HDU1097

A hard puzzle

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19131    Accepted Submission(s): 6844

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

求a的b次方的最后一个数字。

数论公式，b = b%4+4。直接暴力。不需要快速幂。

# include <stdio.h>

int qpow(int a, int b)

{

    int i, mul = 1 ;

    for (i = 0 ; i < b ; i++)

        mul = mul * a % 10 ;

    return mul ;

}

int main ()

{

    int a, b ;

    while (~scanf ("%d%d", &a, &b))

    {

        if (b > 4) b = b%4 + 4;

        a%= 10 ;

        printf ("%d\n", qpow(a,b)) ;

    }

    return 0 ;

}