1096 Consecutive Factors-PAT甲级

最新推荐文章于 2022-12-02 22:14:11 发布

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最新推荐文章于 2022-12-02 22:14:11 发布

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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，并列出最小的连续因子序列。通过遍历从2到根号N范围内的连续数，找到最长的连续因子序列。对于质数，直接输出其本身。

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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.

输入描述:

Each input file contains one test case, which gives the integer N (131).

输出描述:

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.

示例1

输入

输出

3
5*6*7

我采用的是暴力解法，遍历2-根号n范围内连续的数，截至根号n即可，后面不会有再比前面更多的连续数

第二个注意点在于质数，质数因子只有1和它本身，题目中明确1不在内，所以只需要输出n即可

满分代码如下：

#include<bits/stdc++.h>
using namespace std;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	int n;
	cin>>n;
	int max_num=0,sum=0,v,i=2;
	for(;i<=(int)(sqrt(n*1.0));i++){
		sum=0;
		if(n%i==0){
			for(int j=i;n%j==0;j*=(i+sum)){
				sum++;
			}
			if(sum>max_num){
				max_num=sum;
				v=i;
			}
		}	
	}
	if(max_num==0){
		cout<<1<<endl;
		cout<<n;
	}else{
		cout<<max_num<<endl;
		int flag=1;
		for(int i=v,j=1;j<=max_num;i++,j++){
			if(!flag){
				cout<<'*';
			}
			cout<<i;
			flag=0;
		}
	}
	
	return 0;
}