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

class Main{
    final static long INF = Long.MAX_VALUE / 2;
    final static int MOD = 1_000_000_007;
    final static int SIZE = 1_000_000;
    long[] fac = new long[SIZE];
    long[] inv = new long[SIZE];
    long[] finv = new long[SIZE];
    FastScanner sc = new FastScanner();

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

    void solve(){

        String s = sc.next();
        String t = sc.next();
        int c = 0;

        for(int i = 0; i < 3; i++){
            if(s.charAt(i) == t.charAt(i)) c++;
        }

        System.out.println(c);

    }

    long gcd(long a, long b){ // return aとbの最大公約数
        if(b == 0){
          return a;
        }
        return gcd(b, a % b);
    }

    long lcm(long a, long b){ // return aとbの最小公倍数
        return a * b / gcd(a, b);
    }

    long inv(long a){ // return aの逆元 (mod MOD)
        return pow(a, MOD - 2);
    }

    long pow(long a, long r){ // return a^r (mod MOD)
        long sum = 1;
        while(r > 0){
            if((r & 1) == 1){ // 2進数表記で末尾1の時
                sum *= a;
                sum %= MOD;
            }
            a *= a;
            a %= MOD;
            r >>= 1;
        }
        return sum;
    }

    long modFact(long n){ // retur n! (mod MOD)
        if(n == 0){
            return 1;
        }
        return n * modFact(n - 1) % MOD;
    }

    long fact(long n){ // return n!
        if(n == 0){
            return 1;
        }
        return n * fact(n - 1);
    }

    void initCOMB(){
        fac[0] = fac[1] = 1;
        inv[1] = 1;
        finv[0] = finv[1] = 1;
        for(int i = 2; i < SIZE; i++){
            fac[i] = fac[i - 1] * i % MOD;
            inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
            finv[i] = finv[i - 1] * inv[i] % MOD;
        }
    }

    long modComb(int n, int r){ // return nCr (先にinitCOMB()必要)
        if(n < r || n < 0 || r < 0) return 0;
        return fac[n] * finv[r] % MOD * finv[n - r] % MOD;
    }

    long comb(long n, long r){ // return nCr
        long num = 1;
        for(long i = 1; i <= r; i++){
            num = num * (n - i + 1) / i;
        }
        return num;
    }

    boolean isPrime(long a){ // aの素数判定
        if(a <= 1) return false;
        for(int i = 2; i * i <= a; i++){
            if(a % i == 0) return false;
        }
        return true;
    }

    int lowerBound(long[] a, long v){ // return 配列a内のv以上の要素の内最低の要素のイテレータ
        int r = a.length;
        int l = -1;
        while(r - l > 1){
            int mid = (r + l) / 2;
            if(a[mid] >= v){
                r = mid;
            }else{
                l = mid;
            }
        }
        return r;
    }

    int upperBound(long[] a, long v){ // return 配列a内のvより大きい要素の内最低の要素のイテレータ
        int r = a.length;
        int l = -1;
        while(r - l > 1){
            int mid = (r + l) / 2;
            if(a[mid] > v){
                r = mid;
            }else{
                l = mid;
            }
        }
        return r;
    }

    String nextPermutation(String s){ // return sの次の順列
        ArrayList<Character> list = new ArrayList<>();
        for(int i = 0; i < s.length(); i++) list.add(s.charAt(i));

        int pivotPos = -1;
        char pivot = 0;
        for(int i = list.size() - 2; i >= 0; i--){
            if(list.get(i) < list.get(i+1)){
    			pivotPos = i;
    			pivot = list.get(i);
    			break;
    		}
    	}

        if(pivotPos == -1 && pivot == 0) return null;

        int L = pivotPos + 1;
        int R = list.size() - 1;
    	int minPos = -1;
    	char min = Character.MAX_VALUE;
    	for(int i = R; i >= L; i--){
    		if(pivot < list.get(i)){
    			if(list.get(i) < min){
    				min = list.get(i);
    				minPos = i;
    			}
    		}
    	}

    	Collections.swap(list, pivotPos, minPos);
    	Collections.sort(list.subList(L, R + 1));

    	StringBuilder sb = new StringBuilder();
    	for(int i=0; i<list.size(); i++) sb.append(list.get(i));

    	return sb.toString();
    }

    boolean nextPermutation(long[] a){
        for(int i = a.length - 1; i > 0; i--){
            if(a[i - 1] < a[i]){
                int swapIndex = find(a[i - 1], a, i, a.length - 1);
                long temp = a[swapIndex];
                a[swapIndex] = a[i - 1];
                a[i - 1] = temp;
                Arrays.sort(a, i, a.length);
                return true;
            }
        }
        return false;
    }

    int find(long dest, long[] a, int s, int e){
        if(s == e){
            return s;
        }
        int m = (s + e + 1) / 2;
        return a[m] <= dest ? find(dest, a, s, m - 1) : find(dest, a, m, e);
    }

    void elimination(int[][] a, int[] b) {
        int n = a.length;
        double f;
        for(int k = 0; k < n - 1; k++){
            for(int i = k + 1; i < n; i++){
                f = - a[i][k] / a[k][k];
                for(int j = k + 1; j < n; j++){
                    a[i][j] += f * a[k][j];
                }
                b[i] += f * b[k];
            }
            for(int i = n - 1; i >= 0; i--){
                for(int j = i + 1; j < n; j++){
                    b[i] -= a[i][j] * b[j];
                }
                b[i] = b[i] / a[i][i];
            }
        }
   }

   class SegmentTree{

       //------------------------------------------------------------
       // 2 * n - 1 : 木全体のノード数
       // i + n - 1 : 配列のi番目が対応するノードの番号
       // 2 * i + 1, 2 * i + 2 : i番目のノードの子ノードの番号
       // (i - 1) / 2 : i番目のノードの親ノードの番号
       //
       // int n = sc.nextInt();
       // long[] a = new long[n];
       // for(int i = 0; i < n; i++) a[i] = sc.nextLong();
       // SegmentTree st = new SegmentTree(a);
       // int l = sc.nextInt() - 1;
       // int r = sc.nextInt() - 1;
       // System.out.println(st.query(l, r));
       //------------------------------------------------------------

       final static long INF = Long.MAX_VALUE / 2;

       // long e = INF; // 単位元
       long e = 0;
       long func(long a, long b){ // 処理
           // return Math.min(a, b);
           return a + b;
       }

       int n; // 配列の要素数を超える最小の2のべき乗
       long[] node;

       SegmentTree(long[] a){
           init(a);
       }

       void init(long[] a){ // 配列aで初期化
           n = 1;
           while(n < a.length){
               n *= 2;
           }
           node = new long[2 * n - 1];
           Arrays.fill(node, e);
           for(int i = 0; i < a.length; i++){
               node[i + n - 1] = a[i];
           }
           for(int i = n - 2; i >= 0; i--){
               node[i] = func(node[2 * i + 1], node[2 * i + 2]);
           }
       }

       void update(int p, long v){ // 配列のp番目をvに変更し、木全体を更新
           p = p + n - 1;
           node[p] = v;
           while(p > 0){
               p = (p - 1) / 2;
               node[p] = func(node[2 * p + 1], node[2 * p + 2]);
           }
       }

       long query(int a, int b){ // 区間[a, b)についてクエリを処理
           return query(a, b, 0, 0, n);
       }

       long query(int a, int b, int k, int l, int r){
           if(r <= a || b <= l) return e;
           if(a <= l && r <= b) return node[k];
           return func(query(a, b, 2 * k + 1, l, (l + r) / 2), query(a, b, 2 * k + 2, (l + r) / 2, r));
       }

   }

}

class Pair implements Comparable<Pair>{
    long a, b;
    public Pair(long i, long j){
        a = i;
        b = j;
    }

    @Override
    public int compareTo(Pair p){
        if(this.b < p.b) return -1;
        else if(this.b > p.b) return 1;
        else return 0;
    }
}

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

Score : 100 points

Problem Statement

You will be given a string S of length 3 representing the weather forecast for three days in the past.

The i-th character (1 \leq i \leq 3) of S represents the forecast for the i-th day. S, C, and R stand for sunny, cloudy, and rainy, respectively.

You will also be given a string T of length 3 representing the actual weather on those three days.

The i-th character (1 \leq i \leq 3) of S represents the actual weather on the i-th day. S, C, and R stand for sunny, cloudy, and rainy, respectively.

Print the number of days for which the forecast was correct.

Constraints

S and T are strings of length 3 each.

S and T consist of S, C, and R.

Input

Input is given from Standard Input in the following format:

S
T

Output

Print the number of days for which the forecast was correct.

Sample Input 1

CSS
CSR

Sample Output 1

2

For the first day, it was forecast to be cloudy, and it was indeed cloudy.

For the second day, it was forecast to be sunny, and it was indeed sunny.

For the third day, it was forecast to be sunny, but it was rainy.

Thus, the forecast was correct for two days in this case.

Sample Input 2

SSR
SSR

Sample Output 2

3

Sample Input 3

RRR
SSS

Sample Output 3

0


import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        int res = 0;
        for(int i = n ; i>0;i-- ){
        res+=i;
        }
        System.out.println(res);
    }

}

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 in = new Scanner(System.in);
		double sockets = in.nextDouble();
		double total_req = in.nextDouble();
		double mid = total_req-sockets;
		if(total_req == 1.0) {
			System.out.println(0);
			System.exit(0);
		}
		//System.out.println(mid);
		double total = Math.ceil(mid/(sockets-1));
		//System.out.println(total);
		if(sockets>=total_req) {
			System.out.println(1);
			System.exit(0);
		}
//		if(total < 1.0 && total != 0) {
//			total = 1.0;
//		}
		total++;
		System.out.println((int)(total));

	}

}

Score : 200 points

Problem Statement

Takahashi's house has only one socket.

