hdoj-1061-Rightmost Digit

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. 
    
    
    
    
     
   
   
   
   

水题水题。题意就是求n^n最右边的那一位数。

仔细一想可以发现，求最后一位完全只和最后一位有关系。有的就是本身，而有的则有循环节。

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;

int main()
{
    int t,digit,r,a;
    scanf("%d",&t);
    while(t--)
    {
     scanf("%d",&digit);
     r=digit%10;
     if(r==1){
        printf("1\n");
     }
     else if(r==2){
         a=digit%4;
         if(a==0) printf("6\n");
         else if(a==2)  printf("4\n");
     }
     else if(r==3){
        a=digit%4;
        if(a==3) printf("7\n");
        else if(a==1)  printf("3\n");
     }
     else if(r==4){
            printf("6\n");
     }
     else if(r==5){
        printf("5\n");
     }
     else if(r==6){
        printf("6\n");
     }
     else if(r==7){
            a=digit%4;
     if(a==1)  printf("7\n");
     else if(a==3) printf("3\n");
     }
     else if(r==8){
        a=digit%4;
        if(a==0) printf("6\n");
        else if(a==2)  printf("4\n");
     }
     else if(r==9){
        printf("9\n");
     }
     else if(r==0){
        printf("0\n");
     }
    }
}