矩阵快速幂

矩阵快速幂的学习：

https://blog.csdn.net/wust_zzwh/article/details/52058209

n>>1是指n的二进制数向右移一位

n&1等价n%2

https://vjudge.net/contest/231312#problem/F

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
typedef long long ll;
using namespace std;
ll ans=1;
void quickpow(ll x,ll n)
{
    ll res=x;
    while(n)
    {
        if(n%2==1)
        {
            ans*=res;
            res*=res;
            if(ans>9)
                ans=ans%10;
            if(res>9)
                res=res%10;
        }
        else
        {
            res*=res;
            if(res>9)
                res=res%10;
        }
        n=n>>1;
    }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ans=1;
        ll n;
        scanf("%I64d",&n);
        quickpow(n,n);
        printf("%I64d\n",ans);
    }
    return 0;
}