put(str, 1);
            }
        }
        int M = sc.nextInt();
        for(int i = 0; i < M; i++) {
            String str = sc.next();
            if( cp.containsKey( str ) ) {
                int x = cp.get(str);
                cp.put( str, --x );
            }
        }

        int max = 0;
        for( Map.Entry<String, Integer> entry : cp.entrySet() ) {
            max = Math.max( max, entry.getValue() );
        }
        System.out.println( max );
    }
}

Score : 200 points

Problem Statement

Takahashi has N blue cards and M red cards.
A string is written on each card. The string written on the i-th blue card is s_i, and the string written on the i-th red card is t_i.

Takahashi will now announce a string, and then check every card. Each time he finds a blue card with the string announced by him, he will earn 1 yen (the currency of Japan); each time he finds a red card with that string, he will lose 1 yen.

Here, we only consider the case where the string announced by Takahashi and the string on the card are exactly the same. For example, if he announces atcoder, he will not earn money even if there are blue cards with atcoderr, atcode, btcoder, and so on. (On the other hand, he will not lose money even if there are red cards with such strings, either.)

At most how much can he earn on balance?

Note that the same string may be written on multiple cards.

Constraints

N and M are integers.

1 \leq N, M \leq 100

s_1, s_2, ..., s_N, t_1, t_2, ..., t_M are all strings of lengths between 1 and 10 (inclusive) consisting of lowercase English letters.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N
M
t_1
t_2
:
t_M

Output

If Takahashi can earn at most X yen on balance, print X.

Sample Input 1

3
apple
orange
apple
1
grape

Sample Output 1

2

He can earn 2 yen by announcing apple.

Sample Input 2

3
apple
orange
apple
5
apple
apple
apple
apple
apple

Sample Output 2

1

If he announces apple, he will lose 3 yen. If he announces orange, he can earn 1 yen.

Sample Input 3

1
voldemort
10
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort
voldemort

Sample Output 3

0

If he announces voldemort, he will lose 9 yen. If he announces orange, for example, he can avoid losing a yen.

Sample Input 4

6
red
red
blue
yellow
yellow
red
5
red
red
yellow
green
blue

Sample Output 4

1


import java.awt.Point;
import java.util.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
		Point[] red = new Point[N];
		for (int i = 0; i < N; i++)
			red[i] = new Point(sc.nextInt(), sc.nextInt());
		Point[] blue = new Point[N];
		for (int i = 0; i < N; i++)
			blue[i] = new Point(sc.nextInt(), sc.nextInt());
		sc.close();

		Arrays.sort(red, (pt1, pt2) -> (pt2.y - pt1.y));
		Arrays.sort(blue, (pt1, pt2) -> (pt1.x - pt2.x));

		int ans = 0;

		for (int i = 0; i < N; i++) {
			for (int j = 0; j < N; j++) {
				if (red[j] == null)
					continue;
				if (red[j].x < blue[i].x && red[j].y < blue[i].y) {
					red[j] = null;
					ans++;
					break;
				}
			}
		}

		System.out.println(ans);
	}
}

Score : 400 points

Problem Statement

On a two-dimensional plane, there are N red points and N blue points.
The coordinates of the i-th red point are (a_i, b_i), and the coordinates of the i-th blue point are (c_i, d_i).

A red point and a blue point can form a friendly pair when, the x-coordinate of the red point is smaller than that of the blue point, and the y-coordinate of the red point is also smaller than that of the blue point.

At most how many friendly pairs can you form? Note that a point cannot belong to multiple pairs.

Constraints

All input values are integers.

1 \leq N \leq 100

0 \leq a_i, b_i, c_i, d_i < 2N

a_1, a_2, ..., a_N, c_1, c_2, ..., c_N are all different.

b_1, b_2, ..., b_N, d_1, d_2, ..., d_N are all different.

Input

Input is given from Standard Input in the following format:

N
a_1 b_1
a_2 b_2
:
a_N b_N
c_1 d_1
c_2 d_2
:
c_N d_N

Output

Print the maximum number of friendly pairs.

Sample Input 1

3
2 0
3 1
1 3
4 2
0 4
5 5

Sample Output 1

2

For example, you can pair (2, 0) and (4, 2), then (3, 1) and (5, 5).

Sample Input 2

3
0 0
1 1
5 2
2 3
3 4
4 5

Sample Output 2

2

For example, you can pair (0, 0) and (2, 3), then (1, 1) and (3, 4).

Sample Input 3

2
2 2
3 3
0 0
1 1

Sample Output 3

0

It is possible that no pair can be formed.

Sample Input 4

5
0 0
7 3
2 2
4 8
1 6
8 5
6 9
5 4
9 1
3 7

Sample Output 4

5

Sample Input 5

5
0 0
1 1
5 5
6 6
7 7
2 2
3 3
4 4
8 8
9 9

Sample Output 5

4


import java.util.*;
public class Main{
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int M = sc.nextInt();
    int answer = (Math.abs(N-2))*(Math.abs(M-2));
    System.out.println(answer);
  }

  //最大公約数・最小公倍数（セットで使う）
static int gcd (int a, int b) {return b>0?gcd(b,a%b):a;}
static int lcm (int a, int b) {return a*b/gcd(a,b);}

//素数判定
static boolean isPrime (int n) {
	if (n==2) return true;
	if (n<2 || n%2==0) return false;
	double d = Math.sqrt(n);
	for (int i=3; i<=d; i+=2) if(n%i==0){return false;}
	return true;
}

//倍数判定（10進数以外のときに有用）
static boolean isMultiple (String s, int base, int m) {
	int temp = 0;
	for (int i=0; i<s.length(); i++) {
		temp = (temp*base+Character.getNumericValue(s.charAt(i)))%m;
	}
	if (temp==0) {return true;}
	return false;
}

//階乗
static long factorial (int i) {
	if (i==1) {return 1;}
	else {return i*factorial(i-1);}
}

//進数変換
static String toNbase (String sm, int m, int n) {
	return Long.toString(Long.parseLong(sm,m),n);
}

}

Score : 300 points

Problem Statement

There is a grid with infinitely many rows and columns. In this grid, there is a rectangular region with consecutive N rows and M columns, and a card is placed in each square in this region.
The front and back sides of these cards can be distinguished, and initially every card faces up.

We will perform the following operation once for each square contains a card:

For each of the following nine squares, flip the card in it if it exists: the target square itself and the eight squares that shares a corner or a side with the target square.

It can be proved that, whether each card faces up or down after all the operations does not depend on the order the operations are performed.
Find the number of cards that face down after all the operations.

Constraints

1 \leq N,M \leq 10^9

All input values are integers.

Input

Input is given from Standard Input in the following format:

N M

Output

Print the number of cards that face down after all the operations.