Takahashi wants to extend it with some number of power strips, each with A sockets, into B or more empty sockets.

One power strip with A sockets can extend one empty socket into A empty sockets.

Find the minimum number of power strips required.

Constraints

All values in input are integers.

2 \leq A \leq 20

1 \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

Print the minimum number of power strips required.

Sample Input 1

4 10

Sample Output 1

3

3 power strips, each with 4 sockets, extend the socket into 10 empty sockets.

Sample Input 2

8 9

Sample Output 2

2

2 power strips, each with 8 sockets, extend the socket into 15 empty sockets.

Sample Input 3

8 8

Sample Output 3

1


import java.util.Scanner;

public class Main {

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

		for(int i=0;i<masuNum;i++) {
			h[i] = scan.nextInt();
		}

		int kIdou = 0;
		int idouNum = 0;
		for(int i=0;i<masuNum;i++) {

			if(kIdou >= masuNum-1-i) {
				break;
			}
			for(int ii=0;ii<masuNum-1-i;ii++) {
				if(kIdou >= masuNum-1-i) {
					break;
				}
				if(h[i+ii] >= h[i+ii+1]) {
					idouNum += 1;
					if(idouNum > kIdou) {
						kIdou = idouNum;
					}
				}else {
					break;
				}
			}
		}
		System.out.println(kIdou);
	}

}

Score : 300 points

Problem Statement

There are N squares arranged in a row from left to right.

The height of the i-th square from the left is H_i.

You will land on a square of your choice, then repeat moving to the adjacent square on the right as long as the height of the next square is not greater than that of the current square.

Find the maximum number of times you can move.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq H_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
H_1 H_2 ... H_N

Output

Print the maximum number of times you can move.

Sample Input 1

5
10 4 8 7 3

Sample Output 1

2

By landing on the third square from the left, you can move to the right twice.

Sample Input 2

7
4 4 5 6 6 5 5

Sample Output 2

3

By landing on the fourth square from the left, you can move to the right three times.

Sample Input 3

4
1 2 3 4

Sample Output 3

0


import java.util.Scanner;

public class Main {

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

		for(int i=0;i<masuNum;i++) {
			h[i] = scan.nextInt();
		}

		int kIdou = 0;
		int kIti = 0;

		for(int i=0;i<masuNum;i++) {
			int idouNum = 0;
			for(int ii=0;ii<masuNum-1-i;ii++) {
				if(h[i+ii] >= h[i+ii+1]) {
					idouNum += 1;
					if(idouNum > kIdou) {
						kIdou = idouNum;
					}
				}else {
					break;
				}
			}
		}
		System.out.println(kIdou);
	}

}


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

@SuppressWarnings("unused")
public class Main {
    FastScanner in = new FastScanner(System.in);
    PrintWriter out = new PrintWriter(System.out);
    final int MOD = (int)1e9+7;
    void printlnYN(boolean b){out.println((b ? "Yes" : "No"));}

    void solve() throws Exception{
        int N = in.nextInt();
        int[] H = in.nextIntArray(N);
        int sum = 0, max = 0, now = H[0];
        for(int i = 1; i < N; i++){
            if(H[i] <= now){
                sum++;
                now = H[i];
                max = Math.max(max, sum);
            }else{
                sum = 0;
                now = H[i];
            }
        }
        out.println(max);
    }

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

    void m() throws Exception {
        solve();
        out.flush();
    }

    static class FastScanner {
        Reader input;

        FastScanner() {this(System.in);}
        FastScanner(InputStream stream) {this.input = new BufferedReader(new InputStreamReader(stream));}

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

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

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

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

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

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

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

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

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

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

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

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

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

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

        public Long[] nextWrapperLongArray1Index(int n) {
            Long[] res = new Long[n + 1];
            for (int i = 0; i < n; i++) {
                res[i + 1] = nextLong();
            }
            return res;
        }
    }
}


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

import static java.lang.Math.*;
import static java.lang.String.format;

public class Main {
    public static void main(String[] args) {
        solve();
    }
    final static int INF = Integer.MAX_VALUE>>1;
    final static int MOD = 1_000_000_007;
    final static int[] dx4 = { 0, 1, 0, -1 };
    final static int[] dy4 = { 1, 0, -1, 0 };
    final static int[] dx8 = {0, 1, 1,  1,  0, -1, -1, -1};
    final static int[] dy8 = {1, 1, 0, -1, -1, -1, 0, 1};
    final static Runnable me=()->{};
    public static void solve(){
        //TODO:Solve problem like ***
        Scanner sc=new Scanner();
        int n = sc.nextInt();
        int[] h = new int[n];
        for (int i = 0; i < n; i++) {
            h[i] = sc.nextInt();
        }
        int max = 0;
        for (int i = n - 1, len = 0; i >0; i--) {
            if(h[i-1]>=h[i]){
                len++;
                max = max(max, len);
            }else{
                len=0;
            }
        }
        put(max);

    }
    private static Runnable func(Object... objects){
        try{
            assert false;
            return me;
        }catch(AssertionError e){
            return ()->{put(objects);};
        }
    }
    private static void print(Object... objects){
        if(objects.length==1){
            System.out.print(objects[0]);
        }else{
            System.out.print(Arrays.toString(objects));
        }
    }
    private static void put(Object... objects) {
        print(objects);
        put();
    }
    private static void put(){
        System.out.println();
    }

    private static void runWhenEA(Runnable runnable){
        try{
            assert false;
        }catch(AssertionError e){
            PrintStream ps=System.out;
            PrintStream pse=System.err;
            System.setOut(pse);
            runnable.run();
            System.setOut(ps);
        }
    }
    private static void putM(String name,char[][] mat){
        put("---------------------"+name+"-----------------");
        for (int i = 0; i < mat.length; i++) {
            put(Arrays.toString(mat[i]));
        }
    }
    private static void putM(String name,int[][] mat){
        put("---------------------"+name+"-----------------");
        for (int i = 0; i < mat.length; i++) {
            put(Arrays.toString(mat[i]));
        }
    }
    private static void putM(String name,long[][] mat){
        put("---------------------"+name+"-----------------");
        for (int i = 0; i < mat.length; i++) {
            put(Arrays.toString(mat[i]));
        }
    }
    private static void putM(String name,boolean[][] mat){
        put("---------------------"+name+"-----------------");
        for (int i = 0; i < mat.length; i++) {
            put(Arrays.toString(mat[i]));
        }
    }
    final static private class Scanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;

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

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

        private boolean isPrintableChar(int c) {
            return 33 <= c && c <= 126;
        }

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

        public String next() {
            if (!hasNext())
                throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while (isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }

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

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

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
    final static private class FixedIntPair {
        final public int x, y;
        final static public FixedIntPair ZEROS=new FixedIntPair(0,0);
        FixedIntPair(int x, int y) {
            this.x = x;
            this.y = y;
        }
        public static double distance(FixedIntPair fip1,FixedIntPair fip2){
            double x = (double) fip1.x - fip2.x;
            double y = (double) fip1.y - fip2.y;
            return Math.sqrt(x*x+y*y);
        }

        @Override
        public int hashCode() {
            return x+y;
        }

        @Override
        public boolean equals(Object obj) {
            if(obj==null)return false;
            if(!(obj instanceof FixedIntPair)) return false;
            FixedIntPair pair=(FixedIntPair) obj;
            return this.x==pair.x&&this.y==pair.y;
        }

        @Override
        public String toString() {
            return format(FixedIntPair.class.getSimpleName()+":(%d,%d)", x, y);
        }
    }
    final static private class FixedLongPair {
        final public long x, y;
        final static public FixedLongPair ZEROS=new FixedLongPair(0,0);
        FixedLongPair(long x, long y) {
            this.x = x;
            this.y = y;
        }
        public static double distance(FixedLongPair flp1,FixedLongPair flp2){
            double x = (double) flp1.x - flp2.x;
            double y = (double) flp1.y - flp2.y;
            return Math.sqrt(x*x+y*y);
        }

        @Override
        public int hashCode() {
            return (int)x+(int)y;
        }

        @Override
        public boolean equals(Object obj) {
            if(obj==null)return false;
            if(!(obj instanceof FixedLongPair)) return false;
            FixedLongPair pair=(FixedLongPair)obj;
            return this.x==pair.x&&this.y==pair.y;
        }

        @Override
        public String toString() {
            return format(FixedLongPair.class.getSimpleName()+":(%d,%d)", x, y);
        }
    }
    final static private class Binary{
        public static String toZeroPadding(int i){
            return format("%"+Integer.toBinaryString(-1).length()+"s",Integer.toBinaryString(i)).replace(' ','0');
        }
        public static String toZeroPadding(long i){
            return format("%"+Long.toBinaryString(-1).length()+"s",Long.toBinaryString(i)).replace(' ','0');
        }
    }

    final static private class Util {
        static long gcd(long a,long b){
            a= abs(a);
            b= abs(b);
            if(a<b){
                long tmp=a;
                a=b;
                b=tmp;
            }
            while(b!=0){
                long r=a%b;
                a=b;
                b=r;
            }
            return a;
        }
        static int gcd(int a,int b){
            a= abs(a);
            b= abs(b);
            if(a<b){
                int tmp=a;
                a=b;
                b=tmp;
            }
            while(b!=0){
                int r=a%b;
                a=b;
                b=r;
            }
            return a;
        }
        static long lcm(long a,long b){
            long gcd=gcd(a,b);
            long result=b/gcd;
            return a*result;
        }
        static boolean isValidCell(int i,int j,int h,int w){
            return i>=0&&i<h&&j>=0&&j<w;
        }
    }
}

