HDOJ1076 An Easy Task

An Easy Task

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

Total Submission(s): 24613    Accepted Submission(s): 15866


[align=left]Problem Description[/align]
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.

 

[align=left]Input[/align]
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).

 

[align=left]Output[/align]
For each test case, you should output the Nth leap year from year Y.

 

[align=left]Sample Input[/align]

3
2005 25
1855 12
2004 10000

 

[align=left]Sample Output[/align]

2108
1904
43236

Hint
We call year Y a leap year only if (Y%4==0 && Y%100!=0) or Y%400==0.

 给出其实年份，求该年份的第N个闰年，计算闰年的方法都给了

import java.util.Scanner;

public class Main{
private static Scanner scanner;
public static void main(String[] args) {
scanner = new Scanner(System.in);
int cases = scanner.nextInt();
while(cases-->0){
int year = scanner.nextInt();
int m = scanner.nextInt();
int count = 0,i = 0;
while(count<m){
if(judgeYear(year+i)){
count++;
}
i++;
}
System.out.println(year+i-1);
}
}
private static boolean judgeYear(int i) {
if((i%4==0 && i%100!=0 )|| i%400 == 0){
return true;
}
return false;
}
}