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

234

 

Sample Output

76
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;}