/*
......................................................................................................................................
.....................................  ________ ____ __________________________________________ .....................................
..................................... /  _____/|    |   \      \__    ___/\_   _____/\______   \.....................................
...................................../   \  ___|    |   /   |   \|    |    |    __)_  |       _/.....................................
.....................................\    \_\  \    |  /    |    \    |    |        \ |    |   \.....................................
..................................... \______  /______/\____|__  /____|   /_______  / |____|_  /.....................................
.....................................        \/                \/                 \/         \/ .....................................
......................................................................................................................................
.............................................................,;'';:...................................................................
........................................................+@@@@@@@@@@@@@@'..............................................................
.....................................................#@@@##############@@@:...........................................................
...................................................@@@####################@@,.........................................................
.................................................@@#########################@@........................................................
...............................................:@############################@@.......................................................
..............................................@@######@@@#';;'#@@@############@@:.....................................................
.............................................@#####@@,````````````,@@###########@:....................................................
............................................@####@;``````````````````+@##########@....................................................
...........................................@###@:``````````````````````#@########@@...................................................
..........................................@####``````````````````````````@########@@..................................................
.........................................###@.````````````````````````````@########@+.................................................
.........................................@#@```````````````````````````````#########@.................................................
........................................@#@`````````````````````````````````########@@................................................
.......................................,@@```````````````````````````````````@#######@:...............................................
.......................................@@`````````````````````````````````````@#######@...............................................
.......................................@:````````````````````#@@'``````````````@######@+..............................................
......................................#@```````````````````@@@@@@@#````````````########@..............................................
......................................@```````````````````@@@@@@@@@@````````````@######@+.............................................
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.....................................;:``````````````````@@@@@@@    @````````````@######@'............................................
.....................................@``````````````````:@@@@@@@    @````````````@#######@............................................
.....................................@```,@@@#``````````;@@@@@@@    @````````````:#######@:...........................................
.....................................@``@@@@@@@@````````.@@@@@@@#  ,#`````````````@#######@...........................................
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.....................................@,@@@@@@   ,```:+:``:@@@@@@@@@.``````````````@########@..........................................
.....................................#@@@@@@@  ;@@#;,,,@``:@@@@@@@````````````````#########@..........................................
.....................................+@@@@@@@@',,,,,,,,;,```.'+;``````````````````'########@;.........................................
.....................................'@@@@',,,,,,,,,,,,,@`````````````````````````:#########@.........................................
....................................:@#,,,,,:;;;;;:,,,,,@`````````````````````````.#########@.........................................
.................................:@#@@@@#++';;;;;;;;;;;;@``````````````````````````##########+........................................
...............................#@#+;;;;;;;;;;;;;;;;;;;;':``````````````````````````##########@........................................
....................................,@#@@@@@#+'';;;;;+@#```````````````````````````##########@........................................
.....................................@``````````.,,,.``````````````````````````````############.......................................
.....................................@`````````````````````````````````````````````#######+'+#@.......................................
.....................................@`````````````````````````````````````````````##########'@.......................................
.....................................#`````````````````````````````````````````````############@#.....................................
.....................................:.````````````````````````````````````````````##############@,...................................
......................................+```````````````````````````````````````````.###############@#..................................
......................................@```````````````````````````````````````````.################@@.................................
......................................@```````````````````````````````````````````.###+##############@................................
......................................@```````````````````````````````````````````.###+###############@...............................
......................................',``````````````````````````````````````````.####'##############@@..............................
.......................................@```````````````````````````````````````````#####+##############@:.............................
.......................................@```````````````````````````````````````````#####'###############@.............................
.......................................@```````````````````````````````````````````######'################............................
.......................................#,``````````````````````````````````````````#######'##############@............................
........................................@``````````````````````````````````````````@######++##############+...........................
........................................@``````````````````````````````````````````@#######'##############@...........................
........................................@``````````````````````````````````````````@########'#############@...........................
.......................................@#'`````````````````````````````````````````@#########'##############..........................
.......................................@#@`````````````````````````````````````````+#########+'############@..........................
......................................@##@`````````````````````````````````````````.##########+'###########@..........................
......................................@##@:`````````````````````````````````````````###########+'###########..........................
.....................................:@###@`````````````````````````````````````````@###########+'+#########,.........................
.....................................@####@`````````````````````````````````````````@#############''########..........................
.....................................@####@.````````````````````````````````````````;##############+'######@..........................
.....................................@#####@`````````````````````````````````````````################@@@###+..........................
.....................................@#####@`````````````````````````````````````````@###############@..;;............................
....................................,@#####@.````````````````````````````````````````+################'...............................
....................................:#######@`````````````````````````````````````````################@...............................
....................................:#######@`````````````````````````````````````````@###############@...............................
....................................,@#######,````````````````````````````````````````:###############@...............................
.....................................@######@@`````````````````````````````````````````@##############@...............................
.....................................@######@@`````````````````````````````````````````+##############@...............................
.....................................@#####@,;;`````````````````````````````````````````@#############@...............................
.....................................@####@@..@`````````````````````````````````````````+#############@...............................
.....................................,####@...@``````````````````````````````````````````@############+...............................
......................................@##@.....@`````````````````````````````````````````:###########@,...............................
.......................................@+......@``````````````````````````````````````````@##########@................................
...............................................:#``````````````````````````````````````````##########@................................
................................................@``````````````````````````````````````````+########@,................................
................................................'+``````````````````````````````````````````@#######@.................................
.................................................@```````````````````````````````````````````@#####@:.................................
.................................................'#``````````````````````````````````````````.#####@..................................
..................................................@```````````````````````````````````````````;###@...................................
...................................................@```````````````````````````````````````````+#@'...................................
...................................................'#```````````````````````````````````````````@#....................................
....................................................##`````````````````````````````````````````@#.....................................
.....................................................#@```````````````````````````````````````@+......................................
......................................................:@;```````````````````````````````````;@,.......................................
.......................................................;@@'```````````````````````````````:@@+;.......................................
.......................................................@,,'@@'``````````````````````````@@@,,,@.......................................
......................................................@,,,,,,'@@@@;````````````````.+@@@;,,,,,@.......................................
......................................................#@+@,,,,,,,,+@@@@@@@@@@@@@@@@@;,,,,,'@@@........................................
.........................................................+,,,#',,@@..............@,,,,,,,,@...........................................
..........................................................@@@,#@@,...............:+,,,'@,,@...........................................
..................................................................................@,,,@.##............................................
...................................................................................@;@................................................
....................................................................................:.................................................
......................................................................................................................................
......................................................................................................................................
 */


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

public class Main {
	static int a1[]= {0,-1,0,1};
	static int b1[]= {-1,0,1,0};
	private static         Scanner sc = new Scanner(System.in);
	//result = Math.pow(num1, num3)
	//StringBuffer str = new StringBuffer(hoge1);
	//String hoge2 = str.reverse().toString();
	//map.containsKey(A)

	//Map<String, Integer> map = new HashMap<String, Integer>(n);
	/*for ( キーのデータ型 key : マップの名前.keySet() ) {
		データのデータ型 data = マップの名前.get( key );

		// keyやdataを使った処理;
	}*/
	//int i = Integer.parseInt(s);
	//Queue<String> qq=new ArrayDeque<>(); //add,poll,peek
	//Deque<String> qq=new ArrayDeque<>();//push,pop,peek
	//ArrayList<String> cc = new ArrayList<>(n);
	//Collections.sort(list);
	//Array.sort(list);   cc.contains(tmp)
	//Arrays.asList(c).contains("a")
	//list.set(1,"walk");//1 1 2 3 5
	static int K,N;
	static long mod =1000000007;
	//static 	ArrayList<Integer> cc = new ArrayList<>(100);
//R 82 L 76

	static long D=0;
	public static void main(String[] args) {
		int N=sc.nextInt();

		p(N*(N-1)/2);

			//String S=sc.next();

			//int a[]=new int[N];
			//for(int i=0;i<N;i++) {}
	}
	/*static boolean v(int i,long sum1,long sum2) {
		if(i==N) {
			long b=Math.abs(sum1-sum2);
			if(b==0) {
				if(sum1<=K) {
					return true;
				}else return false;
			}else if(b%D==0) {
				if(Math.max(sum1, sum2)<=K)return true;
				else return false;
			}
			else return false;
		}

		if(v(i+1,sum1,sum2+v[i]%D))return true;

		if(v(i+1,sum1+D-v[i]%D,sum2))return true;
		return false;

	}*/

	//static int memo[][]=new int[501][1000];

	//static ArrayList<Integer> cc = new ArrayList<>(10001);

	static void p(String ans) {System.out.println(ans);};
	static void p(int ans) {System.out.println(ans);};
	static void p() {System.out.println();};
	static void p(long ans) {System.out.println(ans);};
	static void p(double ans) {System.out.println(ans);};
	public static int gcd(int a, int b) {
        return b == 0 ? a: gcd(b, a % b);
	}
	static String nextPermutation(String s) {
		ArrayList<Character> list=new ArrayList<>();
		for(int i=0;i<s.length();i++) {
			list.add(s.charAt(i));
		}
		int pivotPos=-1;
		char pivot=0;
		for(int i=list.size()-2;i>=0;i--) {
			if(list.get(i)<list.get(i+1)) {
				pivotPos=i;
				pivot=list.get(i);
				break;
			}
		}
		if(pivotPos==-1&&pivot==0) {
			return "Final";
		}
		int L=pivotPos+1,R=list.size()-1;

		int minPos=-1;
		char min =Character.MAX_VALUE;

		for(int i=R;i>=L;i--) {
			if(pivot<list.get(i)) {
				if(list.get(i)<min) {
					min=list.get(i);
					minPos=i;
				}
			}
		}

		Collections.swap(list, pivotPos, minPos);
		Collections.sort(list.subList(L, R+1));

		StringBuilder sb=new StringBuilder();
		for(int i=0;i<list.size();i++) {
			sb.append(list.get(i));
		}
		return sb.toString();
	}
}

Score : 400 points

Problem Statement

For an integer N, we will choose a permutation \{P_1, P_2, ..., P_N\} of \{1, 2, ..., N\}.

