Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) 
Total Submission(s): 56655 Accepted Submission(s): 21400

Problem Description 
Given a positive integer N, you should output the most right digit of N^N.

Input 
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow. 
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output 
For each test case, you should output the rightmost digit of N^N.

Sample Input 
2 
3 
4

Sample Output 
7 
6

Hint

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7. 
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6. 
解题思路:本题有规律可寻个位数相乘mod10会出现循环

if (ans == 0 || ans == 1 || ans == 5 || ans == 6)     return ans; if (ans == 2 || ans == 3 || ans == 7 || ans == 8)     // T = 4 if (ans == 4 || ans == 9)       // T = 2

具体实现

#include 
    
      int main() {
       int num;     scanf("%d",&num);     while(num--){
           long long value;         scanf("%lld",&value);         int ans = value % 10;         if (ans == 0 || ans == 1 || ans == 5 || ans == 6)             printf("%d\n",ans);         else         {
               int a[10][10] = {
   0};             for (int i = 2; i < 10; ++i){
                   int k = 1;                 for (int j = 1; j < 10; ++j){
                       k *= i;                     a[i][j] = k%10;                 }             }             if (ans == 4 || ans == 9)                 printf("%d\n",a[ans][value % 2 + 2]);             else                 printf("%d\n",a[ans][value % 4 + 4]);         }     }     return 0; }