869B - The Eternal Immortality 暴力 模拟

本文介绍了一个基于凤凰重生周期的算法问题,探讨了如何计算在特定的时间范围内凤凰重生的次数,并给出了一个高效的算法实现方案。

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B. The Eternal Immortality
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

Even if the world is full of counterfeits, I still regard it as wonderful.

Pile up herbs and incense, and arise again from the flames and ashes of its predecessor — as is known to many, the phoenix does it like this.

The phoenix has a rather long lifespan, and reincarnates itself once every a! years. Here a! denotes the factorial of integer a, that is, a! = 1 × 2 × ... × a. Specifically, 0! = 1.

Koyomi doesn't care much about this, but before he gets into another mess with oddities, he is interested in the number of times the phoenix will reincarnate in a timespan of b! years, that is, . Note that when b ≥ a this value is always integer.

As the answer can be quite large, it would be enough for Koyomi just to know the last digit of the answer in decimal representation. And you're here to provide Koyomi with this knowledge.

Input

The first and only line of input contains two space-separated integers a and b (0 ≤ a ≤ b ≤ 1018).

Output

Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.

Examples
input
2 4
output
2
input
0 10
output
0
input
107 109
output
2
Note

In the first example, the last digit of  is 2;

In the second example, the last digit of  is 0;

In the third example, the last digit of  is 2.



题意很清楚,水题一发,肯定不能一个一个去乘,因为有很多末尾是0的情况,所以判断如果b-a>5直接输出0,否则用循环判断一下即可:
#include <stdio.h>
#include <string.h>
int main() {
	long long a , b , i;
	int sum = 1;
	scanf("%lld%lld",&a,&b);
	if(b - a < 5)
	for(i = a+1; i <= b; i++) {
		sum = (sum*(i%10))%10;
	}
	else sum = 0;
	printf("%d\n",sum);
	return 0;
}


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