杭电oj-1076-An Easy Task

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.

这道题的意思是给你一个数T，代表有T组测试数据，每组测试数据有两个整数，分别代表当前的年份m以及第n个闰年，让你计算从当前年份开始第n个闰年的年份y是多少。

思路:这一题咋一看感觉没啥思路，因为像2100年这样非闰年的数没有什么办法判断出来。因此只能一年一年的判断到最后一个闰年的年份，所幸此题数据不大，可暴力强过。

代码如下：

#include <stdio.h>
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int a,b;
        scanf("%d %d",&a,&b);
        int i=0;

//判断当前年份是否为闰年
        if((a%4==0&&a%100!=0)||(a%400==0))
            i=1;

//暴力判断是否为闰年
        while(i!=b)
        {
            a+=1;
            if((a%4==0&&a%100!=0)||(a%400==0))
                i++;
        }
        printf("%d\n",a);
    }
    return 0;
}