HDU 1097 A hard puzzle

最新推荐文章于 2021-04-01 22:58:00 发布

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A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 24376 Accepted Submission(s): 8639

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

题目很简单　主要求Ａ^Ｂ的最后一位数值，找规律可找到方次再大，位值只有四个数值，用数组存起来，再对Ｂ％４找到数组中对应元素即可

#include<iostream>
using namespace std;
int main()
{
    long long a,b;
    long long i,s,c[4];
    while(cin>>a>>b)
    {
        for(i=s=1;i<=4;i++)
        {
             s=s*(a%10);
             s%=10;
             c[i%4]=s;
        }
        cout<<c[b%4]<<endl;
    }
    return 0;
}