1096. Consecutive Factors

最新推荐文章于 2021-07-31 14:30:39 发布

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最新推荐文章于 2021-07-31 14:30:39 发布

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分类专栏： PAT A C语言实现

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针对给定的正整数N，本程序旨在找出其最大的连续因子数量，并列出这些连续因子的最小序列。输入一个正整数N（1<N<2^31），程序将输出连续因子的最大数量及对应的连续因子序列。

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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

#include<stdio.h>
#include<vector>
#include<math.h>
using namespace std;

int main()
{
	int n = 0;
	scanf("%d", &n);
	vector<int> v;
	if(n == 2 || n == 3 || n == 5)
	{
		printf("1\n%d", n);
		return 0;
	}
	//case6：超时在于计算整除的时候，判断条件，如果是i*i<=n，这样会超时的 
	int num = (int)sqrt(n);
	for(int i = 2; i <= num; i ++)
		if(n % i == 0)
			v.push_back(i);
	v.push_back(n);
	int maxLen = 0, maxFactor = n;
	int tmpLen = 0, tmpFactor = 0;
	
	for(int i = 0; i < v.size(); i ++)
	{
		num = n;
		tmpFactor = v[i];
		tmpLen = 0;
		
		while(num%tmpFactor == 0)
		{
			tmpLen ++;
			num /= tmpFactor;
			tmpFactor ++;
		}
		
		if(tmpLen > maxLen)
		{
			maxLen = tmpLen;
			maxFactor = v[i];
		}
	}
	
	printf("%d\n", maxLen);
	for(int i = 0; i < maxLen; i ++)
	{
		if(i)
			printf("*%d", maxFactor);
		else
			printf("%d", maxFactor);
		maxFactor++;
	} 
	printf("\n");
	
	return 0;
}