[Nowcoder 213804] Another GCD problem

Another GCD question

Title link: nowcoder 213804

To Niu Ke:

——>Click me to jump<——

Topic

Give you an array and ask you to find 2 ∼ n 2\sim n from them$2\sim$The maximum value of$n$ gcds.

Ideas

Seeing this question, we consider how many numbers have xxThe factor of$x$ , let it be$f_x$。

Then we decompose the prime factors (use prime number screening and floor(sqrt()) to speed up), and then dfs, you can get each factor of a number, and then get $f_x$。

With this, we can ask to choose $What$ is the maximum gcd of the$y$ number?
(That is to search the firstfi f_ifrom back to$f_i$Greater than or equal to $y$ number)

Then we can first figure out which is just $y$ number, and then run the maximum value from back to front.

Then you can get the answer.

The specific implementation can look at the code.

Code

#include<cmath>
#include<cstdio>
#include<iostream>

using namespace std;

int n, a[100001], su[50001], all[100001], tmpn, tmp, ans[100001], maxn, zyz[1001], numm[1001];
bool susu[100100];

void getsusu() {
    
    //素数筛选法筛出素数
	susu[1] = susu[0] = 1;
	for (int i = 2; i <= maxn + 100; i++)
		if (!susu[i]) {
    
    
			su[++su[0]] = i;
			for (int j = i + i; j <= maxn + 100; j += i)
				susu[j] = 1;
		}
}

void dfs(int now, int sum) {
    
    
	if (now == zyz[0] + 1) {
    
    
		all[sum]++;
		return ;
	}
	int tmpp = 1;
	for (int i = 0; i <= numm[now]; i++) {
    
    //对于这一个因数，选多少个乘
		dfs(now + 1, sum * tmpp);
		tmpp *= zyz[now];
	}
}

int main() {
    
    
	scanf("%d", &n);
	
	for (int i = 1; i <= n; i++) {
    
    
		scanf("%d", &a[i]);
		maxn = max(maxn, a[i]);
		ans[i] = 1;
	}
	
	getsusu();
	
	for (int i = 1; i <= n; i++){
    
    
		zyz[0] = 0;//分解质因数
		tmpn = a[i];
		tmp = floor(sqrt(a[i]));
		for (int j = 1; su[j] <= tmpn && j <= tmp; j++) {
    
    
			if (tmpn % su[j] == 0) {
    
    
				zyz[++zyz[0]] = su[j];
				numm[zyz[0]] = 0;
				while (tmpn % su[j] == 0) {
    
    
					numm[zyz[0]]++;
					tmpn /= su[j];
				}
			}
		}
		
		if (tmpn != 1) {
    
    
			zyz[++zyz[0]] = tmpn;
			numm[zyz[0]] = 1;
		}
		
		dfs(1, 1);//标记每一个因数
	}
	
	for (int i = maxn; i >= 1; i--) {
    
    
		if (ans[all[i]] == 1) ans[all[i]] = i;
	}
	
	for (int i = n - 1; i >= 2; i--)//选 i 个也可以在 i+x 个中选 i 个，不一定要刚刚好
		ans[i] = max(ans[i + 1], ans[i]);
	
	for (int i = 2; i <= n; i++) printf("%d ", ans[i]);
	
	return 0;
}