PAT (Advanced Level) Practice 1096 Consecutive Factors (20分)

1.题目

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<2​31​​).

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:

630

Sample Output:

3
5*6*7

2.题目分析

先判断是不是素数，是的话就输出个数为1，因子为素数本身，不是的话就从2开始2 3 4……判断是否能被n取余得0，到不能为止记录个数与位置，再从3开始，4开始……，当相乘的总因子>n/2时退出循环。

3.代码

#include<iostream>
#include<cmath>
using namespace std;
bool isprime(int x)
{
	if (x == 0 || x == 1)return false;
	for (int i = 2; i <= sqrt(x); i++)
		if (x%i == 0)return false;
	return true;
}

int main()
{
	int n, max = -1, maxi;
	scanf("%d", &n);
	int start = 2,mul=2;
	int count = 0;
    if(isprime(n)){printf("1\n");printf("%d",n);return 0;}
	while (1)
	{
		 count = 0;
		int temp = start;
		int temp2 = mul;
		while (n%temp2 == 0)
		{
			count++;
			temp2 = temp2*(temp + 1); temp++;
		}
		if (max < count-1) { max = count-1; maxi = temp-1; }
		start++; mul++;
		if (temp2 > n/2)break;
	}
	printf("%d\n", max+1);
	for (int i = maxi-count; i<maxi; i++)
	{
		printf("%s%d", i == maxi - count ? "" : "*",  i+1);
	}
}