Codeforces 566 F. Clique in the Divisibility Graph

Codeforces 566F 的传送门

As you must know, the maximum clique problem in an arbitrary graph is NP-hard. Nevertheless, for some graphs of specific kinds it can be solved effectively.

Just in case, let us remind you that a clique in a non-directed graph is a subset of the vertices of a graph, such that any two vertices of this subset are connected by an edge. In particular, an empty set of vertexes and a set consisting of a single vertex, are cliques.

Let’s define a divisibility graph for a set of positive integers A = {a1, a2, …, an} as follows. The vertices of the given graph are numbers from set A, and two numbers ai and aj (i ≠ j) are connected by an edge if and only if either ai is divisible by aj, or aj is divisible by ai.

You are given a set of non-negative integers A. Determine the size of a maximum clique in a divisibility graph for set A.

Input

The first line contains integer n (1 ≤ n ≤ 106), that sets the size of set A.

The second line contains n distinct positive integers a1, a2, …, an (1 ≤ ai ≤ 106) — elements of subset A. The numbers in the line follow in the ascending order.

Output

Print a single number — the maximum size of a clique in a divisibility graph for set A.

Sample test(s)

Input
8
3 4 6 8 10 18 21 24

Output
3

题目大意：，要求，给出一个数组序列，要求最长的成倍增长的序列如 3 6 18等。

解题思路：dp值，，，详见代码：

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
const int N = 1e6+10;
using namespace std;

int dp[N], a;

int main()
{
    int n, maxn;
    while(~scanf("%d", &n))
    {
        memset(dp, 0, sizeof(dp));
        maxn = 0;
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a);
            dp[a]++;
            maxn = max(maxn, dp[a]);
            for(int j = a * 2; j <= N; j += a)
                dp[j] = max(dp[j], dp[a]);
        }
        printf("%d\n", maxn);
    }
}