A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 36871    Accepted Submission(s): 13154

Problem Description

lcy gives a hard puzzle to feng5166,lwg,JGShining and Ignatius: gave a and b,how to know the a^b.everybody objects to this BT problem,so lcy makes the problem easier than begin.

this puzzle describes that: gave a and b,how to know the a^b's the last digit number.But everybody is too lazy to slove this problem,so they remit to you who is wise.

 

 

Input

There are mutiple test cases. Each test cases consists of two numbers a and b(0<a,b<=2^30)

 

 

Output

For each test case, you should output the a^b's last digit number.

 

 

Sample Input

7 66 8 800

 

 

Sample Output

9 6

 

今天第一次写博客，好激动啊！

这道题就是求a^b的最后一位是什么，懒得用循环求解，就用打表写了这道题；

规律很好总结，很水的题~~~~~~~~~~~

#include
     
      #include
      
       int main(){    __int64 l,a,b,len;    char num[40];    int u[9][4]={
   {
   1},{
   6,2,4,8},{
   1,3,9,7},{
   6,4},{
   5},{
   6},{
   1,7,9,3},{
   6,8,4,2},{
   1,9}};    while(scanf("%s%I64d",num,&b)!=EOF)    {        len=strlen(num);        a=num[len-1]-'0';        if(a==0&&b==0) continue;        if(b==0) {            printf("1\n");            continue;        }        switch(a)        {            case 1: l=1; break;            case 2: l=u[1][b%4]; break;            case 3: l=u[2][b%4]; break;            case 4: l=u[3][b%2]; break;            case 5: l=5; break;            case 6: l=6; break;            case 7: l=u[6][b%4]; break;            case 8: l=u[7][b%4]; break;            case 9: l=u[8][b%2]; break;            case 0: l=0;        }        printf("%I64d\n",l);    }    return 0;}