An Easy Task


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 14727    Accepted Submission(s): 9404

Problem Description


Ignatius was born in a leap year, so he want to know when he could hold his birthday party. Can you tell him?
Given a positive integers Y which indicate the start year, and a positive integer N, your task is to tell the Nth leap year from year Y.
Note: if year Y is a leap year, then the 1st leap year is year Y.


 


Input


The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains two positive integers Y and N(1<=N<=10000).


 


Output


For each test case, you should output the Nth leap year from year Y.


 


Sample Input


3 2005 25 1855 12 2004 10000


 


Sample Output


2108 1904 43236


​​题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1076​​

/*
给定起始年 计算第N个闰年 
*/
#include<iostream>
using namespace std;
int main(){
    int m,n,y,i,k;
    cin>>m;
    while(m--)
    {
        cin>>y>>n;
        k=0;
        for(i=y;;i++)
        {
            if((i%4==0&&i%100!=0)||i%400==0) k++;
            if (k==n) 
            {
                printf("%d\n",i);
                break;
            }
        }
    }
    return 0;
}


import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int t = input.nextInt();
        while (t-- > 0) {
            int year = input.nextInt();
            int n = input.nextInt();
            int k = 0;
            for (int i = year;; i++) {
                if ((i % 4 == 0 && i % 100 != 0) || (i % 400 == 0))
                    k++;
                if (k == n) {
                    System.out.println(i);
                    break;
                }
            }
        }
    }
}