您的位置：首页 > 其它

ACM--steps--dyx--1.2.2--

2015-03-21 08:54 239 查看

An Easy Task

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3165 Accepted Submission(s): 1843
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

#include<iostream>
using namespace std;
bool dyx(int year)
{
if((year%4==0&&year%100!=0)||year%400==0)
return true;
else
return false;
}
int main()
{
int n;
cin>>n;
while(n--)
{
int start_year,cont;//cont表示第几个闰年；
cin>>start_year>>cont;
int sum=0;
for(int i=start_year;;i++)
{
if(cont==sum)
{
cout<<i-1<<endl;
break;
}
if(dyx(i))
sum++;
}
}
return  0;
}


内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理 点击这里给我发消息
标签： 
相关文章推荐
新的分享
章节导航