求n^n的最后一位,快速幂,主要是留下整理的头文件
#include <bits/stdc++.h>
typedef long long ll;
typedef double db;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define sspeed ios::sync_with_stdio(false);cin.tie(0)
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod = int(1e9) + 7;
const int md=10;
ll mypow(ll a,ll e){
if(e==0)return 1;
return e&1?mypow(a,e-1)*a%md:mypow(a*a%md,e>>1);
}
inline ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
inline int read(){
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
//-------- head
int main()
{
ll a,b;
while(cin>>a>>b) cout<<mypow(a,b)<<endl;
return 0;
}