快速幂取模

原理：

相当于还是用了状态压缩。

B - A hard puzzle

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

代码：

#include<bits/stdc++.h>
using namespace std;

int _pow(int a,int b)
{
    a%=10;
    int ans=1;

    while(b>0)
    {
        if(b&1) ans=ans*a%10;
        
        b>>=1;
        a=a*a%10;
    }
    return ans;
}

int main()
{
    int a,b;
    
    while(~scanf("%d%d",&a,&b))
    {
        printf("%d\n",_pow(a,b));
    }
    return 0;
}