杭电1047 An Easy Task – java 解读
最后更新于：2022-04-01 09:48:42
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# 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
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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
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3
2005 25
1855 12
2004 10000
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Sample Output
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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。
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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));
}
}
}
~~~