【PAT】【Advanced Level】1096. Consecutive Factors (20)

最新推荐文章于 2022-01-23 08:52:18 发布

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最新推荐文章于 2022-01-23 08:52:18 发布

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针对给定的正整数N，本篇博客介绍了一种高效算法来找出N的最大连续因子序列，并通过示例展示了如何使用该算法求解具体问题。

1096. Consecutive Factors (20)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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

原题链接：
https://www.patest.cn/contests/pat-a-practise/1096

https://www.nowcoder.com/pat/5/problem/4320

思路：

T1竟然卡了很久，把简单的问题复杂化了

因式分解，遇到一个因子进行检查、递增、统计个数即可

CODE：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
	int n;
	cin>>n;
	int st=0;
	int sum=0;
	for (int i=2;i<=sqrt(n)+1;i++)
	{
		if (n%i!=0) continue;
		///cout<<i<<endl;
		int num=0;
		int t=i;
		int op=n;
		while (op%t==0)
		{
			//cout<<op<<" "<<t<<endl;
			num++;
			op/=t;
			t++;
			//cout<<op<<" "<<t<<endl;
		}
		//cout<<num<<" "<<t<<endl;
		if (num>sum)
		{
			st=t-num;
			sum=num;
		}
	}
	if (sum==0)
	{
		cout<<"1"<<endl<<n;
	}
	else
	{
		cout<<sum<<endl;
		for(int i=st;i<st+sum;i++)
		{
			if (i!=st) cout<<"*";
			cout<<i;
		}
	}
	return 0;
}