贪心算法—杭电1076 An Easy Task

http://acm.hdu.edu.cn/showproblem.php?pid=1076

An Easy Task

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 13386    Accepted Submission(s): 8514


[align=left]Problem Description[/align]
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.

 

 

[align=left]Input[/align]
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).

 

 

[align=left]Output[/align]
For each test case, you should output the Nth leap year from year Y.

 

 

[align=left]Sample Input[/align]

3
2005 25
1855 12
2004 10000

 

 

[align=left]Sample Output[/align]

2108
1904
43236

Hint
We call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.

 

 

#include <iostream>
using namespace std;
int main()
{
int cishu,k;
cin>>cishu;
for(k=1;k<=cishu;k++)
{
int y,num,n;
cin>>y>>num;
int i;
n=0;
for(i=y;;i++)
{
if( (i%4==0&&i%100!=0) || i%400==0)
n++;
if(n==num)
break;
}
cout<<i<<endl;
}
return 0;
}