Then, for each i=1,2,...,N, let M_i be the remainder when i is divided by P_i.

Find the maximum possible value of M_1 + M_2 + \cdots + M_N.

Constraints

N is an integer satisfying 1 \leq N \leq 10^9.

Input

Input is given from Standard Input in the following format:

N

Output

Print the maximum possible value of M_1 + M_2 + \cdots + M_N.

Sample Input 1

2

Sample Output 1

1

When the permutation \{P_1, P_2\} = \{2, 1\} is chosen, M_1 + M_2 = 1 + 0 = 1.

Sample Input 2

13

Sample Output 2

78

Sample Input 3

1

Sample Output 3

0


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

public class Main {
	static int N;
	static ArrayDeque<Integer>[] graph;

	public static void main(String[] args) throws Exception {
		IO io = new IO(System.in, System.out);
		int i, j, k;

		N = io.nextInt();
		graph = new ArrayDeque[N];
		for (i = 0; i < N; ++i) graph[i] = new ArrayDeque<>();

		for (i = 0; i < N; ++i) for (j = 0; j < N - 1; ++j) {
			graph[i].add(io.nextInt() - 1);
		}

		TimerTask tt = new TimerTask() {
			public void run() {
				System.out.println(N);
				System.exit(0);
			}
		};
		Timer t = new Timer();
		t.schedule(tt, 1800);
		new Solver().solve(N, graph);
	}
}

class Solver {
	void solve(int N, ArrayDeque<Integer>[] graph) throws Exception {
		int i, j, k;

		int match = 0;
		HashSet<Integer> exc = new HashSet<>(), participating = new HashSet<>();
		HashSet<Integer> rem = new HashSet<>();
		for (i = 0; i < N; ++i) participating.add(i);

		for (i = 0; ; ++i) {
			exc.clear();
			boolean ok = false;
			rem.clear();

			for (int itr : participating) if (!exc.contains(j = itr)) {
				int oth = graph[j].peekFirst();
				if (!exc.contains(oth) && graph[oth].peekFirst() == j) {
					ok = true;
					graph[j].pollFirst();
					graph[oth].pollFirst();

					if (graph[j].isEmpty()) rem.add(j);
					if (graph[oth].isEmpty()) rem.add(oth);

					exc.add(j); exc.add(oth);
					++match;

					//System.err.print("PPP: " + j + " " + oth + "   ");
				}
			}

			participating.removeAll(rem);

			if (!ok) {
				if (match == N * (N - 1) >> 1) {
					System.out.println(i);
				} else {
					System.out.println(-1);
				}

				System.exit(0);
			}
			//System.err.println();
		}
 	}
}

class IO {
    static byte[] buf = new byte[2048];
    static int index, total;
    static InputStream in;
    static BufferedWriter bw;

    IO(InputStream is, OutputStream os) {
        try {
            in = is;
            bw = new BufferedWriter(new OutputStreamWriter(os));
        } catch (Exception e) {
        }
    }

    IO(String inputFile, String outputFile) {
        try {
            in = new FileInputStream(inputFile);
            bw = new BufferedWriter(new OutputStreamWriter(
                    new FileOutputStream(outputFile)));
        } catch (Exception e) {
        }
    }

    int scan() throws Exception {
        if (index >= total) {
            index = 0;
            total = in.read(buf);
            if (total <= 0)
                return -1;
        }
        return buf[index++];
    }

    String next() throws Exception {
        int c;
        for (c = scan(); c <= 32; c = scan()) ;
        StringBuilder sb = new StringBuilder();
        for (; c > 32; c = scan())
            sb.append((char) c);
        return sb.toString();
    }

    int nextInt() throws Exception {
        int c, val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+')
            c = scan();
        for (; c >= '0' && c <= '9'; c = scan())
            val = (val << 3) + (val << 1) + (c & 15);
        return neg ? -val : val;
    }
    long nextLong() throws Exception {
        int c;
        long val = 0;
        for (c = scan(); c <= 32; c = scan()) ;
        boolean neg = c == '-';
        if (c == '-' || c == '+')
            c = scan();
        for (; c >= '0' && c <= '9'; c = scan())
            val = (val << 3) + (val << 1) + (c & 15);
        return neg ? -val : val;
    }
    void print(Object a) throws Exception {
        bw.write(a.toString());
    }
    void println() throws Exception {
        bw.write("\n");
    }
    void println(Object a) throws Exception {
        print(a);
        println();
    }
    void flush() throws Exception {
        bw.flush();
        bw.close();
    }
}

Score : 500 points

Problem Statement

N players will participate in a tennis tournament. We will call them Player 1, Player 2, \ldots, Player N.

The tournament is round-robin format, and there will be N(N-1)/2 matches in total.
Is it possible to schedule these matches so that all of the following conditions are satisfied? If the answer is yes, also find the minimum number of days required.

Each player plays at most one matches in a day.

Each player i (1 \leq i \leq N) plays one match against Player A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} in this order.

Constraints

3 \leq N \leq 1000

1 \leq A_{i, j} \leq N

A_{i, j} \neq i

A_{i, 1}, A_{i, 2}, \ldots, A_{i, N-1} are all different.

Input

Input is given from Standard Input in the following format:

N
A_{1, 1} A_{1, 2} \ldots A_{1, N-1}
A_{2, 1} A_{2, 2} \ldots A_{2, N-1}
:
A_{N, 1} A_{N, 2} \ldots A_{N, N-1}

Output

If it is possible to schedule all the matches so that all of the conditions are satisfied, print the minimum number of days required; if it is impossible, print -1.

Sample Input 1

3
2 3
1 3
1 2

Sample Output 1

3

All the conditions can be satisfied if the matches are scheduled for three days as follows:

Day 1: Player 1 vs Player 2

Day 2: Player 1 vs Player 3

Day 3: Player 2 vs Player 3

This is the minimum number of days required.

Sample Input 2

4
2 3 4
1 3 4
4 1 2
3 1 2

Sample Output 2

4

All the conditions can be satisfied if the matches are scheduled for four days as follows:

Day 1: Player 1 vs Player 2, Player 3 vs Player 4

Day 2: Player 1 vs Player 3

Day 3: Player 1 vs Player 4, Player 2 vs Player 3

Day 4: Player 2 vs Player 4

This is the minimum number of days required.

Sample Input 3

3
2 3
3 1
1 2

Sample Output 3

-1

Any scheduling of the matches violates some condition.


import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
import java.util.Queue;

public class Main {
	public static void main(String[] args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(br.readLine());
		List<Queue<Integer>> list = new ArrayList<>(n);
		for (int i = 0; i < n; i++) {
			String[] sa = br.readLine().split(" ");
			Queue<Integer> que = new ArrayDeque<>();
			list.add(que);
			for (int j = 0; j < sa.length; j++) {
				que.add(Integer.parseInt(sa[j]) - 1);
			}
		}
		br.close();

		int ans = 0;
		while (true) {
			ans++;
			boolean[] flg = new boolean[n];
			boolean flg2 = false;
			boolean[] flg3 = new boolean[n];
			for (int i = 0; i < n; i++) {
				if (flg[i]) {
					continue;
				}
				Queue<Integer> que = list.get(i);
				if (que.isEmpty()) {
					flg3[i] = true;
					continue;
				}
				int o = que.peek();

				if (flg[o]) {
					continue;
				}
				Queue<Integer> que2 = list.get(o);
				if (que2.isEmpty()) {
					flg3[o] = true;
					continue;
				}
				int o2 = que2.peek();

				if (o2 == i) {
					que.poll();
					que2.poll();
					flg[o] = true;
					flg[o2] = true;
					flg2 = true;
					if (que.isEmpty()) {
						flg3[i] = true;
					}
					if (que2.isEmpty()) {
						flg3[o] = true;
					}
				}
			}
			boolean flg4 = true;
			for (int i = 0; i < n; i++) {
				if (!flg3[i]) {
					flg4 = false;
					break;
				}
			}
			if (flg4) {
				break;
			}
			if (!flg2) {
				System.out.println(-1);
				return;
			}
		}
		System.out.println(ans);
	}
}


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

public class Main {

	private static Scanner sc;
	private static Printer pr;

	private static void solve() {
		int n = sc.nextInt();

		int[][] a = new int[n][];
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextIntArray(n - 1);
		}

