An Easy Task HDU 1076

An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11906    Accepted Submission(s): 7509

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

/*
 * Author:  
 * Created Time:  2013/10/8 12:31:09
 * File Name: C.cpp
 * solve: C.cpp
 */
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<stack>
#include<set>
#include<iostream>
#include<vector>
#include<queue>
//ios_base::sync_with_stdio(false);
//#pragma comment(linker, "/STACK:1024000000,1024000000")

using namespace std;
#define sz(v) ((int)(v).size())
#define rep(i, a, b) for (int i = (a); i < (b); ++i)
#define repf(i, a, b) for (int i = (a); i <= (b); ++i)
#define repd(i, a, b) for (int i = (a); i >= (b); --i)
#define clr(x) memset(x,0,sizeof(x))
#define clrs( x , y ) memset(x,y,sizeof(x))
#define out(x) printf(#x" %d
", x)
#define sqr(x) ((x) * (x))
typedef long long LL;

const int INF = 1000000000;
const double eps = 1e-8;
const int maxn = 30000;

int sgn(const double &x) {  return (x > eps) - (x < -eps); }

int main() 
{
    //freopen("in.txt","r",stdin);
    int T;
    scanf("%d",&T);
    while(T--)
    {
        LL a,b;
        scanf("%I64d%I64d",&a,&b);
        int ans = 0;
        LL Y;
        repf(i,a,INF)
        {
           if ((i%4==0 && i%100!=0) || i%400==0)
           {
               ans++;
               if(ans == b)
               {
                   Y = i;
                   break;
               }
           }
        }
        cout<<Y<<endl;
    }
    return 0;
}