HDU 1076 An Easy Task

An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17145    Accepted Submission(s): 10946

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

 

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#include<math.h>
#include<stdio.h>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;

bool judge(int Y)
{
    if( (Y%4==0 && Y%100!=0) || Y%400==0 )return 1;
    else return 0;
}

int n,y;

int main()
{
    int t;cin>>t;
    while(t--)
    {
        scanf("%d%d",&y,&n);
        int i=0;
        while(1)
        {

            if(judge(y))i++;
            if(i==n){printf("%d
",y);break;}
            y++;
        }
    }
    return 0;
}

//freopen("1.txt", "r", stdin);
    //freopen("2.txt", "w", stdout);
//**************************************