atcoder 它February 29th

Time limit : 2sec / Stack limit : 256MB / Memory limit : 256MB

Problem

Charlie was born January 1st of the year A, on the Earth. He will leave the Earth on December 31st of the year B.

He wants to know how many times he passes February 29th on the Earth.

February 29th is a leap day. A year that contains a leap day is called a leap year. You can determine if a year is leap year or not by following rules.

• If the year is divisible by 4, it is a leap year except the following case.
• If the year is divisible by 100, it is NOT a leap year except the following case.
• If the year is divisible by 400, it is a leap year.

Output how many times Charlie passes February 29th on the Earth. Note that Charlie lives very long.

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Input

The input will be given in the following format from the Standard Input.

A B

• On the first line, you will be given the year A(1≦A≦2,000,000,000), when Charlie born, followed by a space and the year B(A≦B≦2,000,000,000), when he leaves the Earth.

Achievements and Points

• When you pass every test case where 1≦A≦B≦3,000 , you will be awarded 25 points.
• In addition, if you pass all the rest test cases you will be awarded 25 more points.

Output

Output how many times Charlie passes February 29th on the Earth in one line. Make sure to insert a line break at the end of the output.

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Inout Example 1

1988 2014

Output Example 1

7

Charlie can pass February 29th of 1988, 1992, 1996, 2000, 2004, 2008, 2012. The total is 7 times.

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Input Example 2

997 1003

Output Example 2

0

Note that the year 1000 is NOT a leap year.

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Input Example 3

1 2000000000

Output Example 3

485000000

Note that Charlie lives very long.

思路:这题就是求两个年份之间的闰年数目。简单题。直接上代码。

import java.util.Scanner;
 
public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		while (sc.hasNext()) {
			int a = sc.nextInt();
			int b = sc.nextInt();
			int temp1=b/4-(a-1)/4;
			int temp2=b/100-(a-1)/100;
			int temp3=b/400-(a-1)/400;
			System.out.println(temp1-temp2+temp3);
		}
	}
}

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