Codeforce 题目264B Good Sequences（DP，相邻两数不互质的LIS）

B. Good Sequences

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinks n integers a1, a2, ..., an are good.

Now she is interested in good sequences. A sequence x1, x2, ..., xk is called good if it satisfies the following three conditions:

• The sequence is strictly increasing, i.e. xi < xi + 1 for each i (1 ≤ i ≤ k - 1).
• No two adjacent elements are coprime, i.e. gcd(xi, xi + 1) > 1 for each i (1 ≤ i ≤ k - 1) (where gcd(p, q) denotes the greatest common divisor of the integers p and q).
• All elements of the sequence are good integers.

Find the length of the longest good sequence.

Input

The input consists of two lines. The first line contains a single integer n (1 ≤ n ≤ 105) — the number of good integers. The second line contains a single-space separated list of good integers a1, a2, ..., an in strictly increasing order (1 ≤ ai ≤ 105; ai < ai + 1).

Output

Print a single integer — the length of the longest good sequence.

Sample test(s)

input

52 3 4 6 9

output

4

input

91 2 3 5 6 7 8 9 10

output

4

Note

In the first example, the following sequences are examples of good sequences: [2; 4; 6; 9], [2; 4; 6], [3; 9], [6]. The length of the longest good sequence is 4.
ac代码

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<stdlib.h>#include<vector>#define maxn 100005using namespace std;vector<int>vc[maxn];void init(){    int i,j;    for(i=2;i<=maxn;i++)    {        for(j=i;j<=maxn;j+=i)            vc[j].push_back(i);    }}int a[maxn],dp[maxn];int main(){    init();    int n;    while(scanf("%d",&n)!=EOF)    {        int i;        memset(dp,0,sizeof(dp));        for(i=1;i<=n;i++)            scanf("%d",&a[i]);        vector<int>:: iterator it;        for(i=1;i<=n;i++)        {            int temp=0;            for(it=vc[a[i]].begin();it!=vc[a[i]].end();it++)            {                temp=max(temp,dp[*it]);            }            for(it=vc[a[i]].begin();it!=vc[a[i]].end();it++)            {                dp[*it]=temp+1;            }        }        int ans=1;        for(i=0;i<=maxn;i++)            ans=max(ans,dp[i]);        printf("%d\n",ans);    }}