Pat (advanced level) 1096 - continuous

Subject portal

• Data range use int that will do
• The factor is sqrt Level of complexity , Simple violence is enough
• Prime numbers require special treatment

I unconsciously wrote the code for finding the factor at the beginning （ Number of each Judge alone Is it divisible ）
But the title means , These series of factors need to be multiplied
That is, when recalculating the longest continuous factor sequence , It is necessary to accumulate the influence of various factors
a /= (cnt + i);

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I also take it for granted that , The continuity factor has such a property ： If tested $[a,b]$ This sequence , Then we just need to start from $b+1$ Just start the test
This is not at all wrong ！
So we have to be right from 2 To sqrt Each number of i, All calculated in i The longest sequence of continuous factors starting from

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
#include<cmath>
using namespace std;

int x;

int main()
{
    
	scanf("%d", &x);
	int cnt = 0;
	int ans_cnt = 0;
	int begin;
	for (int i = 2; i <= sqrt(x);++i) {
    
		if (x % i == 0) {
    
			int a = x / i;
			cnt = 1;
			while (a % (cnt + i) == 0) {
    
				a /= (cnt + i);
				cnt++;
			}
			if (cnt > ans_cnt) {
    
				ans_cnt = cnt;
				begin = i;
			}
		}
	}
	if (ans_cnt == 0) {
    
		printf("1\n%d",x);
	}
	else {
    
		printf("%d\n", ans_cnt); 
		for (int i = 0; i < ans_cnt; ++i) {
    
			printf("%d", begin + i);
			if (i < ans_cnt - 1) printf("*");
		}
	}
	return 0;
}