杭电OJ--1076 An Easy Task

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. 
    
    
    
    
     
   
   
   
   

 

大水题啊！这里仅仅记录一个知识点，什么是闰年？

闰年判断方法：
1.能被400整除的年份；
2.能被4整除但同时不能被100整除的年份。
满足上述两个条件之一的即为闰年。

为什么整千年要除以400？
因为整千年一定可以被4整除，用能否被4整除的方法无法准确判别。除以400可以消除整千年末尾的00带来的影响撒。

发代码吧！

#include<iostream>
using namespace std;

int main()
{
	int cases;
	int k;
	cin>>cases;
	while(cases--)
	{
		int y,n;
		cin>>y>>n;
		int num=0;
		for(k=y;;k++)
		{
			if(num==n) break;
			//判断是不是闰年
			if(k%4==0 && k%100!=0 || k%400==0)
				num++;
		}
		cout<<k-1<<endl;
	}
	system("pause");
	return 0;
}