Rightmost Digit（快速幂）

 1 #include <iostream>
 2 using namespace std; 
 3 int a[25]={0,1,4,7,6,5,6,3,6,9,0,1,6,3,6,5,6,7,4,9,0};
 4 int main()
 5 {
 6     int b,n; 
 7     cin>>b;
 8     while(b--)
 9     {
10        cin>>n;
11        cout<<a[n%20]<<endl; 
12     }
13     return 0 ;
14 }

 1 #include<stdio.h>
 2 int my_power(int m, int n); // 求m的n次方的尾数
 3 int main()
 4 {
 5     int cases, n;
 6     scanf("%d", &cases);
 7     while(cases--)
 8     {
 9         scanf("%d", &n);
10         printf("%d\n", my_power(n, n));
11     }
12     
13     return 0;
14 }
15 
16 int my_power(int m, int n)
17 {
18     m = m%10;
19     if(n == 1)
20         return m;
21     if(n%2 == 0)
22         return ( my_power(m*m, n/2) ) % 10;
23     else
24         return ( my_power(m*m, n/2)*m ) % 10;
25 }