HDU-1061-Rightmost Digit

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33177    Accepted Submission(s): 12701

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. 
    
    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    以前做的，之前没做出来，直接用的循环，TLE了，应该是常理之中吧，然后现在想想这种题就是要去找规律的，突然发现自己进步了不少，这题放在以前都不知道会从这种方面想。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    要输出最右边一位，并且是N^N的最右一位，
   
   
   
   
   
   
   
   

    
    
    
     
   
   
   
   
   
   
   
   

    
    
    
    先举出一下例子，如11， 以1结尾，乘多少次都是1， 同样的还有0， 5 ， 6。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    其他的也有规律，他们乘以好多次都会回到原来的数，如7 ： 7 -〉 9 -〉 3 -〉 1 -〉 7   就是这个规律了。
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

    
    
    
    这题还是很水的，贴个代码：
   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   


   
   
   
   
   
   
   
   

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

int main()
{
    int T, s, N;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&N);
        s=N%10;
        if(s == 0 || s == 1 || s == 5 || s == 6)
        {
        	printf("%d\n", s);
        }
        else if(s == 2)
        {
        	if(N%4==1)printf("2\n");
        	if(N%4==2)printf("4\n");
        	if(N%4==3)printf("8\n");
        	if(N%4==0)printf("6\n");
        }
        else if(s == 3)
        {
        	if(N%4==1)printf("3\n");
        	if(N%4==2)printf("9\n");
        	if(N%4==3)printf("7\n");
        	if(N%4==0)printf("1\n");
        }
        else if(s == 4)
        {
        	if(N%2==1)printf("4\n");if(N%2==0)printf("6\n");
        }
        else if(s == 7)
        {
        	if(N%4==1)printf("7\n");if(N%4==2)printf("9\n");
        	if(N%4==3)printf("3\n");if(N%4==0)printf("1\n");
        }
        else if(s == 8)
        {
        	if(N%4==1)printf("8\n");if(N%4==2)printf("4\n");
        	if(N%4==3)printf("2\n");if(N%4==0)printf("6\n");
        }
        else if(s == 9)
        {
        	if(N%2==1)printf("9\n");
        	if(N%2==0)printf("1\n");
        }
    }
    return 0;
}