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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
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