1096. Consecutive Factors (20)

最新推荐文章于 2021-07-31 14:30:39 发布

kakitgogogo

最新推荐文章于 2021-07-31 14:30:39 发布

阅读量541

点赞数

CC 4.0 BY-SA版权

分类专栏： pat-advanced 文章标签： math

本文链接：https://blog.youkuaiyun.com/kakitgogogo/article/details/52065529

pat-advanced 专栏收录该内容

115 篇文章

订阅专栏

本文介绍了一种算法，用于找出正整数N的最大连续因子序列，并确保这些因子的数量最多且尽量小。输入一个正整数N，算法将输出连续因子的最大数量及这些连续因子的具体序列。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Among all the factors of a positive integer N, there may exist several consecutive numbers. For example, 630 can be factored as 3*5*6*7, where 5, 6, and 7 are the three consecutive numbers. Now given any positive N, you are supposed to find the maximum number of consecutive factors, and list the smallest sequence of the consecutive factors.

Input Specification:

Each input file contains one test case, which gives the integer N (1<N<2³¹).

Output Specification:

For each test case, print in the first line the maximum number of consecutive factors. Then in the second line, print the smallest sequence of the consecutive factors in the format "factor[1]*factor[2]*...*factor[k]", where the factors are listed in increasing order, and 1 is NOT included.

Sample Input:

Sample Output:

3
5*6*7

求一个数n连续的因子，要求使得因子数最多且使因子尽量小。2的31次方等于2 147 483 648，而12的阶乘为479 001 600，13的阶乘是6 227 020 800，故最大的连续因子数为12，因此可以暴力破之。需要注意的是连续因子数小于2时，即没有因子是连续的情况，如果数n是合数就输出最小的因子；如果n是素数就输出该数（因为1不能是答案）。

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
	long long n;
	cin>>n;
	
	for(int num=11;num>1;num--)
	{
		for(int i=2;i+num<=13;i++)
		{
			long long m=1;
			for(int j=i;j<i+num;j++)
			{
				m*=j;
			}
			if(n%m==0)
			{
				printf("%d\n%d",num,i);
				for(int j=i+1;j<i+num;j++)
				{
					printf("*%d",j);
				}
				return 0;
			}
		}
	}
	for(int i=2;i<=sqrt(n);i++)
	{
		if(n%i==0) 
		{
			printf("1\n%d",i);
			return 0;
		}
	}
	printf("1\n%d",n);
}