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): 37341    Accepted Submission(s): 13361

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

7 668 800

 

Sample Output

96

#include<iostream>
#include<map>
#include<string>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<algorithm>
using namespace std;


int main()
{
    long long a,b;
    while(cin>>a>>b)
    {
        a%=10,b%=4;
        int ans=1;
        if(b==0)    / /要另外判断b是否为0

            b=4;
        for(int i=1;i<=b;i++)
        {
            ans=ans*a%10;
        }
        cout<<ans<<endl;
    }
    return 0;
}

注：

按常规的快速幂取模会超时

先打表找规律，发现：（每4个一循环）

a%10  b :1 2 3 4| 5 6 7 8

0               0 0 0 0| 0 0 0 0

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

5               5 5 5 5| 5 5 5 5

6               6 6 6 6| 6 6 6 6

7               7 9 3 1| 7 9 3 1

8               8 4 2 6| 8 4 2 6

9               9 1 9 1| 9 1 9 1

再利用快速幂取模来算

（新手上路，请大神指点~）