HDU-1076An 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.

 

AC代码：

#include<iostream>
#include<stdio.h>
using namespace std;
bool judge(int n) {
 return (n%4==0&&n%100!=0||n%400==0);
}
int Y,n;
int main() {
 int T;
 scanf("%d",&T);
 while(T--) {
  scanf("%d%d",&Y,&n);
  while(n) {
   if(judge(Y)) {
    n--;
    if(n!=0)
    Y+=4;
   }
   else
   Y+=1;
  }
  printf("%d\n",Y);
 }
 return 0;
}