CF1614D1 Divan and Kostomuksha (easy version)

Divan and Kostomuksha (easy version)

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题面翻译

给定一个序列 $a$ ，定义其权值为：

$${\Sigma }_{i=1}^n\gcd (a_1,a_2,...,a_i)$$

现在你可以重排 $a$ ，求这个权值的最大值。

本题与同场 D2 的唯一差别在于 $a_i$ 的范围。

题目描述

This is the easy version of the problem. The only difference is maximum value of $a_i$.

Once in Kostomuksha Divan found an array $a$ consisting of positive integers. Now he wants to reorder the elements of $a$ to maximize the value of the following function:
$${\Sigma }_{i=1}^n\gcd (a_1,a_2,\dots ,a_i),$$
where $\gcd (x_1,x_2,\dots ,x_k)$ denotes the greatest common divisor of integers $x_1,x_2,\dots ,x_k$ , and $\gcd (x)=x$ for any integer $x$.

Reordering elements of an array means changing the order of elements in the array arbitrary, or leaving the initial order.

Of course, Divan can solve this problem. However, he found it interesting, so he decided to share it with you.

输入格式

The first line contains a single integer $n$ ( $1\le n\le 10^5$ ) — the size of the array $a$ .

The second line contains $n$ integers $a_1,a_2,\dots ,a_n$ ( $1\le a_i\le 5\cdot 10^6$ ) — the array $a$ .

输出格式

Output the maximum value of the function that you can get by reordering elements of the array $a$ .

样例 #1

样例输入 #1

6
2 3 1 2 6 2

样例输出 #1

14

样例 #2

样例输入 #2

10
5 7 10 3 1 10 100 3 42 54

样例输出 #2

131

提示

In the first example, it’s optimal to rearrange the elements of the given array in the following order: $[6,2,2,2,3,1]$ :

$$\begin{gathered} \gcd (a_1)+\gcd (a_1,a_2)+\gcd (a_1,a_2,a_3)+\gcd (a_1,a_2,a_3,a_4) \\ +\gcd (a_1,a_2,a_3,a_4,a_5)+\gcd (a_1,a_2,a_3,a_4,a_5,a_6) \\ =6+2+2+2+1+1=14. \end{gathered}$$
It can be shown that it is impossible to get a better answer.

In the second example, it’s optimal to rearrange the elements of a given array in the following order: $[100,10,10,5,1,3,3,7,42,54]$.

$以上来自洛谷$

解题思路

设重排 $\gcd (a_1,a_2,\dots ,a_i)=su{m}_i$ ，有一条很重要的性质：
$$su{m}_{i+1}\mid su{m}_i$$
根据这一条我们这样写出方程，设 $f_i$ 表示目前到某个 $su{m}_j$，$su{m}_j=i$时，${\Sigma }_{k=1}^{j}su{m}_k$的最大值。不难得到：
$$f_i=\max (f_j+i\times (cn{t}_i-cn{t}_j))$$
其中 $cn{t}_i$ 表示 $i$ 出现的次数。
时间复杂度为 $O(n+p\log \log p)$，$p$ 为值域。

AC Code

#include 
using namespace std;
#define int long long
const int Maxn = 3e5 + 5, Maxv = 5e6 + 5;
int n, a[Maxn];
int maxx, f[Maxv], cnt[Maxv], tot[Maxv];
inline void work() {
	maxx = INT_MIN;
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
	}
	for (int i = 1; i <= n; i++) {
		maxx = max(maxx, a[i]);
		tot[a[i]] += 1;
	}
	cnt[1] = n;
	for (int i = 2; i <= maxx; i++) {
		for (int j = i; j <= maxx; j += i) {
			cnt[i] += tot[j];
		}
	}
	for (int i = maxx; i >= 1; i--) {
		f[i] = cnt[i] * i;
		for (int j = i + i; j <= maxx; j += i) {
			f[i] = max(f[j] + (cnt[i] - cnt[j]) * i, f[i]);
		}
	}
	cout << f[1] << endl;
}
signed main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	work();
	return 0;
}

写完了，去看看进阶版吧。