hdu 1061

这一题我有两种解法；

1：打表

#include<iostream>
using namespace std;
int main()
{
int T;
cin>>T;
int a[10][4] = {{0},{1},{6,2,4,8},{1,3,9,7},{6,4},{5},{6},{1,7,9,3},{6,8,4,2},{1,9}};
while(T--)
{
int n;
cin>>n;
   int d=n%10;
   if(d == 0||d == 1||d == 5||d == 6)    
              printf("%d\n",d);
else if(d==4 || d==9)
 printf("%d\n",a[d][n%2]);
else if(d==2||d==3||d==7||d==8)
 printf("%d\n",a[d][n%4]);         //cout<<n<<endl;
}
return 0;
}

2.幂取模：

#include<iostream>
using namespace std;
int pow_mod(int a,int n,int m)
{
if(n==0) return 1;
int x = pow_mod(a,n/2,m);
long long ans =(long long)x*x%m;
if(n%2==1) ans = ans *a % m;
return (int)ans;
}
int main(){
int T;
cin>>T;
while(T--)
{
int n;
cin>>n;
cout<<pow_mod(n,n,10)<<endl;
}
return 0;
}