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HDU-1097,A hard puzzle(快速幂)

作者:互联网

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

 解题思路:

这道题就是求a的b次方的最后一位数字,典型的快速幂,因为是要求最后一位,所以在进行快速幂之前,我们可以对原数进行取个位数字之后再运算,因为决定最后一位数字的始终就是个位数字,也就是%10。 

程序代码: 

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int quickmul(int a,int b)
{
	int ans=1;
	while(b)
	{
		if(b&1)
			ans=(ans*a)%10;
		a=(a*a)%10;
		b>>=1;
	}
	return ans%10;
}
int main()
{
	int n,m;
	while(cin>>n>>m)
	{
		cout<<quickmul(n%10,m)<<endl;//关键n%10
	}
	return 0;
}

 

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标签:10,HDU,1097,int,puzzle,hard,ans,problem,include
来源: https://blog.csdn.net/weixin_43823808/article/details/104075568