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An Easy Task

描述

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.

输入

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).

输出

For each test case, you should output the Nth leap year from year Y.

样例输入

3
2005 25
1855 12
2004 10000

样例输出

2108
1904
43236

 

code :

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;bool isLoop(int y){    if((y%4==0 && y%100!=0) || y%400==0){      return true;    }    return false;}int main(){    int t, y, n;    cin>>t;    while(t--){       cin>>y>>n;       while(n){            if(isLoop(y)){                n--;            }            y++;         }         cout<<y-1<<endl;    } return 0;}

  

An Easy Task