hdu 1076 An Easy Task(判断闰年)

An Easy Task

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

Total Submission(s): 16139 Accepted Submission(s): 10309



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

#include <cstdio>

using namespace std;

int t,a,b;

bool check ( int a )
{
    if ( a%100 ) return a%4 == 0;
    else return a%400 == 0;
}

int main ( )
{
    scanf ( "%d" , &t );
    while ( t-- )
    {
        scanf ( "%d%d" , &a , &b );
        int j = 0 , i = a;
        for ( ; j < b ; i++ )
            if ( check ( i ) )
                j++;
        printf ( "%d\n" , i-1 );
    }
}