Sample Input 1

2 2

Sample Output 1

0

We will flip every card in any of the four operations. Thus, after all the operations, all cards face up.

Sample Input 2

1 7

Sample Output 2

5

After all the operations, all cards except at both ends face down.

Sample Input 3

314 1592

Sample Output 3

496080


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

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

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

        System.out.println(reminder(n,0,k,k+1));
    }

    private static long reminder(int n, long val, int k, int cur){
        if(cur > n) return val;
        if(k == 0) return n*n;

        while(cur <= n){
            int volume = cur - k;
            int cycle = n/cur;
            val += cycle * volume;

            int rest = n % cur;
            val += rest - k >= 0 ? rest - k + 1 : 0;

            cur++;
        }

        return val;
    }
}

Score : 400 points

Problem Statement

Takahashi had a pair of two positive integers not exceeding N, (a,b), which he has forgotten.
He remembers that the remainder of a divided by b was greater than or equal to K.
Find the number of possible pairs that he may have had.

Constraints

1 \leq N \leq 10^5

0 \leq K \leq N-1

All input values are integers.

Input

Input is given from Standard Input in the following format:

N K

Output

Print the number of possible pairs that he may have had.

Sample Input 1

5 2

Sample Output 1

7

There are seven possible pairs: (2,3),(5,3),(2,4),(3,4),(2,5),(3,5) and (4,5).

Sample Input 2

10 0

Sample Output 2

100

Sample Input 3

31415 9265

Sample Output 3

287927211


import java.util.*;
import java.*;
class Main {
   public static void main(String[] args)
   {
      Scanner scan = new Scanner(System.in);
      int     N    = scan.nextInt();
      int     K    = scan.nextInt();
      long    coun = 0;

      if(K == 0){
         coun = N * N;
         System.out.println(coun);
         return;
         }
      for(int i = 1; i <= N; i++){
          if(i > K){
             coun += (N / i) * (i - K) >= 0 ? (N / i) * (i - K) : 0;

             coun += (N % i) - K + 1 >= 0 ? (N % i) - K + 1 : 0;
             }
          }
      System.out.println(coun);
   }
}


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) {
            int tmp = n;
            n = m;
            m = tmp;
        }

        long ans = 0;

        if (n == 1) {
            if (m == 1)
                ans = 1;
            else
                ans = m - 2;
        } else {
            ans = (n - 2) * (m - 2);
        }

        System.out.println(ans);
    }
}


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==1&&m==1) System.out.println(1);
		else if(n==1||m==1) System.out.println(Math.max(n, m)-2);
		else System.out.println((n-2)*(m-2));
	}
}


import java.util.*;
public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int N = sc.nextInt();
                String S[] = new String [N];
                String Str = "Two";
                 for(int i=0;i<N;i++){
                  S[i] = sc.next();
                 }
                 for(int i=0;i<N;i++){
                    if(S[i].equals("Y")){
                      Str = "Four";
                      break;
                    }
                    else
                      Str = "three";
                  }
                System.out.println(Str);
	}
}

Score : 200 points

Problem Statement

In Japan, people make offerings called hina arare, colorful crackers, on March 3.

We have a bag that contains N hina arare. (From here, we call them arare.)

It is known that the bag either contains arare in three colors: pink, white and green, or contains arare in four colors: pink, white, green and yellow.

We have taken out the arare in the bag one by one, and the color of the i-th arare was S_i, where colors are represented as follows - pink: P, white: W, green: G, yellow: Y.

If the number of colors of the arare in the bag was three, print Three; if the number of colors was four, print Four.

Constraints

1 \leq N \leq 100

S_i is P, W, G or Y.

There always exist i, j and k such that S_i=P, S_j=W and S_k=G.

Input

Input is given from Standard Input in the following format:

N
S_1 S_2 ... S_N

Output

If the number of colors of the arare in the bag was three, print Three; if the number of colors was four, print Four.

Sample Input 1

6
G W Y P Y W

Sample Output 1

Four

The bag contained arare in four colors, so you should print Four.

Sample Input 2

9
G W W G P W P G G

Sample Output 2

Three

The bag contained arare in three colors, so you should print Three.

Sample Input 3

8
P Y W G Y W Y Y

Sample Output 3

Four


import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        long n = sc.nextLong();
        char[] ch = String.valueOf(n).toCharArray();
        System.out.println((ch.length - 1) * 9 + (ch[0] - '0') - 1);
    }

}

Score : 300 points

Problem Statement

Find the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Constraints

1\leq N \leq 10^{16}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible sum of the digits (in base 10) of a positive integer not greater than N.

Sample Input 1

100

Sample Output 1

18

For example, the sum of the digits in 99 is 18, which turns out to be the maximum value.

Sample Input 2

9995

Sample Output 2

35

For example, the sum of the digits in 9989 is 35, which turns out to be the maximum value.

Sample Input 3

3141592653589793

Sample Output 3

137


import java.util.Scanner;

public class Main {

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

        String s = sc.nextLine();

        String[] str = s.split("");
        int top = Integer.parseInt(str[0]);
        int ans = 0;

        for (int i = 0; i < str.length; i++) {
            ans += Integer.parseInt(str[i]);
        }

        if (ans < top - 1 + 9 * (str.length - 1)) {
            ans = top - 1 + 9 * (str.length - 1);
        }

        System.out.println(ans);

    }

}


import java.util.Scanner;

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

		int n = scan.nextInt();

		int ans = Integer.parseInt(String.valueOf(n).substring(0, 1)) - 1;

		int nLen = String.valueOf(n).length();

		for (int i = 1; i < nLen; i++) {
			ans = ans + 9;
		}

		if ( ans == -1) {
			ans = 0;
		}
		System.out.println(ans);
	}
}


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

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Egor Kulikov (egor@egork.net)
 */
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);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        int[][][] answer;
        String s;

        public void solve(int testNumber, InputReader in, OutputWriter out) {
            s = in.readString();
            int n = s.length();
            int k = in.readInt();
            answer = new int[n][n][k + 1];
            ArrayUtils.fill(answer, -1);
            out.printLine(go(0, n - 1, k));
        }

        private int go(int from, int to, int k) {
            if (from > to) {
                return 0;
            }
            if (from == to) {
                return 1;
            }
            if (answer[from][to][k] != -1) {
                return answer[from][to][k];
            }
            if (s.charAt(from) == s.charAt(to)) {
                answer[from][to][k] = go(from + 1, to - 1, k) + 2;
            } else {
                answer[from][to][k] = Math.max(go(from, to - 1, k), go(from + 1, to, k));
                if (k != 0) {
                    answer[from][to][k] = Math.max(answer[from][to][k], go(from + 1, to - 1, k - 1) + 2);
                }
            }
            return answer[from][to][k];
        }

    }

    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 printLine(int i) {
            writer.println(i);
        }

    }

    static class ArrayUtils {
        public static void fill(int[][] array, int value) {
            for (int[] row : array) {
                Arrays.fill(row, value);
            }
        }

        public static void fill(int[][][] array, int value) {
            for (int[][] subArray : array) {
                fill(subArray, value);
            }
        }

    }

    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 String readString() {
            int c = read();
            while (isSpaceChar(c)) {
                c = read();
            }
            StringBuilder res = new StringBuilder();
            do {
                if (Character.isValidCodePoint(c)) {
                    res.appendCodePoint(c);
                }
                c = read();
            } while (!isSpaceChar(c));
            return res.toString();
        }

        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 : 900 points

Problem Statement

Takahashi has decided to give a string to his mother.

