1097 A hard puzzle

最新推荐文章于 2021-04-01 22:58:00 发布

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最新推荐文章于 2021-04-01 22:58:00 发布

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分类专栏： HDU基础

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本文提供了一种高效的算法来计算任意整数a的b次幂的最后一位数字，包括如何处理大数乘法可能引起的溢出问题。

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A hard puzzle

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 30644 Accepted Submission(s): 10999

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

注意溢出情况。当最大值乘最大值出现溢出

#include<cstdio>
#include<iostream>
using namespace std;
int main()
{
    int y,a[5],n,m;
    while(scanf("%d%d",&n,&m)!=EOF){
        a[1]=n%10;
        for(int i=2; i <= 4;i++)a[i]=a[1]*a[i-1]%10;               
        int x=m%4;
        if(x==0)y=a[4];
        else
            y=a[x];
        printf("%d\n",y);
    }
    return 0;
}