CodeForces - 869B The Eternal Immortality

题意：已知a,b，求的最后一位。

分析：

1、若b-a>=5，则尾数一定为0，因为连续5个数的尾数要么同时包括一个5和一个偶数，要么包括一个0。

2、若b-a<5，直接暴力求即可。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 2000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int main(){
    LL a, b;
    scanf("%lld%lld", &a, &b);
    if(b - a >= 5){
        printf("0\n");
    }
    else{
        LL ans = 1;
        for(LL i = a + 1; i <= b; ++i){
            (ans *= (i % 10)) %= 10;
        }
        printf("%lld\n", ans);
    }
    return 0;
}