The value of a string T is the length of the longest common subsequence of T and T', where T' is the string obtained by reversing T.
That is, the value is the longest length of the following two strings that are equal: a subsequence of T (possibly non-contiguous), and a subsequence of T' (possibly non-contiguous).

Takahashi has a string S. He wants to give her mother a string of the highest possible value, so he would like to change at most K characters in S to any other characters in order to obtain a string of the highest possible value.
Find the highest possible value achievable.

Constraints

1 \leq |S| \leq 300

0 \leq K \leq |S|

S consists of lowercase English letters.

K is an integer.

Input

Input is given from Standard Input in the following format:

S
K

Output

Print the highest possible value achievable.

Sample Input 1

abcabcabc
1

Sample Output 1

7

Changing the first character to c results in cbcabcabc.
Let this tring be T, then one longest common subsequence of T and T' is cbabcbc, whose length is 7.

Sample Input 2

atcodergrandcontest
3

Sample Output 2

15


import java.util.Scanner;

public class Main {

    public static void main(String[] args){

        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int a = sc.nextInt();
        System.out.println(n % 500 <= a ? "YES" : "NO");
    }

}

Score: 100 points

Problem Statement

E869120 has A 1-yen coins and infinitely many 500-yen coins.

Determine if he can pay exactly N yen using only these coins.

Constraints

N is an integer between 1 and 10000 (inclusive).

A is an integer between 0 and 1000 (inclusive).

Input

Input is given from Standard Input in the following format:

N
A

Output

If E869120 can pay exactly N yen using only his 1-yen and 500-yen coins, print Yes; otherwise, print No.

Sample Input 1

2018
218

Sample Output 1

Yes

We can pay 2018 yen with four 500-yen coins and 18 1-yen coins, so the answer is Yes.

Sample Input 2

2763
0

Sample Output 2

No

When we have no 1-yen coins, we can only pay a multiple of 500 yen using only 500-yen coins. Since 2763 is not a multiple of 500, we cannot pay this amount.

Sample Input 3

37
514

Sample Output 3

Yes


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];
		for(int i=0;i<N;i++){
			a[i]=sc.nextInt();
		}
		int max = a[0];
		int min = a[0];
		for(int i=0;i<N;i++) {
			if(max<a[i]) {
				max=a[i];
			}
		}
		for(int i=0;i<N;i++) {
			if(min>a[i]){
				min=a[i];
			}
		}
		System.out.println(max-min);
	}
}

Score: 200 points

Problem Statement

We have N cards. A number a_i is written on the i-th card.

Alice and Bob will play a game using these cards. In this game, Alice and Bob alternately take one card. Alice goes first.

The game ends when all the cards are taken by the two players, and the score of each player is the sum of the numbers written on the cards he/she has taken. When both players take the optimal strategy to maximize their scores, find Alice's score minus Bob's score.

Constraints

N is an integer between 1 and 100 (inclusive).

a_i \ (1 \leq i \leq N) is an integer between 1 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 a_3 ... a_N

Output

Print Alice's score minus Bob's score when both players take the optimal strategy to maximize their scores.

Sample Input 1

2
3 1

Sample Output 1

2

First, Alice will take the card with 3. Then, Bob will take the card with 1.
The difference of their scores will be 3 - 1 = 2.

Sample Input 2

3
2 7 4

Sample Output 2

5

First, Alice will take the card with 7. Then, Bob will take the card with 4. Lastly, Alice will take the card with 2. The difference of their scores will be 7 - 4 + 2 = 5. The difference of their scores will be 3 - 1 = 2.

Sample Input 3

4
20 18 2 18

Sample Output 3

18


import java.util.*;

public class Main{
  public static void main(String[]args){
    Scanner sc = new Scanner(System.in);
    int[][] c = new int[3][3];
    c[0][0] = Integer.parseInt(sc.next());
    c[0][1] = Integer.parseInt(sc.next());
    c[0][2] = Integer.parseInt(sc.next());
    c[1][0] = Integer.parseInt(sc.next());
    c[1][1] = Integer.parseInt(sc.next());
    c[1][2] = Integer.parseInt(sc.next());
    c[2][0] = Integer.parseInt(sc.next());
    c[2][1] = Integer.parseInt(sc.next());
    c[2][2] = Integer.parseInt(sc.next());

    int[] a = new int[3];
    int[] b = new int[3];
    a[0] = 0;
    b[0] = c[0][0] - a[0];
    b[1] = c[0][1] - a[0];
    b[2] = c[0][2] - a[0];
    a[1] = c[1][1] - b[1];
    a[2] = c[2][1] - b[1];

    String ans = "Yes";
    for(int i = 0; i < 3; i++){
      for(int j = 0; j < 3; j++){
        if(c[i][j] != a[i] + b[j]){
          ans = "No";
          break;
        }
      }
    }
    System.out.println(ans);
  }
}

Score: 300 points

Problem Statement

We have a 3 \times 3 grid. A number c_{i, j} is written in the square (i, j), where (i, j) denotes the square at the i-th row from the top and the j-th column from the left.

According to Takahashi, there are six integers a_1, a_2, a_3, b_1, b_2, b_3 whose values are fixed, and the number written in the square (i, j) is equal to a_i + b_j.

Determine if he is correct.

Constraints

c_{i, j} \ (1 \leq i \leq 3, 1 \leq j \leq 3) is an integer between 0 and 100 (inclusive).

Input

Input is given from Standard Input in the following format:

c_{1,1} c_{1,2} c_{1,3}
c_{2,1} c_{2,2} c_{2,3}
c_{3,1} c_{3,2} c_{3,3}

Output

If Takahashi's statement is correct, print Yes; otherwise, print No.

Sample Input 1

1 0 1
2 1 2
1 0 1

Sample Output 1

Yes

Takahashi is correct, since there are possible sets of integers such as: a_1=0,a_2=1,a_3=0,b_1=1,b_2=0,b_3=1.

