最原始的方法超時，Wa了幾次，原因是 0^0 = 0 ?

#include <iostream>
using namespace std;

int  main()
{
	int t;
	cin >> t;
	while(t--)
	{
		int n; 
		cin >> n;
		int a = n % 10;
		int b = a;
		int mod = 0;
		for(int i = 1; i < n; i++)
		{
			b = (b * a) % 10;
			if(b == a)
			{
				mod = i;
				break;
			}
		}
		if(1 == mod)
		{
			cout << a << endl;
			continue;
		}
		if(mod != 0)
			n =  n % mod + mod;
		b = a;
		for(int i = 1; i < n; i++)
				b = (b * a) % 10;
		cout << b << endl;
	}
	return 0;
}
 

打表方法肯定是最快，附上打表程式碼：

#include<stdio.h>  
int main()  
{  
 int n;  
 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}},d,num;  
 scanf("%d",&num);  
 while(num--)  
 {  
    scanf("%ld",&n);  
    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]);  
 }  
 return 0;  
}