A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24658 Accepted Submission(s): 8778

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

取a的最后一位0-9，发现乘方个位数有周期关系（可以证明不大于4），找到周期和周期里对应的数，对于b直接查找即可

#include<stdio.h>
int T[5];
int main()
{
    int a,b,i,j,temp,t,d;
    while(~scanf("%d%d",&a,&b))
    {
        t = 0;
        temp = 1;
        a %= 10;
        if(a==1)
        {
            printf("1
");
            continue ;
        }
        d = a;
        while(temp != a)
        {
            T[t++] = d;
            temp = (d*a)%10;
            d = temp;
        }
        t = (b%t==0?t-1:b%t-1);
        printf("%d
",T[t]);
    }
    return 0;
}