# An Easy Task
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 17866    Accepted Submission(s): 11383  

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.
 
题目大意是：输入一个整数控制输入的数据的组数，以下输入哪一年 第几个闰年，闰年的判断标准是：年数除以4余数为0并且除以100余数不为0，或者年数是400的整数倍，到达第几个闰年的时候输出年数。

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main{

	public static void main(String[] args) {

		Scanner scanner = new Scanner(System.in);
		int num = scanner.nextInt();
		int i = 0, num_year = 0, num_times = 0, count = 0;
		List<Integer> list = new ArrayList<Integer>();
		while(i < num)
		{
			count = 0;
			num_year = scanner.nextInt();
			num_times = scanner.nextInt();
			while(count < num_times)
			{
				 
				 if (num_year%4==0 && num_year%100!=0 || num_year%400==0)
				 {
					 count++;
				 }
				 
				 if(count == num_times)
				 {
					 list.add(num_year);
				 }
				 num_year ++;
			}
			i++;
		}
		for(i = 0; i < list.size(); i++)
		{
			System.out.println(list.get(i));
		}
	}

}