（HDOJ 1076）An Easy Task

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原文链接：http://www.cnblogs.com/cpoint/archive/2011/04/13/2015320.html

本文介绍了一个简单的算法任务——确定从给定起始年份开始的第N个闰年。通过一段C语言代码实现这一功能，利用了闰年的定义进行判断。该问题通过输入起始年份和目标闰年的序号，输出相应的闰年。

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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

43236Hint
We call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.

Author

Ignatius.L

AC code：

#include<stdio.h>

int leapyear(int Y)
{

    if((Y%4==0 && Y%100!=0) || (Y%400==0))
      return 1;
    else
      return 0;
}

int main()
{
    int Y,N,n,count;
    scanf("%d",&n);
    while(n--)
    {
        count=0;
        scanf("%d%d",&Y,&N);
        while(count<N)
        {
            if(leapyear(Y))
            {
              count++;
              Y++;
            }
            else
            {
                Y++;
            }
        }
        printf("%d\n",Y-1);
    }

}

转载于:https://www.cnblogs.com/cpoint/archive/2011/04/13/2015320.html