Sample Input 2

2 2 2
2 1 2
2 2 2

Sample Output 2

No

Takahashi is incorrect in this case.

Sample Input 3

0 8 8
0 8 8
0 8 8

Sample Output 3

Yes

Sample Input 4

1 8 6
2 9 7
0 7 7

Sample Output 4

No


import java.util.*;

class Main {
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N = sc.nextInt();
    int A[][] = new int[2][N];
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < N; j++) {
        A[i][j] = sc.nextInt();
      }
    }
    int max = 0;
    for (int i = 0; i < N; i++) {
      int c = A[0][0];
      for (int j = 1; j <= i; j++) {
        c += A[0][j];
      }
      for (int j = i; j < N; j++) {
        c += A[1][j];
      }
      if (c > max) max = c;
    }
    System.out.println(max);
  }
}

Score : 300 points

Problem Statement

We have a 2 \times N grid. We will denote the square at the i-th row and j-th column (1 \leq i \leq 2, 1 \leq j \leq N) as (i, j).

You are initially in the top-left square, (1, 1).
You will travel to the bottom-right square, (2, N), by repeatedly moving right or down.

The square (i, j) contains A_{i, j} candies.
You will collect all the candies you visit during the travel.
The top-left and bottom-right squares also contain candies, and you will also collect them.

At most how many candies can you collect when you choose the best way to travel?

Constraints

1 \leq N \leq 100

1 \leq A_{i, j} \leq 100 (1 \leq i \leq 2, 1 \leq j \leq N)

Input

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} ... A_{1, N}
A_{2, 1} A_{2, 2} ... A_{2, N}

Output

Print the maximum number of candies that can be collected.

Sample Input 1

5
3 2 2 4 1
1 2 2 2 1

Sample Output 1

14

The number of collected candies will be maximized when you:

move right three times, then move down once, then move right once.

Sample Input 2

4
1 1 1 1
1 1 1 1

Sample Output 2

5

You will always collect the same number of candies, regardless of how you travel.

Sample Input 3

7
3 3 4 5 4 5 3
5 3 4 4 2 3 2

Sample Output 3

29

Sample Input 4

1
2
3

Sample Output 4

5


import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

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

    static class PeopleLine {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            WeightedUnionFind uf = new WeightedUnionFind(n + 1);
            for (int i = 0; i < m; i++) {
                int l = in.nextInt();
                int r = in.nextInt();
                int w = in.nextInt();
                if (uf.same(l, r)) {
                    int d = uf.diff(l, r);
                    if (d != w) {
                        out.println("No");
                        return;
                    }
                } else {
                    uf.unite(l, r, w);
                }
            }
            out.println("Yes");
        }

        public class WeightedUnionFind {
            private int[] rank;
            private int[] parents;
            private int[] weight;

            public WeightedUnionFind() {
            }

            public WeightedUnionFind(int size) {
                this.rank = new int[size];
                this.parents = new int[size];
                this.weight = new int[size];

                for (int i = 0; i < size; i++) {
                    makeSet(i);
                }
            }

            public void makeSet(int x) {
                parents[x] = x;
                rank[x] = 0;
            }

            public int findSet(int x) {
                if (x != parents[x]) {
                    int r = findSet(parents[x]);
                    weight[x] += weight[parents[x]];
                    parents[x] = r;
                }
                return parents[x];
            }

            public int weight(int x) {
                findSet(x);
                return weight[x];
            }

            public int diff(int x, int y) {
                return weight(x) - weight(y);
            }

            public boolean unite(int x, int y, int w) {
                if (same(x, y)) {
                    return false;
                }
                int px = findSet(x);
                int py = findSet(y);

                if (rank[px] > rank[py]) {
                    // xにくっつける
                    parents[py] = px;
                    weight[py] = -w - weight[y] + weight[x];
                } else {
                    // yにくっつける
                    parents[px] = py;
                    weight[px] = w + weight[y] - weight[x];
                    if (rank[px] == rank[py]) {
                        // 接点の高さの上階.こちらによりくっつくように++
                        rank[py]++;
                    }
                }
                return true;
            }

            public boolean same(int x, int y) {
                return findSet(x) == findSet(y);
            }

        }

    }
}

Score : 400 points

Problem Statement

There are N people standing on the x-axis.
Let the coordinate of Person i be x_i.
For every i, x_i is an integer between 0 and 10^9 (inclusive).
It is possible that more than one person is standing at the same coordinate.

You will given M pieces of information regarding the positions of these people.
The i-th piece of information has the form (L_i, R_i, D_i).
This means that Person R_i is to the right of Person L_i by D_i units of distance, that is, x_{R_i} - x_{L_i} = D_i holds.

It turns out that some of these M pieces of information may be incorrect.
Determine if there exists a set of values (x_1, x_2, ..., x_N) that is consistent with the given pieces of information.

Constraints

1 \leq N \leq 100 000

0 \leq M \leq 200 000

1 \leq L_i, R_i \leq N (1 \leq i \leq M)

0 \leq D_i \leq 10 000 (1 \leq i \leq M)

L_i \neq R_i (1 \leq i \leq M)

If i \neq j, then (L_i, R_i) \neq (L_j, R_j) and (L_i, R_i) \neq (R_j, L_j).

D_i are integers.

Input

Input is given from Standard Input in the following format:

N M
L_1 R_1 D_1
L_2 R_2 D_2
:
L_M R_M D_M

Output

If there exists a set of values (x_1, x_2, ..., x_N) that is consistent with all given pieces of information, print Yes; if it does not exist, print No.

Sample Input 1

3 3
1 2 1
2 3 1
1 3 2

Sample Output 1

Yes

Some possible sets of values (x_1, x_2, x_3) are (0, 1, 2) and (101, 102, 103).

Sample Input 2

3 3
1 2 1
2 3 1
1 3 5

Sample Output 2

No

If the first two pieces of information are correct, x_3 - x_1 = 2 holds, which is contradictory to the last piece of information.

