hdu.1076.An Easy Task

An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21625 Accepted Submission(s): 13860

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.
注意闰年不是四年一次，闰年规律如下：
4年一润，100年不润，400年再润，3200年再不润
即年份能被4整除，但不能被100整除的是闰年
能被100整除，但不能被400整除的不是闰年
能被400整除但不能被3200整除的是闰年
能被3200整除的不是闰年。
我也是刚发现，不知道是不是真的。。。。。。

#include <stdio.h>
#include <stdlib.h>

int main()
{
    int t,y,n,i,m;
    while(scanf("%d",&t)!=EOF)
    {
        while(t--)
        {
            scanf("%d%d",&y,&n);
            i=0;
            while(n!=0)
            {
                m=y+i;
                i++;
                if((m%4==0&&m%100!=0)||m%400==0)
                {
                    n--;
                }
            }
            printf("%d\n",m);
        }
    }
    return 0;
}