hdu 1076

An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10133    Accepted Submission(s): 6291

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

 

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>

using namespace std;

bool leap(int n)
{
    if (n%400 == 0 || (n%4==0 && n%100!=0))
        return true;
    else return false;
}

int main(void)
{
    int t;
#ifndef ONLINE_JUDGE
    freopen("1076.in", "r", stdin);
#endif
    scanf("%d", &t);
    while (t--)
    {
        int n, y;
        scanf("%d%d", &y, &n);
        if (leap(y))
        {
            int cnt = 1, end;
            for (int i = y+4; ; ++i)
            {
                if (leap(i))
                    cnt++;
                if (cnt == n)
                {
                    end = i;
                    break;
                }
            }
            printf("%d\n", end);
        }
        else
        {
            int cnt = 0, end;
            for (int i = y; ; ++i)
            {
                if (leap(i))
                    cnt++;
                if (cnt == n)
                {
                    end = i;
                    break;
                }
            }
            printf("%d\n", end);
        }
    }

    return 0;
}

这题不难……