Sample Input 3

4 3
2 1 1
2 3 5
3 4 2

Sample Output 3

Yes

Sample Input 4

10 3
8 7 100
7 9 100
9 8 100

Sample Output 4

No

Sample Input 5

100 0

Sample Output 5

Yes


import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        DJSet dj = new DJSet(n);
        int[][] e = new int[m][3];
        for (int i = 0; i < m; i++) {
            int l = sc.nextInt()-1;
            int r = sc.nextInt()-1;
            int d = sc.nextInt();
            e[i][0] = l; e[i][1] = r; e[i][2] = d;
            dj.union(l, r);
        }
        int[][][] g = buildWeightedGraph(n, e);
        Set<Integer> set = new HashSet<>();
        for (int i = 0; i < n; i++) {
            int r = dj.root(i);
            if(r < 0) r = i;
            set.add(r);
        }
        int[] p = new int[n];
        boolean[] used = new boolean[n];
        boolean ans = true;
        for(int r : set){
            ArrayDeque<Integer> q = new ArrayDeque<>();
            q.add(r);
            used[r] = true;
            while(!q.isEmpty()){
                int v = q.poll();
                used[v] = true;
                for(int[] na : g[v]){
                    int nv = na[0], nd = na[1];
                    if(used[nv]){
                        if(p[nv] != p[v] + nd){
                            ans = false;
                            break;
                        }
                    }else{
                        p[nv] = p[v] + nd;
                        q.add(nv);
                    }
                }
                if(!ans)break;
            }
            if(!ans)break;
        }
        System.out.println(ans ? "Yes" : "No");
        sc.close();

    }

    public static int[][][] buildWeightedGraph(int n, int[][] edges){
        int m = edges.length;
        int[][][] graph = new int[n][][];
        int[] deg = new int[n];
        for (int i = 0; i < m; i++) {
            deg[ edges[i][0] ]++;
            deg[ edges[i][1] ]++;
        }
        for (int i = 0; i < n; i++) {
            graph[i] = new int[deg[i]][2];
        }
        for (int i = 0; i < m; i++) {
            int a = edges[i][0];
            int b = edges[i][1];
            int c = edges[i][2];
            graph[a][--deg[a]][0] = b;
            graph[a][deg[a]][1] = c;
            graph[b][--deg[b]][0] = a;
            graph[b][deg[b]][1] -= c;
        }

        return graph;
    }

}

class DJSet {
    /**
     * 各要素の親を保持する配列。
     * 自信が根である要素は負の数を持つ。
     * （負の数の場合、絶対値がその集合の要素数）
     */
    public int[] upper;

    /**
     * コンストラクタ。
     * @param n 対象となる要素数
     */
    public DJSet(int n) {
        upper = new int[n];
        Arrays.fill(upper, -1);
    }

    /**
     * 要素xが含まれる集合のrootを求める。
     * rootを求める過程で経路圧縮しrootが親になるようにする。
     * @param x 要素x
     * @return 要素xが含まれる集合のroot
     */
    public int root(int x) {
        return upper[x] < 0 ? x : (upper[x] = root(upper[x]));
    }

    /**
     * 要素xと要素yが同じ集合か判定する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yが同じ集合であればtrue、そうでなければfalseを返す。
     */
    public boolean same(int x, int y) {
        return root(x) == root(y);
    }

    /**
     * 要素xと要素yを含むそれぞれの集合を合併する。
     * 要素数の多い集合のrootを残し、少ないほうのrootをその下に合併する。
     * @param x 要素x
     * @param y 要素y
     * @return 要素xと要素yを含むそれぞれの集合を合併できたらtrue、もともと同じ集合で合併しなかったらfalseを返す。
     */
    public boolean union(int x, int y) {
        x = root(x);
        y = root(y);
        if (x != y) {
            if (upper[y] < upper[x]) {
                int t = x;
                x = y;
                y = t;
            }
            upper[x] += upper[y];
            upper[y] = x;
        }
        return x != y;
    }

    /**
     * 集合の数を数える。
     * @return 集合の数
     */
    public int countSet() {
        int c = 0;
        for (int u : upper) {
            if (u < 0)
                c++;
        }
        return c;
    }

    /**
     * 要素ｘが含まれる集合の要素数を数える。
     * @param x 要素ｘ
     * @return 要素ｘが含まれる集合の要素数
     */
    public int countElement(int x) {
        return upper[root(x)] * -1;
    }
}


import java.io.*;
import java.util.*;
class pers
{
	boolean isr=true;
	int par;
	int dist=0;
}
public class Main
{
	static void pars(int par, int chi, int dist)
	{
		if(pe[par].isr&&pe[chi].isr)
		{
	//		System.out.println("hola");
			pe[chi].isr=false;
	//		System.out.println(chi+" "+pe[chi].isr);
			pe[chi].par=par;
			pe[chi].dist=dist;
		}
		else if(fp(par)==fp(chi))
		{
			if(dist!=pe[chi].dist-pe[par].dist)
			{
	//			System.out.println(pe[chi].dist+" "+pe[par].dist);
				poss=false;
			}
		}
		else if(pe[par].isr)
		{
			pe[pe[chi].par].par=par;
			pe[pe[chi].par].isr=false;
			pe[pe[chi].par].dist=dist-pe[chi].dist;
		}
		else if(pe[chi].isr)
		{
			pe[chi].isr=false;
			pe[chi].par=par;
			pe[chi].dist=dist;
		}

		else
		{
			if(pe[chi].dist>dist+pe[par].dist)
			{
				pe[pe[par].par].dist=pe[chi].dist-dist-pe[chi].dist;
				pe[pe[par].par].par=pe[chi].par;
			}
			else
			{
				pe[pe[chi].par].dist=dist+pe[par].dist-pe[chi].dist;
				pe[pe[chi].par].par=pe[par].par;
			}
		}
	}
	static boolean poss;
	static int fp(int a)
	{
		if(pe[a].isr)
			return a;
		else
		{
			int te=fp(pe[a].par);
		//	pe[a].par=fp(pe[a].par);
			pe[a].dist=pe[a].dist+pe[pe[a].par].dist;
			pe[a].par=te;
			return pe[a].par;
		}
	}
	static pers[] pe;
	public static void main(String[] args) throws IOException
	{
		Reader in=new Reader();
		poss=true;
		int n=in.nextInt();
		int m=in.nextInt();
		int a[]=new int[n];
		pe=new pers[n+1];
		for(int i=0; i<=n; i++)
			pe[i]=new pers();
		int l,r,d;
		for(int i=0; i<m; i++)
		{
			l=in.nextInt();
			r=in.nextInt();
			d=in.nextInt();
			pars(l, r, d);
		//	for(int j=1; j<=n; j++)
		//	{
		//		System.out.println(pe[j].isr+" "+pe[j].dist+" "+pe[j].par);
		//	}
		}
		if(poss)
		{
			System.out.println("Yes");
		}
		else
			System.out.println("No");
	}
	static StringBuilder ans=new StringBuilder();
	static class Reader
    {
        final private int BUFFER_SIZE = 1 << 16;
        private DataInputStream din;
        private byte[] buffer;
        private int bufferPointer, bytesRead;
        StringTokenizer st;

        public Reader()
        {
            din = new DataInputStream(System.in);
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public Reader(String file_name) throws IOException
        {
            din = new DataInputStream(new FileInputStream(file_name));
            buffer = new byte[BUFFER_SIZE];
            bufferPointer = bytesRead = 0;
        }