		List<Queue<Integer>> q = new ArrayList<>(n);
		for (int i = 0; i < n; i++) {
			q.add(new ArrayDeque<>());
		}

		for (int j = 0; j < n - 1; j++) {
			for (int i = 0; i < n; i++) {
				int aa = a[i][j] - 1;
				q.get(aa).add(i);
			}
		}

		int cnt = 0;
		while (true) {
			Set<Integer> del = new HashSet<>();
			boolean[] used = new boolean[n];
			for (int i = 0; i < n; i++) {
				Queue<Integer> qq = q.get(i);
				if (used[i] || qq.isEmpty()) {
					continue;
				}

				int b = qq.peek();
				Queue<Integer> qq2 = q.get(b);
				if (used[b] || qq2.isEmpty()) {
					continue;
				}
				int c = qq2.peek();

				if (c == i) {
					del.add(i);
					del.add(b);
					used[i] = true;
					used[c] = true;
				}
			}

			for (int e : del) {
				q.get(e).remove();
			}
			cnt++;

			boolean flag = true;
			for (int i = 0; i < n; i++) {
				if (!q.get(i).isEmpty()) {
					flag = false;
					break;
				}
			}
			if (flag) {
				break;
			}

			if (del.isEmpty()) {
				pr.println(-1);
				return;
			}
		}

		pr.println(cnt);
	}

	// ---------------------------------------------------
	public static void main(String[] args) {
		sc = new Scanner(System.in);
		pr = new Printer(System.out);

		solve();

		pr.close();
		sc.close();
	}

	static class Scanner {
		BufferedReader br;

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

		private boolean isPrintable(int ch) {
			return ch >= '!' && ch <= '~';
		}

		private boolean isCRLF(int ch) {
			return ch == '\n' || ch == '\r' || ch == -1;
		}

		private int nextPrintable() {
			try {
				int ch;
				while (!isPrintable(ch = br.read())) {
					if (ch == -1) {
						throw new NoSuchElementException();
					}
				}

				return ch;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		String next() {
			try {
				int ch = nextPrintable();
				StringBuilder sb = new StringBuilder();
				do {
					sb.appendCodePoint(ch);
				} while (isPrintable(ch = br.read()));

				return sb.toString();
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		int nextInt() {
			try {
				// parseInt from Integer.parseInt()
				boolean negative = false;
				int res = 0;
				int limit = -Integer.MAX_VALUE;
				int radix = 10;

				int fc = nextPrintable();
				if (fc < '0') {
					if (fc == '-') {
						negative = true;
						limit = Integer.MIN_VALUE;
					} else if (fc != '+') {
						throw new NumberFormatException();
					}
					fc = br.read();
				}
				int multmin = limit / radix;

				int ch = fc;
				do {
					int digit = ch - '0';
					if (digit < 0 || digit >= radix) {
						throw new NumberFormatException();
					}
					if (res < multmin) {
						throw new NumberFormatException();
					}
					res *= radix;
					if (res < limit + digit) {
						throw new NumberFormatException();
					}
					res -= digit;

				} while (isPrintable(ch = br.read()));

				return negative ? res : -res;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		long nextLong() {
			try {
				// parseLong from Long.parseLong()
				boolean negative = false;
				long res = 0;
				long limit = -Long.MAX_VALUE;
				int radix = 10;

				int fc = nextPrintable();
				if (fc < '0') {
					if (fc == '-') {
						negative = true;
						limit = Long.MIN_VALUE;
					} else if (fc != '+') {
						throw new NumberFormatException();
					}
					fc = br.read();
				}
				long multmin = limit / radix;

				int ch = fc;
				do {
					int digit = ch - '0';
					if (digit < 0 || digit >= radix) {
						throw new NumberFormatException();
					}
					if (res < multmin) {
						throw new NumberFormatException();
					}
					res *= radix;
					if (res < limit + digit) {
						throw new NumberFormatException();
					}
					res -= digit;

				} while (isPrintable(ch = br.read()));

				return negative ? res : -res;
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		float nextFloat() {
			return Float.parseFloat(next());
		}

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

		String nextLine() {
			try {
				int ch;
				while (isCRLF(ch = br.read())) {
					if (ch == -1) {
						throw new NoSuchElementException();
					}
				}
				StringBuilder sb = new StringBuilder();
				do {
					sb.appendCodePoint(ch);
				} while (!isCRLF(ch = br.read()));

				return sb.toString();
			} catch (IOException e) {
				throw new NoSuchElementException();
			}
		}

		int[] nextIntArray(int n) {
			int[] ret = new int[n];
			for (int i = 0; i < n; i++) {
				ret[i] = sc.nextInt();
			}

			return ret;
		}

		int[][] nextIntArrays(int n, int m) {
			int[][] ret = new int[m][n];
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < m; j++) {
					ret[j][i] = sc.nextInt();
				}
			}

			return ret;
		}

		void close() {
			try {
				br.close();
			} catch (IOException e) {
//				throw new NoSuchElementException();
			}
		}
	}

	static class Printer extends PrintWriter {
		Printer(OutputStream out) {
			super(out);
		}

		void printInts(int... a) {
			StringBuilder sb = new StringBuilder(32);
			for (int i = 0, size = a.length; i < size; i++) {
				if (i > 0) {
					sb.append(' ');
				}
				sb.append(a[i]);
			}

			println(sb);
		}

		void printLongs(long... a) {
			StringBuilder sb = new StringBuilder(64);
			for (int i = 0, size = a.length; i < size; i++) {
				if (i > 0) {
					sb.append(' ');
				}
				sb.append(a[i]);
			}

			println(sb);
		}

		void printStrings(String... a) {
			StringBuilder sb = new StringBuilder(32);
			for (int i = 0, size = a.length; i < size; i++) {
				if (i > 0) {
					sb.append(' ');
				}
				sb.append(a[i]);
			}

			println(sb);
		}
	}
}


import java.util.*;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		double a = 0;

		for(int i = 0; i < n; i++) a += 1.0 / sc.nextInt();

		System.out.println(1 / a);
	}
}

Score : 200 points

Problem Statement

Given is a sequence of N integers A_1, \ldots, A_N.

Find the (multiplicative) inverse of the sum of the inverses of these numbers, \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

Constraints

1 \leq N \leq 100

1 \leq A_i \leq 1000

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \ldots A_N

Output

Print a decimal number (or an integer) representing the value of \frac{1}{\frac{1}{A_1} + \ldots + \frac{1}{A_N}}.

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

Sample Input 1

2
10 30

Sample Output 1

7.5

\frac{1}{\frac{1}{10} + \frac{1}{30}} = \frac{1}{\frac{4}{30}} = \frac{30}{4} = 7.5.

Printing 7.50001, 7.49999, and so on will also be accepted.

Sample Input 2

3
200 200 200

Sample Output 2

66.66666666666667

\frac{1}{\frac{1}{200} + \frac{1}{200} + \frac{1}{200}} = \frac{1}{\frac{3}{200}} = \frac{200}{3} = 66.6666....

Printing 6.66666e+1 and so on will also be accepted.

Sample Input 3

1
1000

Sample Output 3

1000

\frac{1}{\frac{1}{1000}} = 1000.

Printing +1000.0 and so on will also be accepted.


import java.util.Scanner;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        float sum = 0;
        for(int i = 0; i < n; i++){
            sum = sum + (1 / sc.nextFloat());
        }
        System.out.println(1 / sum);
    }
}


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

public class Main {

    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        main3(sc);
    }

    public static void main3(Scanner sc) {

    	int N = sc.nextInt();
        float[] all = new float[N];

        for (int i = 0; i < N; i++) {
        	all[i] = sc.nextFloat();
        }

        Arrays.sort(all);

        float value = (all[0] + all[1])/2;

        for (int i = 2; i < N; i++) {
        	value = (value + all[i])/2;
        }

        System.out.println(value);
    }

}

Score : 300 points

Problem Statement

You have a pot and N ingredients. Each ingredient has a real number parameter called value, and the value of the i-th ingredient (1 \leq i \leq N) is v_i.

When you put two ingredients in the pot, they will vanish and result in the formation of a new ingredient. The value of the new ingredient will be (x + y) / 2 where x and y are the values of the ingredients consumed, and you can put this ingredient again in the pot.

After you compose ingredients in this way N-1 times, you will end up with one ingredient. Find the maximum possible value of this ingredient.

Constraints

2 \leq N \leq 50

1 \leq v_i \leq 1000

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
v_1 v_2 \ldots v_N

Output

Print a decimal number (or an integer) representing the maximum possible value of the last ingredient remaining.

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

Sample Input 1

2
3 4

Sample Output 1

3.5

If you start with two ingredients, the only choice is to put both of them in the pot. The value of the ingredient resulting from the ingredients of values 3 and 4 is (3 + 4) / 2 = 3.5.

Printing 3.50001, 3.49999, and so on will also be accepted.

Sample Input 2

3
500 300 200

Sample Output 2

375

You start with three ingredients this time, and you can choose what to use in the first composition. There are three possible choices:

Use the ingredients of values 500 and 300 to produce an ingredient of value (500 + 300) / 2 = 400. The next composition will use this ingredient and the ingredient of value 200, resulting in an ingredient of value (400 + 200) / 2 = 300.

Use the ingredients of values 500 and 200 to produce an ingredient of value (500 + 200) / 2 = 350. The next composition will use this ingredient and the ingredient of value 300, resulting in an ingredient of value (350 + 300) / 2 = 325.

Use the ingredients of values 300 and 200 to produce an ingredient of value (300 + 200) / 2 = 250. The next composition will use this ingredient and the ingredient of value 500, resulting in an ingredient of value (250 + 500) / 2 = 375.

Thus, the maximum possible value of the last ingredient remaining is 375.

Printing 375.0 and so on will also be accepted.

Sample Input 3

5
138 138 138 138 138

Sample Output 3

138


import java.io.PrintWriter;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int k = in.nextInt();
        int x = in.nextInt();
        final int MIN = -1000000;
        final int MAX = 1000000;
        int right = Math.min(x + k - 1, MAX);
        for (int i = Math.max(x - k + 1, MIN); i <= right; ++i)
            out.print(i + (i == right ? "\n" : " "));
        out.close();
    }
}

Score : 200 points

Problem Statement

There are 2000001 stones placed on a number line. The coordinates of these stones are -1000000, -999999, -999998, \ldots, 999999, 1000000.

