An Easy Task


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19908    Accepted Submission(s): 12725


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 


Hint


Author

Ignatius.L

 

Recommend


We have carefully selected several similar problems for you:   ​​1097​​​  ​​​1061​​​  ​​​1013​​​  ​​​1032​​​  ​​​1170​​ 


​​Statistic​​ | ​​Submit​​ | ​​Discuss​​ |​​Note​​


Y 开始以后的第N个闰年

闰年为能被4整除不能被100整除 或者被400整除

#include <stdio.h>
int main()
{
  int ncase;
  scanf("%d",&ncase);
  while(ncase--)
  {
    int year,n;
    scanf("%d %d",&year,&n);
    for(int i=year;;i++)
    {
      if((i%4==0&&i%100)||i%400==0)
      n--;
      if(n==0)
      {
        printf("%d\n",i);
        break;
      }
    }
  }
  return 0;
}