杭电 A hard puzzle

最新推荐文章于 2024-08-03 08:00:00 发布

原创最新推荐文章于 2024-08-03 08:00:00 发布 · 287 阅读

CC 4.0 BY-SA版权

本文介绍了一种解决特定数学问题的方法：通过快速幂取模算法来计算 a 的 b 次方对 10 取余的结果。该算法特别适用于求解大整数幂运算的最后一位数字问题。文章提供了完整的 C++ 实现代码，并通过示例说明了如何使用该算法。

                                        A hard puzzle

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

提示

求a^b的最后一位数字

#include<iostream>
#include<cmath>
#include<cstdio>
using namespace std;

int pow(int a,int n,int b)//运用快速幂取模，a的n次方对b取余
{
    int result=1;
    a=a%b;
    while(n>0)
    {
        if(n%2==1)//如果n为奇数
        result=result*a%b;
        n=n/2;
        a=a*a%b;
    }
    return result;
}



int main()
{
    int a,b;
    while(cin>>a>>b)
    {
        cout<<pow(a,b,10)<<endl;
    }
}