思路:
掐住最后一位,快速幂一发就好了
#include<cstdio>
#include <map>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;typedef __int64 LL;int cal(int g,int x)
{int ans=1;while(g){if(g%2)ans=(ans*x)%10;x=(x*x)%10;g>>=1;}return ans%10;
}int main()
{int n;scanf("%d",&n);int ans;ans=cal(n,n%10);printf("%d",ans);return 0;
}