        public String next() throws IOException
        {
        	while(st==null||!st.hasMoreElements())
        	{
        		try
        		{
        			st=new StringTokenizer(readLine());
        		}
        		catch(IOException e)
        		{
        			e.printStackTrace();
        		}
        	}
        	return
        			st.nextToken();
        }
        public String readLine() throws IOException
        {
            byte[] buf = new byte[65535]; // line length
            int cnt = 0, c;
            while ((c = read()) != -1)
            {
                if (c == '\n')
                    break;
                buf[cnt++] = (byte) c;
            }
            return new String(buf, 0, cnt);
        }

        public int nextInt() throws IOException
        {
            int ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do
            {
                ret = ret * 10 + c - '0';
            }  while ((c = read()) >= '0' && c <= '9');

            if (neg)
                return -ret;
            return ret;
        }

        public long l() throws IOException
        {
            long ret = 0;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();
            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');
            if (neg)
                return -ret;
            return ret;
        }

        public double nextDouble() throws IOException
        {
            double ret = 0, div = 1;
            byte c = read();
            while (c <= ' ')
                c = read();
            boolean neg = (c == '-');
            if (neg)
                c = read();

            do {
                ret = ret * 10 + c - '0';
            }
            while ((c = read()) >= '0' && c <= '9');

            if (c == '.')
            {
                while ((c = read()) >= '0' && c <= '9')
                {
                    ret += (c - '0') / (div *= 10);
                }
            }

            if (neg)
                return -ret;
            return ret;
        }

        private void fillBuffer() throws IOException
        {
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
            if (bytesRead == -1)
                buffer[0] = -1;
        }

        private byte read() throws IOException
        {
            if (bufferPointer == bytesRead)
                fillBuffer();
            return buffer[bufferPointer++];
        }

        public void close() throws IOException
        {
            if (din == null)
                return;
            din.close();
        }
    }
}


import java.util.Scanner;

public class Main {

  public static void main(String[] args) {

    Scanner scan = new Scanner(System.in);

    int n = scan.nextInt();

    int[] t = new int[n + 1];
    int[] x = new int[n + 1];
    int[] y = new int[n + 1];

    t[0] = 0;
    x[0] = 0;
    y[0] = 0;

    for (int i = 1; i < n + 1; i++) {
      t[i] = scan.nextInt();
      x[i] = scan.nextInt();
      y[i] = scan.nextInt();

      if ((Math.abs(x[i] - x[i - 1]) + Math.abs(y[i] - y[i - 1]) > t[i] - t[i - 1])
          || (((Math.abs(x[i] - x[i - 1]) + Math.abs(y[i] - y[i - 1])) % 2) != ((t[i] - t[i - 1]) % 2))) {
        System.out.println("No");
        return;
      }

    }

    System.out.println("Yes");

  }

}

Score : 300 points

Problem Statement

AtCoDeer the deer is going on a trip in a two-dimensional plane.
In his plan, he will depart from point (0, 0) at time 0, then for each i between 1 and N (inclusive), he will visit point (x_i,y_i) at time t_i.

If AtCoDeer is at point (x, y) at time t, he can be at one of the following points at time t+1: (x+1,y), (x-1,y), (x,y+1) and (x,y-1).
Note that he cannot stay at his place.
Determine whether he can carry out his plan.

Constraints

1 ≤ N ≤ 10^5

0 ≤ x_i ≤ 10^5

0 ≤ y_i ≤ 10^5

1 ≤ t_i ≤ 10^5

t_i < t_{i+1} (1 ≤ i ≤ N-1)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
t_1 x_1 y_1
t_2 x_2 y_2
:
t_N x_N y_N

Output

If AtCoDeer can carry out his plan, print Yes; if he cannot, print No.

Sample Input 1

2
3 1 2
6 1 1

Sample Output 1

Yes

For example, he can travel as follows: (0,0), (0,1), (1,1), (1,2), (1,1), (1,0), then (1,1).

Sample Input 2

1
2 100 100

Sample Output 2

No

It is impossible to be at (100,100) two seconds after being at (0,0).

Sample Input 3

2
5 1 1
100 1 1

Sample Output 3

No


import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        List<Integer> t = new ArrayList<>();
        List<Integer> x = new ArrayList<>();
        List<Integer> y = new ArrayList<>();
        t.add(0);
        x.add(0);
        y.add(0);
        int[] xt = {-1,0,1,0};
        int[] yt = {0,1,0,-1};
        for(int i = 0;i < n;i++){
            t.add(sc.nextInt());
            x.add(sc.nextInt());
            y.add(sc.nextInt());
        }
        for(int i = 1;i < n;i++){
            int dt = t.get(i) - t.get(i-1);
            int dist = Math.abs(x.get(i)-x.get(i-1)) + Math.abs(y.get(i)-y.get(i-1));
            if(dist <= dt && (dist % 2) == (dt % 2)){
                System.out.println("Yes");
                return;
            }
        }
        System.out.println("No");
    }
}


import java.io.*;
import java.util.*;
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        TaskE solver = new TaskE();
        solver.solve(1, in, out);
        out.flush();out.close();
    }
        static class TaskE {
            boolean chk(int a,int b,int p){
             if(a<0||b<0)return false;
             if(cal(a,b)<=p)return true;return false;
            }
            int cal(int a,int b){
             return (int)Math.ceil((double)Math.max(a,b)/(double)(Math.min(a,b)+1));
            }
         public void solve(int testNumber, InputReader in, PrintWriter out) {
             int n=in.nextInt();
             while(n-->0){
              int a=in.nextInt(),b=in.nextInt(),c=in.nextInt(),d=in.nextInt();
              int k=cal(a,b);//System.out.println(k+" k");
              int l=Math.max(0,c-1-b),r=c-1;int ans=0;
              while(l<=r){
               int m=(l+r)>>1;
               if(chk(m,c-1-m,k)&&chk(a-m,b-(c-1-m),k)){
                ans=m;l=m+1;
               }else r=m-1;
              }//System.out.println(ans+" ans");
              int la=a-ans,lb=b-(c-1-ans);int nb=0,na=0;
              if(chk(la,lb+1,k)&&chk(ans,c-1-ans-1,k)){
               int L=1,R=Math.min(c-1,b);
               while(L<=R){
                int m=(L+R)>>1;
                if(chk(la,lb+m,k)&&chk(ans,c-1-ans-m,k)){
                 nb=m;L=m+1;
                }else R=m-1;
               }
              }else if(chk(la+1,lb,k)&&chk(ans-1,c-1-ans,k)){
               int L=1,R=Math.min(c-1,a);
               while(L<=R){
                int m=(L+R)>>1;
                if(chk(la+m,lb,k)&&chk(ans-m,c-1-ans,k)){
                 na=m;R=m-1;
                }else L=m+1;
               }
               if(ans==c-1)na=ans;
              }//System.out.println(na+" "+nb);
              for(int i=0;i<=d-c;i++){
               if(chk(la-1,lb,k)&&na+1<=k){
                out.print("A");++na;nb=0;--la;
               }else{
                out.print("B");++nb;na=0;--lb;
               }
              }out.println();
             }
         }
    }
    static class InputReader {
        BufferedReader br;
        StringTokenizer st;
        public InputReader(InputStream stream) {
            br = new BufferedReader(new InputStreamReader(stream));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return null;
                st = new StringTokenizer(s);
            }
            return st.nextToken();
        }

