1096 Consecutive Factors (20point(s))

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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的范围，算法能够找到最长的连续因子序列，并在输入为630的情况下，输出了最大连续因子数量为3，具体因子序列为5*6*7。

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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.

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

思路：n不会被除自己以外大于根号n的数整除，所以遍历2 到根号n。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;

int main(int argc, char** argv) {
	long long  n;
	cin >> n;
	int maxlen = 0, maxi;
	long long sqr = (long long)sqrt(1.0 * n);
	for(int i = 2; i <= sqr; i++) {
		long long temp = i, j = i;//temp保存当前的乘积
		while(n % temp == 0) {//除得尽
			j++;
			temp = temp * j;
		}
		if(maxlen < j - i) {
			maxlen = j - i;
			maxi = i;
		}

	}
	if(maxlen == 0) {//如果最长长度没改变则这个数为本身
		cout << 1 << endl;
		cout << n << endl;
	} else {
		cout << maxlen << endl;
		cout << maxi;
		for(int i = maxi + 1; i < maxlen + maxi; i++) {
			cout << "*" << i;
		}
		cout << endl;
	}
	return 0;
}