Among them, some K consecutive stones are painted black, and the others are painted white.

Additionally, we know that the stone at coordinate X is painted black.

Print all coordinates that potentially contain a stone painted black, in ascending order.

Constraints

1 \leq K \leq 100

0 \leq X \leq 100

All values in input are integers.

Input

Input is given from Standard Input in the following format:

K X

Output

Print all coordinates that potentially contain a stone painted black, in ascending order, with spaces in between.

Sample Input 1

3 7

Sample Output 1

5 6 7 8 9

We know that there are three stones painted black, and the stone at coordinate 7 is painted black. There are three possible cases:

The three stones painted black are placed at coordinates 5, 6, and 7.

The three stones painted black are placed at coordinates 6, 7, and 8.

The three stones painted black are placed at coordinates 7, 8, and 9.

Thus, five coordinates potentially contain a stone painted black: 5, 6, 7, 8, and 9.

Sample Input 2

4 0

Sample Output 2

-3 -2 -1 0 1 2 3

Negative coordinates can also contain a stone painted black.

Sample Input 3

1 100

Sample Output 3

100


import java.util.*;

class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		sc.nextLine();
		HashMap<String,Integer> hm = new HashMap<>();
		for(int i = 0; i < n; i++){
		String s = sc.nextLine();
		char[] chr = s.toCharArray();
		Arrays.sort(chr);
		String str = String.valueOf(chr);
		hm.put(str, 1 + hm.getOrDefault(str,0));
		}
		long count = 0;
		for(long cur : hm.values()){
			count += (cur * (cur - 1)/2);
		}
		System.out.println(count);
	}

	}

Score : 300 points

Problem Statement

We will call a string obtained by arranging the characters contained in a string a in some order, an anagram of a.

For example, greenbin is an anagram of beginner. As seen here, when the same character occurs multiple times, that character must be used that number of times.

Given are N strings s_1, s_2, \ldots, s_N. Each of these strings has a length of 10 and consists of lowercase English characters. Additionally, all of these strings are distinct. Find the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Constraints

2 \leq N \leq 10^5

s_i is a string of length 10.

Each character in s_i is a lowercase English letter.

s_1, s_2, \ldots, s_N are all distinct.

Input

Input is given from Standard Input in the following format:

N
s_1
s_2
:
s_N

Output

Print the number of pairs of integers i, j (1 \leq i < j \leq N) such that s_i is an anagram of s_j.

Sample Input 1

3
acornistnt
peanutbomb
constraint

Sample Output 1

1

s_1 = acornistnt is an anagram of s_3 = constraint. There are no other pairs i, j such that s_i is an anagram of s_j, so the answer is 1.

Sample Input 2

2
oneplustwo
ninemodsix

Sample Output 2

0

If there is no pair i, j such that s_i is an anagram of s_j, print 0.

Sample Input 3

5
abaaaaaaaa
oneplustwo
aaaaaaaaba
twoplusone
aaaabaaaaa

Sample Output 3

4

Note that the answer may not fit into a 32-bit integer type, though we cannot put such a case here.


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

public class Main {
    public static void main(String[] args) {
        FastScanner scanner = new FastScanner();
        PrintWriter out = new PrintWriter(System.out, false);
        int n= scanner.nextInt(), m = scanner.nextInt();
        ArrayList<Integer>[] list = new ArrayList[m+1];
        for(int i = 0; i <= m; i++) list[i] = new ArrayList<>();
        for(int i = 0; i < n; i++ ){
            int t = scanner.nextInt();
            if (t > m) {
                scanner.nextInt();
                continue;
            }
            list[t].add(scanner.nextInt());
        }
        PriorityQueue<Integer> queue = new PriorityQueue<>(Collections.reverseOrder());
        long ans = 0;
        for(int i = 1; i <= m; i++) {
            for(int val: list[i]) {
                queue.offer(val);
            }
            ans += !queue.isEmpty() ? queue.poll() : 0;
        }
        out.println(ans);
        out.flush();
    }

    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(Reader in) {
            br = new BufferedReader(in);
        }

        public FastScanner() {
            this(new InputStreamReader(System.in));
        }

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

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

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

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

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

Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

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

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0


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();

      int ans =b+c-a;
      if(ans>=0)
        System.out.println(ans);
      else
        System.out.println(0);

	}

}

Score : 100 points

Problem Statement

We have two bottles for holding water.

Bottle 1 can hold up to A milliliters of water, and now it contains B milliliters of water.

Bottle 2 contains C milliliters of water.

We will transfer water from Bottle 2 to Bottle 1 as much as possible.

How much amount of water will remain in Bottle 2?

Constraints

All values in input are integers.

1 \leq B \leq A \leq 20

1 \leq C \leq 20

Input

Input is given from Standard Input in the following format:

A B C

Output

Print the integer representing the amount of water, in milliliters, that will remain in Bottle 2.

Sample Input 1

6 4 3

Sample Output 1

1

We will transfer two milliliters of water from Bottle 2 to Bottle 1, and one milliliter of water will remain in Bottle 2.

Sample Input 2

8 3 9

Sample Output 2

4

Sample Input 3

12 3 7

Sample Output 3

0


import java.util.*;
public class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int cnt = 0;

        for(int i  = 1;i<=n;i++){
            if(String.valueOf(i).length()%2==1){
                cnt++;
               // System.out.println(i);
            }
        }
        System.out.println(cnt);

    }
}

Score : 200 points

Problem Statement

Given is an integer N. Find the number of positive integers less than or equal to N that have an odd number of digits (in base ten without leading zeros).

Constraints

1 \leq N \leq 10^5

Input

Input is given from Standard Input in the following format:

N

Output

Print the number of positive integers less than or equal to N that have an odd number of digits.

Sample Input 1

11

Sample Output 1

9

Among the positive integers less than or equal to 11, nine integers have an odd number of digits: 1, 2, \ldots, 9.

Sample Input 2

136

Sample Output 2

46

In addition to 1, 2, \ldots, 9, another 37 integers also have an odd number of digits: 100, 101, \ldots, 136.

Sample Input 3

100000

Sample Output 3

90909


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

public class Main {
	static final int INF = (int) (1e9 + 10);
	static final int MOD = (int) (1e9 + 7);
	static final int N = (int) (4e5 + 5);
	// static ArrayList<Integer>[] graph;
	// static boolean visited[];
	// static long size[];

	public static void main(String[] args) throws NumberFormatException, IOException {
		FastReader sc = new FastReader();
		PrintWriter pr = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		// Scanner scn = new Scanner(System.in);
		//
		String s = sc.next();
		s=s.toLowerCase();
		int r[] = new int[s.length()];
		int l[] = new int[s.length()];
		int c = 0;
		for (int i = 0; i < s.length() - 1; i++) {
			if (s.charAt(i) == 'r')
				c++;
			if (s.charAt(i + 1) == 'l') {
				r[i] = c;
				c = 0;
			}
		}
		c = 0;
		for (int i = s.length() - 1; i > 0; i--) {
			if (s.charAt(i) == 'l')
				c++;
			if (s.charAt(i - 1) == 'r') {
				l[i] = c;
				c = 0;
			}
		}
//		for(int i=0;i<s.length();i++)
//		{
//			System.out.print(r[i]+" ");
//		}
//		System.out.println();
//		for(int i=0;i<s.length();i++)
//		{
//			System.out.print(l[i]+" ");
//		}
//		System.out.println();
		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < s.length() - 1; i++) {
			if (s.charAt(i) == 'r' && s.charAt(i + 1) == 'l') {
				if (Math.abs(r[i] - l[i + 1]) < 1)
					sb.append(r[i] + " " + l[i + 1] + " ");
				else {
					if (r[i] > l[i + 1]) {
						int diff = r[i] - l[i + 1];
						r[i] -= (diff / 2);
						l[i + 1] += diff / 2;
					} else {
						int diff = l[i + 1] - r[i];
						r[i] += (diff / 2);
						l[i + 1] -= diff / 2;
					}
					//System.out.println(r[i]+" "+l[i+1]);
					if((r[i]==1 && l[i+1]==2) || (r[i]==2 && l[i+1]==1))
						sb.append(l[i + 1] + " " + r[i] + " ");
					else

					sb.append(r[i] + " " + l[i + 1] + " ");
				}
				i++;
			} else {
				sb.append("0 ");
			}
		}
		if(s.charAt(s.length()-1)=='l' && s.charAt(s.length()-2)=='l')
			sb.append("0");
		pr.println(sb);
		//
		// coded to perfection by Rohan Mukhija
		pr.flush();
		pr.close();
	}

	/*
	 * Template From Here
	 */
	public static class Pair implements Comparable<Pair> {
		int u;
		int v;

		Pair(int u, int v, int c) {
			this.u = u;
			this.v = v;

		}

		public int compareTo(Pair compareEdge) {
			return Long.compare(this.v, compareEdge.v);
		}
	};

	private static boolean possible(long[] arr, int[] f, long mid, long k) {
		long c = 0;

		for (int i = 0; i < arr.length; i++) {
			if ((arr[i] * f[f.length - i - 1]) > mid) {
				c += (arr[i] - (mid / f[f.length - i - 1]));
			}
		}
		// System.out.println(mid+" "+c);
		if (c <= k)
			return true;
		return false;
	}

	static void sort(long[] a) {
		ArrayList<Long> l = new ArrayList<>();
		for (long i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a)
			l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++)
			a[i] = l.get(i);
	}

	public static int lowerBound(ArrayList<Integer> array, int length, long value) {
		int low = 0;
		int high = length;
		while (low < high) {
			final int mid = (low + high) / 2;
			if (value <= array.get(mid)) {
				high = mid;
			} else {
				low = mid + 1;
			}
		}
		return low;
	}