        boolean hasMoreTokens() {
            while (st == null || !st.hasMoreTokens()) {
                String s = null;
                try {
                    s = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                if (s == null)
                    return false;
                st = new StringTokenizer(s);
            }
            return true;
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }
        public long nextLong() {
            return Long.parseLong(next());
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
}

Score : 1100 points

Problem Statement

Let f(A, B), where A and B are positive integers, be the string satisfying the following conditions:

f(A, B) has length A + B;

f(A, B) contains exactly A letters A and exactly B letters B;

The length of the longest substring of f(A, B) consisting of equal letters (ex., AAAAA or BBBB) is as small as possible under the conditions above;

f(A, B) is the lexicographically smallest string satisfying the conditions above.

For example, f(2, 3) = BABAB, and f(6, 4) = AABAABAABB.

Answer Q queries: find the substring of f(A_i, B_i) from position C_i to position D_i (1-based).

Constraints

1 \leq Q \leq 10^3

1 \leq A_i, B_i \leq 5 \times 10^8

1 \leq C_i \leq D_i \leq A_i + B_i

D_i - C_i + 1 \leq 100

All input values are integers.

Partial Score

500 points will be awarded for passing the testset satisfying 1 \leq A_i, B_i \leq 10^3.

Input

Input is given from Standard Input in the following format:

Q
A_1 B_1 C_1 D_1
A_2 B_2 C_2 D_2
:
A_Q B_Q C_Q D_Q

Output

For each query i in order of input, print a line containing the substring of f(A_i, B_i) from position C_i to position D_i (1-based).

Sample Input 1

5
2 3 1 5
6 4 1 10
2 3 4 4
6 4 3 7
8 10 5 8

Sample Output 1

BABAB
AABAABAABB
A
BAABA
ABAB


import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		sc.close();
		StringBuilder sb = new StringBuilder();
		sb.append("2018");
		sb.append(S.substring(4));
		System.out.println(sb.toString());
	}
}

Score : 100 points

Problem Statement

On some day in January 2018, Takaki is writing a document. The document has a column where the current date is written in yyyy/mm/dd format. For example, January 23, 2018 should be written as 2018/01/23.

After finishing the document, she noticed that she had mistakenly wrote 2017 at the beginning of the date column. Write a program that, when the string that Takaki wrote in the date column, S, is given as input, modifies the first four characters in S to 2018 and prints it.

Constraints

S is a string of length 10.

The first eight characters in S are 2017/01/.

The last two characters in S are digits and represent an integer between 1 and 31 (inclusive).

Input

Input is given from Standard Input in the following format:

S

Output

Replace the first four characters in S with 2018 and print it.

Sample Input 1

2017/01/07

Sample Output 1

2018/01/07

Sample Input 2

2017/01/31

Sample Output 2

2018/01/31


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Main {

    public static void main(String[] args) throws IOException {
        MyReader reader = new MyReader();
        int[] ab = reader.readInts(" ");
        int a = ab[0];
        int b = ab[1];
        String s = reader.readString();
        System.out.println(problem084b(a, b, s));
    }

    private static String problem084b(int a, int b, String s) {
        String regex = "[0-9]{" + a + "}-[0-9]{" + b + "}";
        Pattern pattern = Pattern.compile(regex);
        Matcher matcher = pattern.matcher(s);
        return (matcher.matches()) ? "Yes" : "No";
    }

    private static class MyReader {
        private final BufferedReader br;
        private MyReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

        private int readInt() throws IOException {
            return Integer.parseInt(readString());
        }

        private String[] readStrings(String regex) throws IOException {
            return readString().split(regex);
        }

        private int[] readInts(String regex) throws IOException {
            String[] strings = readStrings(regex);
            int[] ints = new int[strings.length];
            for (int i = 0; i < ints.length; i++) {
                ints[i] = Integer.parseInt(strings[i]);
            }
            return ints;
        }
    }
}

Score : 100 points

Problem Statement

How many hours do we have until New Year at M o'clock (24-hour notation) on 30th, December?

Constraints

1≤M≤23

M is an integer.

Input

Input is given from Standard Input in the following format:

M

Output

If we have x hours until New Year at M o'clock on 30th, December, print x.

Sample Input 1

21

Sample Output 1

27

We have 27 hours until New Year at 21 o'clock on 30th, December.

Sample Input 2

12

Sample Output 2

36


import java.util.*;

class Main{
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int m = sc.nextInt();
        int ans = 48-m;
        System.out.println(ans);
    }
}


import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int A = Integer.parseInt(scanner.next());
        int B = Integer.parseInt(scanner.next());
        String S = scanner.next();
        scanner.close();

        if (S.matches(String.format("^[0-9]{%d}-[0-9]{%d}", A, B))) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}

Score : 200 points

Problem Statement

The postal code in Atcoder Kingdom is A+B+1 characters long, its (A+1)-th character is a hyphen -, and the other characters are digits from 0 through 9.

You are given a string S. Determine whether it follows the postal code format in Atcoder Kingdom.

Constraints

1≤A,B≤5

|S|=A+B+1

S consists of - and digits from 0 through 9.

Input

Input is given from Standard Input in the following format:

A B
S

Output

Print Yes if S follows the postal code format in AtCoder Kingdom; print No otherwise.

Sample Input 1

3 4
269-6650

Sample Output 1

Yes

The (A+1)-th character of S is -, and the other characters are digits from 0 through 9, so it follows the format.

Sample Input 2

1 1
---

Sample Output 2

No

S contains unnecessary -s other than the (A+1)-th character, so it does not follow the format.

