HDU 1097 A hard puzzle（快速幂取模）

键盘上的精灵

于 2016-05-24 21:08:46 发布

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分类专栏：数学问题文章标签： HDU解题报告快速幂取模 ACM

本文链接：https://blog.youkuaiyun.com/yangkunpengD/article/details/51493584

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本文介绍了一种快速计算 a 的 b 次方后所得结果的末位数字的算法。该算法通过优化循环的方式降低了计算复杂度，并提供了一个 C++ 实现示例。

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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

题意：求a的b次方的最后一位。

#include<iostream>
using namespace std;
long long pow_mod(long long a,long long b)
{
    long long ans=1;
    a=a%10;
    while(b>0)
    {
        if(b%2==1)
            ans=(ans*a)%10;
        b=b/2;
        a=(a*a)%10;
    }
    return ans;
}
int main()
{
    long long a,b;
    while(cin>>a>>b)
    {
        cout<<pow_mod(a,b)<<endl;
    }
}