小弟在做一个ACM的题目，要求得出数字N的N次方以后的最右边数值，我写了下面这个程序，在C里面运行没有问题，但是传到网站上就显示答案错误了，估计是程序里面有不对的地方，希望各位大大指正，谢谢了。
题目要求：
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


程序如下


[i]#include <stdio.h>
void main ()
{
    int a,s,t,i,b[100],j;
    scanf("%d",&t);
    for(i=0;i<t;i++)
    {
        scanf("%d",&a);
        s=a%10;
        switch(s)
        {
        case 0:b[i]=0;
            break;
        case 1:b[i]=1;
            break;
        case 2:    j=a%4;
            switch(j)
            {
            case 0:b[i]=6;
                break;
            case 1:b[i]=2;
                break;
            case 2:b[i]=4;
                break;
            case 3:b[i]=8;
                break;
            }
            break;
        case 3:j=a%4;
            switch(j)
            {
            case 0:b[i]=1;
                break;
            case 1:b[i]=3;
                break;
            case 2:b[i]=9;
                break;
            case 3:b[i]=7;
                break;
            }
            break;
        case 4:j=a%2;
            switch(j)
            {
            case 0:b[i]=6;
                break;
            case 1:b[i]=4;
                break;
            }
            break;
        case 5:b[i]=5;
            break;
        case 6:b[i]=6;
            break;
        case 7:j=a%4;
            switch(j)
            {
            case 0:b[i]=1;
                break;
            case 1:b[i]=7;
                break;
            case 2:b[i]=9;
                break;
            case 3:b[i]=3;
                break;
            }
            break;
        case 8:
            j=a%4;
            switch(j)
            {
            case 0:b[i]=6;
                break;
            case 1:b[i]=8;
                break;
            case 2:b[i]=4;
                break;
            case 3:b[i]=2;
                break;
            }
            break;
        case 9:
            j=a%2;
            switch(j)
            {
            case 0:b[i]=1;
                break;
            case 1:b[i]=9;
                break;
            }
        }
        
    }
    for(i=0;i<t;i++)printf("%d\n",b);
}

做了好长时间了，还没做起来，很伤心，各位帮帮吧。

都是高手

看错了

[ 本帖最后由 flyingcloude 于 2009-10-10 11:07 编辑 ]

int main(void)   
{   
    int n[10][4] =     
    {   
        { 0, 0, 0, 0 },   
        { 1, 1, 1, 1 },   
        { 6, 2, 4, 8 },   
        { 1, 3, 9, 7 },   
        { 6, 4, 6, 4 },   
        { 5, 5, 5, 5 },   
        { 6, 6, 6, 6 },   
        { 1, 7, 9, 3 },   
        { 6, 8, 4, 2 },   
        { 1, 9, 1, 9 }   
    };   
    int N;   
 
    scanf("%d", &N);      
    printf("%d\n", n[N%10][N%4]);  
   
    return 0;
      
}   

可以更简单的

再说吧

[ 本帖最后由 BlueGuy 于 2009-10-10 11:41 编辑 ]

恩……有道理……

对于一位数可行，两位数以上就不对了。

真的有道理吗？  用 5楼的吧

[ 本帖最后由 BlueGuy 于 2009-10-10 11:43 编辑 ]