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An Easy Task

2017-05-28 21:56 309 查看

An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4555 Accepted Submission(s): 2558
 
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

一个个数下去就行了

#include <iostream>
using namespace std;
int main()
{
int n,i,m;
int t;
cin>>t;
while(t--)
{
cin>>n>>m;
int j=0;
for(i=n;;i++)
{
if(i%4==0&&i%100!=0||i%400==0)
{
j++;
}
if(j==m)break;
}
cout<<i<<endl;
}
return 0;
}


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