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.
The first and only line of input contains two space-separated integers a and b (0 ≤ a ≤ b ≤ 1018).
Output one line containing a single decimal digit — the last digit of the value that interests Koyomi.
2 4
2
0 10
0
107 109
2
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.
#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;
}

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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