HDU1061：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.

 

 

//这道题与HDU1097题基本就是一样的

只需要稍微改一下就可以了

 

#include <stdio.h>

#include<string.h>

int main()

{

    int a,b,n,t;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&a);

        n=a%10;

        b = a;

        if(b==0)

            printf("1\n");

        else

        {

            switch(n)

            {

            case 0:

            case 1:

            case 6:

                break;

            case 2:

                n=b%4;

                switch(n)

                {

                case 1:

                    n=2;

                    break;

                case 2:

                    n=4;

                    break;

                case 3:

                    n=8;

                    break;

                case 0:

                    n=6;

                    break;

                }

                break;

            case 3:

                n=b%4;

                switch(n)

                {

                case 1:

                    n=3;

                    break;

                case 2:

                    n=9;

                    break;

                case 3:

                    n=7;

                    break;

                case 0:

                    n=1;

                    break;

                }

                break;

            case 4:

                n=b%2;

                switch(n)

                {

                case 1:

                    n=4;

                    break;

                case 0:

                    n=6;

                    break;

                }

                break;

            case 7:

                n=b%4;

                switch(n)

                {

                case 1:

                    n=7;

                    break;

                case 2:

                    n=9;

                    break;

                case 3:

                    n=3;

                    break;

                case 0:

                    n=1;

                    break;

                }

                break;

            case 8:

                n=b%4;

                switch(n)

                {

                case 1:

                    n=8;

                    break;

                case 2:

                    n=4;

                    break;

                case 3:

                    n=2;

                    break;

                case 0:

                    n=6;

                    break;

                }

                break;

            case 9:

                n=b%2;

                switch(n)

                {

                case 1:

                    n=9;

                    break;

                case 0:

                    n=1;

                    break;

                }

                break;

            }

        }

        printf("%d\n",n);

    }

    return 0;

}