HDU#1076：An Easy Task

最新推荐文章于 2018-08-25 09:59:35 发布

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最新推荐文章于 2018-08-25 09:59:35 发布

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本文介绍了一个简单的C++程序，该程序能够计算从给定年份开始的第N个闰年的具体年份。通过输入起始年份和目标闰年的序号，程序将输出相应的闰年年份。

Time Limit: 1000MS
Memory Limit: 32768KB
64bit IO Format: %I64d & %I64u

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.

AC代码如下：

#include<iostream>
using namespace std;
int main()
{
    int T,Y,k=0,N;
    cin>>T;
    for(int i=0;i<T;i++)
    {
        cin>>Y>>N;
        for(;;Y++)
        {
            if((Y%4==0&&Y%100!=0)||Y%400==0)
            {
                k++;
            }
            if(k==N)
            {
                cout<<Y<<endl;
                break;
            }
        }
        k=0;
    }
    return 0;
}