An Easy Task

An Easy Task

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 1   Accepted Submission(s) : 1

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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.

题意：

给出起始年份Y，让你求第N个闰年的具体年份。

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

        并不是每4年，一闰年，若成立即： Y%4==0 即可。但要满足这个 (Y%4==0 && Y%100!=0)。

代码：

#include<stdio.h>
int main(){
    int T,n,y;
    int i,count;
    while(scanf("%d",&T)!=EOF){
    while(T--){
        count=0;
        scanf("%d%d",&y,&n);
        for(i=y;count<n;i++)/* 闰年不是隔四年一循环*/            
            if((i%4==0&&i%100!=0)||(i%400==0))
                count++;/*是闰年就++，等到到了第nth时候就停止，i-1就是要求的年份*/
        printf("%d\n",i-1);
    }
    }
}