CF Clique in the Divisibility Graph （DP）

题目链接

As you must know, the maximum clique problem in an arbitrary graph isNP-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 integersA = {a₁, a₂, ..., a_n} as follows. The vertices of the given graph are numbers from set A, and two numbersa_i anda_j (i ≠ j) are connected by an edge if and only if either a_i is divisible bya_j, ora_j is divisible bya_i.

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

Input

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

The second line contains n distinct positive integersa₁, a₂, ..., a_n (1 ≤ a_i ≤ 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 setA.

Sample test(s)

Input

83 4 6 8 10 18 21 24

Output

3

Note

In the first sample test a clique of size 3 is, for example, a subset of vertexes{3, 6, 18}. A clique of a larger size doesn't exist in this graph.

题目大意：给你n个递增的数，叫你选择一个最大的集合，使得在这个集合内任意选出两个数，大的那个数能被小的那个数整除，问你这个集合最大为多大。

分析：由于每个数范围为10^6，我们可以暴力找出能被当前数整除的数，并记录个数即可。

#include<stdio.h>#include<string.h>#include<algorithm>#define maxn 1000002using namespace std;int cnt[maxn];int a[maxn];int main(){    int n,N=-1;    scanf("%d",&n);    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);        N=max(N,a[i]);    }    memset(cnt,0,sizeof(cnt));    int ans=1;    for(int i=1;i<=n;i++)    {        cnt[a[i]]++;//代表当前这个数也满足条件，能被自己整除        ans=max(ans,cnt[a[i]]);        for(int j=a[i]<<1;j<=N;j+=a[i])//暴力找出能被这个数整除的数，并记录个数            cnt[j]=max(cnt[j],cnt[a[i]]);    }    printf("%d\n",ans);    return 0;}