Sample Input 3

1 2
7444

Sample Output 3

No


import java.util.*;
public class Main{
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
  int a = sc.nextInt();
  int b = sc.nextInt();
  String s = sc.next();
  String[] t = s.split("-");
  if(t.length != 2)System.out.println("No");
  else if(t.length == 2 && s.charAt(a) == '-')System.out.println("Yes");
  else System.out.println("No");
}}


import java.util.*;

public class Main {
    static Scanner scanner = new Scanner(System.in);

    public static void main(String[]$) {
        String s = scanner.next();
        int t = Integer.MAX_VALUE;
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i - 1) != s.charAt(i)) {
                t = Math.min(t, Math.max(i, s.length() - i));
            }
        }
        System.out.println(t == Integer.MAX_VALUE ? s.length() : t);
    }
}

Score : 500 points

Problem Statement

You are given a string S consisting of 0 and 1.
Find the maximum integer K not greater than |S| such that we can turn all the characters of S into 0 by repeating the following operation some number of times.

Choose a contiguous segment [l,r] in S whose length is at least K (that is, r-l+1\geq K must be satisfied). For each integer i such that l\leq i\leq r, do the following: if S_i is 0, replace it with 1; if S_i is 1, replace it with 0.

Constraints

1\leq |S|\leq 10^5

S_i(1\leq i\leq N) is either 0 or 1.

Input

Input is given from Standard Input in the following format:

S

Output

Print the maximum integer K such that we can turn all the characters of S into 0 by repeating the operation some number of times.

Sample Input 1

010

Sample Output 1

2

We can turn all the characters of S into 0 by the following operations:

Perform the operation on the segment S[1,3] with length 3. S is now 101.

Perform the operation on the segment S[1,2] with length 2. S is now 011.

Perform the operation on the segment S[2,3] with length 2. S is now 000.

Sample Input 2

100000000

Sample Output 2

8

Sample Input 3

00001111

Sample Output 3

4


import java.util.*;

class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        int n  = s.length();
        int left = n/2-(n%2 ^ 1);
        int right = n/2;
        int width = -n%2;
        char mod = s.charAt(left);
        while(left >= 0 & right < n){
            if (s.charAt(right)!=mod | s.charAt(left)!=mod) {
                break;
            } else {
                width+=2;
            }
            left--;
            right++;
        }
        int k = n/2 + width/2 + n%2;
        System.out.println(k);
    }
}


import java.util.Scanner;

public class Main {

	public static void main(String[] args){

		Scanner sc = new Scanner(System.in);

		String s = sc.next();
		char[] s_chars = s.toCharArray();
		String s_asc = new String(s_chars);

		String t = sc.next();
		char[] t_chars = t.toCharArray();
		String t_asc = new String(t_chars);
		String t_desc = new StringBuilder(t_asc).reverse().toString();

		if (s_asc.compareTo(t_desc) < 0){
			System.out.println("Yes");
		} else {
			System.out.println("No");
		}
	}
}

Score : 200 points

Problem Statement

You are given strings s and t, consisting of lowercase English letters.
You will create a string s' by freely rearranging the characters in s.
You will also create a string t' by freely rearranging the characters in t.
Determine whether it is possible to satisfy s' < t' for the lexicographic order.

Notes

For a string a = a_1 a_2 ... a_N of length N and a string b = b_1 b_2 ... b_M of length M, we say a < b for the lexicographic order if either one of the following two conditions holds true:

N < M and a_1 = b_1, a_2 = b_2, ..., a_N = b_N.

There exists i (1 \leq i \leq N, M) such that a_1 = b_1, a_2 = b_2, ..., a_{i - 1} = b_{i - 1} and a_i < b_i. Here, letters are compared using alphabetical order.

For example, xy < xya and atcoder < atlas.

Constraints

The lengths of s and t are between 1 and 100 (inclusive).

s and t consists of lowercase English letters.

Input

Input is given from Standard Input in the following format:

s
t

Output

If it is possible to satisfy s' < t', print Yes; if it is not, print No.

Sample Input 1

yx
axy

Sample Output 1

Yes

We can, for example, rearrange yx into xy and axy into yxa. Then, xy < yxa.

Sample Input 2

ratcode
atlas

Sample Output 2

Yes

We can, for example, rearrange ratcode into acdeort and atlas into tslaa. Then, acdeort < tslaa.

Sample Input 3

cd
abc

Sample Output 3

No

No matter how we rearrange cd and abc, we cannot achieve our objective.

Sample Input 4

w
ww

Sample Output 4

Yes

Sample Input 5

zzz
zzz

Sample Output 5

No


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

public class Main {
	void run() {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int[] a = new int[n];
		int count = 0;
		for(int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
		}
		while(true) {
			for(int i = 0; i < n; i++) {
				if(a[i] % 2 == 0) {
					a[i] = a[i]/2;
				}else {
					System.out.println(count);
					sc.close();
					return;
				}
			}
			count++;
		}
	}

	void debug(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

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

Score : 200 points

Problem Statement

There are N positive integers written on a blackboard: A_1, ..., A_N.

Snuke can perform the following operation when all integers on the blackboard are even:

Replace each integer X on the blackboard by X divided by 2.

Find the maximum possible number of operations that Snuke can perform.

Constraints

1 \leq N \leq 200

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 maximum possible number of operations that Snuke can perform.

Sample Input 1

3
8 12 40

Sample Output 1

2

Initially, [8, 12, 40] are written on the blackboard.
Since all those integers are even, Snuke can perform the operation.

After the operation is performed once, [4, 6, 20] are written on the blackboard.
Since all those integers are again even, he can perform the operation.

After the operation is performed twice, [2, 3, 10] are written on the blackboard.
Now, there is an odd number 3 on the blackboard, so he cannot perform the operation any more.

Thus, Snuke can perform the operation at most twice.

Sample Input 2

4
5 6 8 10

Sample Output 2

0

Since there is an odd number 5 on the blackboard already in the beginning, Snuke cannot perform the operation at all.

Sample Input 3

6
382253568 723152896 37802240 379425024 404894720 471526144

Sample Output 3

8


import java.util.*;

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

    Scanner sc = new Scanner(System.in);

    String N = sc.next();
    int sum = 0;

    for(int i = 0; i < N.length(); i++){
      int a =Integer.parseInt(N.substring(i, i+1));
      sum += a;
    }

    int total = Integer.parseInt(N);
    String output = null;

    output = (total % sum == 0) ? "Yes":"No";
    System.out.println(output);
  }
}

Score : 200 points

Problem Statement

An integer X is called a Harshad number if X is divisible by f(X), where f(X) is the sum of the digits in X when written in base 10.

Given an integer N, determine whether it is a Harshad number.

Constraints

1?N?10^8

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

Print Yes if N is a Harshad number; print No otherwise.

Sample Input 1

12

Sample Output 1

Yes

f(12)=1+2=3. Since 12 is divisible by 3, 12 is a Harshad number.

Sample Input 2

57

Sample Output 2

No

f(57)=5+7=12. Since 57 is not divisible by 12, 12 is not a Harshad number.

Sample Input 3

148

Sample Output 3

No