HDU 1097 A hard puzzle

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13563 Accepted Submission(s): 4724

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

Author

eddy

Recommend

JGShining

 

又是一个找规律的题目：

当     a    1   2  3  4  5  6  7  8

        1    1   1  1  1  1  1  1  1

　　  2    2   4  8  6  2  4  8  6

        3    3   9  7  1  3  9  7  1

        4    4   6  4  6  4  6  4  6

　　   ...........   每一个数都有以4为周期循环的规律。

找到规律就简单了 ：

#include<stdio.h>
#include<stdlib.h>
int main ()
{
    long long  last[5] , a , b , sum ;
    while ( scanf ( "%I64d%I64d" , &a , &b ) != EOF )
    {
          sum = a ;
          for ( int i = 1 ; i < 5 ; ++ i )
          {
              sum %= 10 ;
              last[i] = sum % 10 ;
              sum *= a ;    
          }    
          last[0] = last[4] ;
          printf ( "%I64d\n" , last[b % 4] ) ;
    }
    return 0 ;
}