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

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. 
    
   
   
   
   

 

 

#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;
}