杭电 1076 An Easy Task

An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10926    Accepted Submission(s): 6818

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.

 

 

Author

Ignatius.L

 

　　这题应该没什么说的，只要判断闰年的函数没有写错，这题应该就能AC！

以下是代码：

#include <stdio.h>
#include <stdlib.h>

int isrui( int y )
{
    if( !(y%400) )
        return 1;
    else if( (y%100)&&!(y%4) )
         return 1;
    return 0;
}

int main(int argc, char *argv[])
{
    int t, y, n, i, cnt;
    scanf( "%d", &t );
    while( t-- )
    {
           scanf( "%d%d", &y, &n );
           cnt = 0;
           for( i = y; cnt < n; i++ )
                if( isrui(i) )
                    cnt++;
           printf( "%d\n", i-1 );
    }
  
  //system("PAUSE");    
  return 0;
}