hdoj.1076 An Easy Task 20140811

An Easy Task

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

Total Submission(s): 14152    Accepted Submission(s): 9041

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<stdio.h>
int runy(int y)
{
    return ((y%4==0&&y%100!=0)||y%400==0);
}
int main()
{
    int T,n,y,i,a;
    while(scanf("%d",&T)!=EOF)
    {
        while(T--)
        {
            scanf("%d %d",&y,&n);
            a=y%4;
            if(a)
                y-=a;
            else
                y-=4;
            for(i=1;i<=n;i++)
            {
                y+=4;
                if(runy(y)==0)
                    y+=4;
            }
            printf("%d\n",y);
        }
    }
    return 0;
}