Hdu 1076 An Easy Task【水】

An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19001 Accepted Submission(s): 12148

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.

   
   
   
    
  
  

给出开始的年份Y，问第N 个闰年是哪一年

唯一能想到的办法就是循环暴力枚举，想想会超时，然后再上网搜解法，继而发现神奇的与闰年相关的结论，没想到...暴力竟然AC了， 然后瞬间感觉自己想多了........

#include<stdio.h>
#include<string.h>
bool leap(int x)
{
	return x%4==0&&x%100!=0||x%400==0;
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int y,n;
		scanf("%d%d",&y,&n);
		while(n)
		{
			if(leap(y))
			{
				--n;
			}
			++y;
		}
		printf("%d\n",y-1);
	}
	return 0;
}