hdu——1061Rightmost Digit（快速幂）

最新推荐文章于 2020-11-06 21:39:56 发布

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最新推荐文章于 2020-11-06 21:39:56 发布

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分类专栏：编程算法文章标签： c++ hdu 乘方运算

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本文介绍了一种使用快速幂算法解决特定数学问题的方法：给定一个正整数N，输出N^N的最右一位数字。通过快速幂算法高效地计算大数幂次方的个位数，避免了直接计算带来的溢出问题。

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Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 45812 Accepted Submission(s): 17250

Problem Description

Given a positive integer N, you should output the most right digit of N^N.

Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case contains a single positive integer N(1<=N<=1,000,000,000).

Output

For each test case, you should output the rightmost digit of N^N.

Sample Input

2
3
4

Sample Output

7
6

Hint

In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.
In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.

Author

Ignatius.L

Recommend

快速幂直接求就好了我有一篇快速幂的精简写法的博客可以看看

#include<iostream>
#include<string>
#include<cmath>
using namespace std;
long long kk(long long n)
{
	long long k=1,m=n;
	while(m!=0)
	{
		if(m&1)
		k*=n%10;
		m>>=1;
		n=n*n%10;
	}
	return k%10;
}
int main()
{
	long long n,m;
	cin>>n;
	while(n--)
	{
		cin>>m;
		cout<<kk(m)<<endl;
	}
	return 0;
}