PTA L1-006 最小连续因子

原题链接

https://pintia.cn/problem-sets/994805046380707840/problems/994805138600869888

N有两种情况 

1. 素数 输出1 和 N

2. 合数 因为我们求的是连续因子,不需要求求全部因子 ,所以只需要保证我们所求的因子的乘积仍然是N的因子即可

 代码 

#include <iostream>
#include<cmath>
using namespace std;

int main() {
	int num;
	while (cin >> num)
	{

		int start = 0, size = 0;
		for (int i = 2, j; i <= sqrt(double(num)) + 1; i++)
			if (num % i == 0) {
				int sum = i;
				for (j = i + 1; j <= sqrt(double(num))+1; j++)
				{
					sum *= j;
					if (num % j == 0 && !(num % sum))
					{
						continue;

					}
					else
						break;

				}
				if (size < (j - i))
				{
					start = i;
					size = j - i;

				}
			}

		
		if (size > 0)
		{
			cout << size << endl;
			for (int i = start; i < start + size; i++)
			{
				if (i != start)
					cout << "*";
				cout << i;
			}
		}
			
		else
			cout <<'1'<<endl<< num;

	}
}

感谢这篇文章

https://blog.csdn.net/qq_31359295/article/details/51598548