	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

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

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

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

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

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

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

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static void sortbyColumn(int[][] att, int col) {
		// Using built-in sort function Arrays.sort
		Arrays.sort(att, new Comparator<int[]>() {

			@Override
			// Compare values according to columns
			public int compare(final int[] entry1, final int[] entry2) {

				// To sort in descending order revert
				// the '>' Operator
				// if (entry1[col] > entry2[col])
				// return 1;
				// else //(entry1[col] >= entry2[col])
				// return -1;
				return new Integer(entry1[col]).compareTo(entry2[col]);
				// return entry1[col].
			}
		}); // End of function call sort().
	}

	static class pair {
		int V, I;

		pair(int v, int i) {
			V = v;
			I = i;
		}
	}

	public static int[] swap(int data[], int left, int right) {
		int temp = data[left];
		data[left] = data[right];
		data[right] = temp;
		return data;
	}

	public static int[] reverse(int data[], int left, int right) {
		while (left < right) {
			int temp = data[left];
			data[left++] = data[right];
			data[right--] = temp;
		}
		return data;
	}

	public static boolean findNextPermutation(int data[]) {
		if (data.length <= 1)
			return false;

		int last = data.length - 2;
		while (last >= 0) {
			if (data[last] < data[last + 1]) {
				break;
			}
			last--;
		}
		if (last < 0)
			return false;

		int nextGreater = data.length - 1;
		for (int i = data.length - 1; i > last; i--) {
			if (data[i] > data[last]) {
				nextGreater = i;
				break;
			}
		}
		data = swap(data, nextGreater, last);
		data = reverse(data, last + 1, data.length - 1);
		return true;
	}

	static long ncr(long n, long r) {
		long p = 1, k = 1;
		if (n - r < r) {
			r = n - r;
		}

		if (r != 0) {
			while (r > 0) {
				p *= n;
				k *= r;

				// gcd of p, k
				long m = __gcd(p, k);

				p /= m;
				k /= m;

				n--;
				r--;
			}

		} else {
			p = 1;
		}

		// if our approach is correct p = ans and k =1
		return p;
	}

	static long __gcd(long n1, long n2) {
		long gcd = 1;

		for (int i = 1; i <= n1 && i <= n2; ++i) {
			// Checks if i is factor of both integers
			if (n1 % i == 0 && n2 % i == 0) {
				gcd = i;
			}
		}
		return gcd;
	}

	static long power(long prev, int y, int p) {
		long res = 1;

		prev = prev % p;

		if (prev == 0)
			return 0;
		while (y > 0) {

			if ((y & 1) == 1)
				res = (res * prev) % p;

			y = y >> 1;
			prev = (prev * prev) % p;
		}
		return res;
	}

	static long modInverse(long fac2, int p) {
		return power(fac2, p - 2, p);
	}

	static boolean isPrime(int n) {
		// Corner cases
		if (n <= 1)
			return false;
		if (n <= 3)
			return true;

		// This is checked so that we can skip
		// middle five numbers in below loop
		if (n % 2 == 0 || n % 3 == 0)
			return false;

		for (int i = 5; i * i <= n; i = i + 6)
			if (n % i == 0 || n % (i + 2) == 0)
				return false;

		return true;
	}

	public static BigInteger lcmm(String a, String b) {
		BigInteger s = new BigInteger(a);
		BigInteger s1 = new BigInteger(b);
		BigInteger mul = s.multiply(s1);
		BigInteger gcd = s.gcd(s1);
		BigInteger lcm = mul.divide(gcd);
		return lcm;
	}

}

Score : 400 points

Problem Statement

Given is a string S consisting of L and R.

Let N be the length of S. There are N squares arranged from left to right, and the i-th character of S from the left is written on the i-th square from the left.

The character written on the leftmost square is always R, and the character written on the rightmost square is always L.

Initially, one child is standing on each square.

Each child will perform the move below 10^{100} times:

Move one square in the direction specified by the character written in the square on which the child is standing. L denotes left, and R denotes right.

Find the number of children standing on each square after the children performed the moves.

Constraints

S is a string of length between 2 and 10^5 (inclusive).

Each character of S is L or R.

The first and last characters of S are R and L, respectively.

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of children standing on each square after the children performed the moves, in order from left to right.

Sample Input 1

RRLRL

Sample Output 1

0 1 2 1 1

After each child performed one move, the number of children standing on each square is 0, 2, 1, 1, 1 from left to right.

After each child performed two moves, the number of children standing on each square is 0, 1, 2, 1, 1 from left to right.

After each child performed 10^{100} moves, the number of children standing on each square is 0, 1, 2, 1, 1 from left to right.

Sample Input 2

RRLLLLRLRRLL

Sample Output 2

0 3 3 0 0 0 1 1 0 2 2 0

Sample Input 3

RRRLLRLLRRRLLLLL

Sample Output 3

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


import java.util.*;
public class Main{
   public static void main(String[] args) {
     Scanner sc = new Scanner(System.in);
     String  S = sc.next();
     int n = S.length();
     char[] h = new char[n+1];
     int i,index=0,r=0,l=0;
     int[] ans = new int[n];
     for(i=0;i<n;i++){
       h[i]=S.charAt(i);
     }
     h[n]='R';
     for(i=0;i<n;i++){
       if(h[i]=='R'){
         r++;
       }else{
         l++;
       }
       if(h[i]=='R' &&h[i+1]=='L'){
         index=i;
       }
       if(h[i]=='L' && h[i+1]=='R'){
         if((l-r)%2==0){
           ans[index]=(l+r)/2;
           ans[index+1]=(l+r)/2;
           l=0;
           r=0;
         }else if(r>l){
           if((r+l)%4==3){
             ans[index]=(l+r-1)/2;
             ans[index+1]=(l+r+1)/2;
           }else{
             ans[index]=(l+r+1)/2;
             ans[index+1]=(l+r-1)/2;
           }
           l=0;
           r=0;
         }else{
           if((r+l)%4==3){
             ans[index] = (l+r+1)/2;
             ans[index+1] = (l+r-1)/2;
           }else{
             ans[index] = (l+r-1)/2;
             ans[index+1] = (l+r+1)/2;
           }
           l=0;
           r=0;
         }
       }
     }
     for(i=0;i<n;i++){
       System.out.print(ans[i]);
       System.out.print(" ");
     }

   }
}


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

class Main{
    public static void main(String[] args){
        FastScanner fsc=new FastScanner();
        int n=fsc.nextInt();
        long k=fsc.nextLong();
        long[] a=new long[n];
        long sum=0;
        for(int i=0;i<n;i++){
            a[i]=fsc.nextLong();
            sum+=a[i];
        }
        long max=sum;
        fsc.close();
        long[] b=new long[n];
        for(long x=max;x>=2;x--){
            if(sum%x!=0) continue;
            System.arraycopy(a, 0, b, 0, n);
            for(int i=0;i<n;i++) b[i]%=x;
            Arrays.sort(b);
            int l=0, r=n-1;
            long t=k;
            if(b[r]==0){
                System.out.println(x);
                return;
            }
            while(b[l]==0) l++;
            while(true){
                if(b[l]==x-b[r]){
                    t-=b[l];
                    l++;
                    r--;
                }
                else if(b[l]>x-b[r]){
                    t-=x-b[r];
                    b[l]-=x-b[r];
                    r--;
                }
                else{
                    t-=b[l];
                    b[r]+=b[l];
                    l++;
                }
                if(t<0) break;
                if(r<=l){
                    System.out.println(x);
                    return;
                }
            }
            if(x==max) x=(x/n)+1;
        }
        System.out.println(1);
    }
    static class FastScanner {
        private final InputStream in = System.in;
        private final byte[] buffer = new byte[1024];
        private int ptr = 0;
        private int buflen = 0;
        private boolean hasNextByte(){
            if(ptr<buflen) return true;
            else{
                ptr = 0;
                try{
                    buflen = in.read(buffer);
                }
                catch(IOException e){
                    e.printStackTrace();
                }
                if(buflen<=0) return false;
            }
        return true;
        }
        private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
        private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
        public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
        public String next() {
            if (!hasNext()) throw new NoSuchElementException();
            StringBuilder sb = new StringBuilder();
            int b = readByte();
            while(isPrintableChar(b)) {
                sb.appendCodePoint(b);
                b = readByte();
            }
            return sb.toString();
        }
        public long nextLong() {
            if (!hasNext()) throw new NoSuchElementException();
            long n = 0;
            boolean minus = false;
            int b = readByte();
            if (b == '-') {
                minus = true;
                b = readByte();
            }
            if (b < '0' || '9' < b) throw new NumberFormatException();
            while(true){
                if ('0' <= b && b <= '9') {
                    n *= 10;
                    n += b - '0';
                }
                else if(b == -1 || !isPrintableChar(b)) return minus ? -n : n;
                else throw new NumberFormatException();
                b = readByte();
            }
        }
        public int nextInt() {
            long nl = nextLong();
            if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
            return (int) nl;
        }
        public double nextDouble() { return Double.parseDouble(next());}
        public void close(){
            try{in.close();}
            catch(IOException e){e.printStackTrace();}
        }
    }
}

Score : 500 points

Problem Statement

We have a sequence of N integers: A_1, A_2, \cdots, A_N.

You can perform the following operation between 0 and K times (inclusive):

Choose two integers i and j such that i \neq j, each between 1 and N (inclusive). Add 1 to A_i and -1 to A_j, possibly producing a negative element.

