An Easy Task

http://acm.hdu.edu.cn/showproblem.php?pid=1076

An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17302    Accepted Submission(s): 11036

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


   

    

     Hint
    
We call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.

   
    
  

 

#include <cstdio>
int main()
{
    int t,y,ans,n,k;
    scanf("%d",&t);
    while(t--){
        ans=0;
        k=0;
        scanf("%d%d",&y,&n);
        for(;!((y%4==0&&y%100!=0)||y%400==0);y++);
        ans++;
        while(ans!=n){
            y+=4;
            if((y%4==0&&y%100!=0)||y%400==0)ans++;
        }
        printf("%d\n",y);
    }
    return 0;
}