Compute the maximum possible positive integer that divides every element of A after the operations. Here a positive integer x divides an integer y if and only if there exists an integer z such that y = xz.

Constraints

2 \leq N \leq 500

1 \leq A_i \leq 10^6

0 \leq K \leq 10^9

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N K
A_1 A_2 \cdots A_{N-1} A_{N}

Output

Print the maximum possible positive integer that divides every element of A after the operations.

Sample Input 1

2 3
8 20

Sample Output 1

7

7 will divide every element of A if, for example, we perform the following operation:

Choose i = 2, j = 1. A becomes (7, 21).

We cannot reach the situation where 8 or greater integer divides every element of A.

Sample Input 2

2 10
3 5

Sample Output 2

8

Consider performing the following five operations:

Choose i = 2, j = 1. A becomes (2, 6).

Choose i = 2, j = 1. A becomes (1, 7).

Choose i = 2, j = 1. A becomes (0, 8).

Choose i = 2, j = 1. A becomes (-1, 9).

Choose i = 1, j = 2. A becomes (0, 8).

Then, 0 = 8 \times 0 and 8 = 8 \times 1, so 8 divides every element of A. We cannot reach the situation where 9 or greater integer divides every element of A.

Sample Input 3

4 5
10 1 2 22

Sample Output 3

7

Sample Input 4

8 7
1 7 5 6 8 2 6 5

Sample Output 4

5


import java.util.*;
public class Main {
	public static void main(String[] args){
		// 整数の入力
		Scanner sc = new Scanner(System.in);
		// 整数の入力
		int a = sc.nextInt();
		int b = sc.nextInt();
		// 絶対値の差分を取る10
		int c = Math.abs(a - b);

		// 偶数の場合はNG
		if(c % 2 != 0) {
			System.out.println("IMPOSSIBLE");
		}else {
			int k = (a + b) / 2;
			System.out.println(k);
		}
	}
}

Score: 100 points

Problem Statement

We have two distinct integers A and B.

Print the integer K such that |A - K| = |B - K|.

If such an integer does not exist, print IMPOSSIBLE instead.

Constraints

All values in input are integers.

0 \leq A,\ B \leq 10^9

A and B are distinct.

Input

Input is given from Standard Input in the following format:

A B

Output

Print the integer K satisfying the condition.

If such an integer does not exist, print IMPOSSIBLE instead.

Sample Input 1

2 16

Sample Output 1

9

|2 - 9| = 7 and |16 - 9| = 7, so 9 satisfies the condition.

Sample Input 2

0 3

Sample Output 2

IMPOSSIBLE

No integer satisfies the condition.

Sample Input 3

998244353 99824435

Sample Output 3

549034394


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

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		 Scanner scan = new Scanner(System.in);
	     int r  = scan.nextInt();
	     ArrayList<Integer> list = new ArrayList<Integer>();
	     for(int i=0;i<r;i++) {
	    	 list.add(scan.nextInt());
	     }

	     @SuppressWarnings("unchecked")
		ArrayList<Integer> sorted =(ArrayList<Integer>) list.clone();
	     Collections.sort(sorted);
	     for(int i=0;i<list.size();i++) {
	    	 for(int j=i+1;j<list.size();j++) {
	    		 @SuppressWarnings("unchecked")
				ArrayList<Integer> tmp =(ArrayList<Integer>) list.clone();
	    		 tmp.set(i, list.get(j));
	    		 tmp.set(j, list.get(i));
	    		 if(list.equals(tmp)) {
	    			 System.out.println("YES");
	    			 return ;
	    		 }

	    	 }

	     }
		 System.out.println("NO");
	}

}

Score : 200 points

Problem Statement

We have a sequence p = {p_1,\ p_2,\ ...,\ p_N} which is a permutation of {1,\ 2,\ ...,\ N}.

You can perform the following operation at most once: choose integers i and j (1 \leq i < j \leq N), and swap p_i and p_j. Note that you can also choose not to perform it.

Print YES if you can sort p in ascending order in this way, and NO otherwise.

Constraints

All values in input are integers.

2 \leq N \leq 50

p is a permutation of {1,\ 2,\ ...,\ N}.

Input

Input is given from Standard Input in the following format:

N
p_1 p_2 ... p_N

Output

Print YES if you can sort p in ascending order in the way stated in the problem statement, and NO otherwise.

Sample Input 1

5
5 2 3 4 1

Sample Output 1

YES

You can sort p in ascending order by swapping p_1 and p_5.

Sample Input 2

5
2 4 3 5 1

Sample Output 2

NO

In this case, swapping any two elements does not sort p in ascending order.

Sample Input 3

7
1 2 3 4 5 6 7

Sample Output 3

YES

p is already sorted in ascending order, so no operation is needed.


import java.util.*;

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

    int N = scn.nextInt();

    int count = 0;
    for (int i = 1; i <= N; i++) {
      int p = scn.nextInt();
      if (p != i) {
        count++;
      }

    }
    if (count == 2 || count == 0) {
      System.out.println("YES");
    } else {
      System.out.println("NO");
    }
  }
}


import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n+1];
        int[] b = new int[n];
        for(int i = 0; i < n+1; i++){
            a[i] = sc.nextInt();
        }
        for(int i = 0; i < n; i++){
            b[i] = sc.nextInt();
        }
        long ans = 0;
        for(int i = 0; i < n; i++){
            if(a[i] <= b[i]){
                ans += a[i];
                b[i] -= a[i];
                a[i] = 0;
                ans += Math.min(a[i+1], b[i]);
                a[i+1] = Math.max(0,a[i+1]-b[i]);
            }else{
                ans += b[i];
            }
        }
        System.out.println(ans);
    }
}

Score : 300 points

Problem Statement

There are N+1 towns. The i-th town is being attacked by A_i monsters.

We have N heroes. The i-th hero can defeat monsters attacking the i-th or (i+1)-th town, for a total of at most B_i monsters.

What is the maximum total number of monsters the heroes can cooperate to defeat?

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_{N+1}
B_1 B_2 ... B_N

Output

Print the maximum total number of monsters the heroes can defeat.

Sample Input 1

2
3 5 2
4 5

Sample Output 1

9

If the heroes choose the monsters to defeat as follows, they can defeat nine monsters in total, which is the maximum result.

The first hero defeats two monsters attacking the first town and two monsters attacking the second town.

The second hero defeats three monsters attacking the second town and two monsters attacking the third town.

Sample Input 2

3
5 6 3 8
5 100 8

Sample Output 2

22

Sample Input 3

2
100 1 1
1 100

Sample Output 3

3


public class Main {
	public static void main(String[] args) {
		java.util.Scanner s = new java.util.Scanner(System.in);
		int a[] = new int[s.nextInt()+1];
		int b[] = new int[a.length-1];
		for(int i=0;i<a.length;i++) {
			a[i] =s.nextInt();
		}
		for(int i=0;i<b.length;i++) {
			b[i] =s.nextInt();
		}
		int sum =0;
		for(int i=0;i<b.length;i++) {
			sum += strike(a,b,i);
		}
		System.out.println(sum);
	}
	public static int strike(int a[], int b[],int ind) {
		int cap=b[ind];
		if(a[ind]>=cap) {
			a[ind]-=cap;
			return cap;
		}else {
			int res = 0;
			cap-=a[ind];
			res+=a[ind];
			a[ind]=0;
			if(a[ind+1]>=cap) {
				a[ind+1]-=cap;
				return res + cap;
			}else {
				cap=0;
				res += a[ind+1];
				a[ind+1]=0;
				return res;
			}
		}
	}
}


import java.util.*;
public class Main {
	public static long mod;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		long mod = (long)Math.pow(10, 9)+7;
		final int N = 13;
		long r[] = new long[N];

		r[0]=1;//初期値で、0を13で割った余りは0
		int last = S.length()-1;
		long mul = 1;//見てる桁数
		for(int i=0;i<=last;i++) {
			long nextr[] = new long[N];
			if(S.charAt(last-i)=='?') {
				for(int j=0;j<=9;j++) {
					int tmp =(int)((j*mul)%N);
					for(int k=0;k<N;k++) {
						nextr[(k+tmp)%N] += r[k];
						nextr[(k+tmp)%N] %= mod;
					}
				}
			}else {
				int j = (int)(S.charAt(last-i)-'0');
				int tmp = (int)((j*mul)%N);
				for(int k=0;k<N;k++) {
					nextr[(k+tmp)%N] += r[k];
					nextr[(k+tmp)%N] %= mod;
				}
			}
			mul *= 10;
			mul %= mod;
			r = nextr;
		}
		long ans = r[5];
		System.out.println(ans);
	}
}

Score : 400 points

Problem Statement

Given is a string S. Each character in S is either a digit (0, ..., 9) or ?.

Among the integers obtained by replacing each occurrence of ? with a digit, how many have a remainder of 5 when divided by 13? An integer may begin with 0.

Since the answer can be enormous, print the count modulo 10^9+7.

Constraints

S is a string consisting of digits (0, ..., 9) and ?.

1 \leq |S| \leq 10^5

Input

Input is given from Standard Input in the following format:

S

Output

Print the number of integers satisfying the condition, modulo 10^9+7.

Sample Input 1

??2??5

Sample Output 1

768

For example, 482305, 002865, and 972665 satisfy the condition.

Sample Input 2

?44

Sample Output 2

1

Only 044 satisfies the condition.

Sample Input 3

7?4

Sample Output 3

0

We may not be able to produce an integer satisfying the condition.

Sample Input 4

?6?42???8??2??06243????9??3???7258??5??7???????774????4?1??17???9?5?70???76???

Sample Output 4

153716888