Codeforces 264B Good Sequences【dp+思维】

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 thinksn integers a₁, a₂, ..., a_n aregood.

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

• The sequence is strictly increasing, i.e. x_i < x_i + 1 for eachi (1 ≤ i ≤ k - 1).
• No two adjacent elements are coprime, i.e. gcd(x_i, x_i + 1) > 1 for eachi (1 ≤ i ≤ k - 1) (wheregcd(p, q) denotes the greatest common divisor of the integersp 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 integersa₁, a₂, ..., a_n in strictly increasing order (1 ≤ a_i ≤ 10⁵; a_i < a_i + 1).

Output

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

Examples

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.

题目大意：

给你一个长度为N的序列，让你找到其中一个严格递增的子序列，使得其长度最长，并且按序相邻的两个元素不互质。（本身序列就是严格递增的）；

思路：

1、思路来源：http://blog.csdn.net/joy_go/article/details/8534940

2、蛮巧妙的一个做法：

①设定dp【i】表示以因子i结尾的最长子序列长度。

②预处理出N个数的所有因子，存入vector中（不处理1）。

③然后O（n）扫这个序列，当扫到第i个数的时候，对应将第i个数的因子都取出来，dp【因子】++，表示以这个因子作为结尾的子序列长度增加了1.然后在这个过程中维护一个最大值，设定为maxn；

④然后再对应第i个数的因子再取出来一遍，对应dp【因子】=maxn，表示现在以第i个数结尾的同时，以任何因子结尾的长度都是当前这个maxn长度。然后进行下一位的处理。

⑤那么此时我们的解就是max（dp【i】）【2<=i<=a【n】】（因为a【n】一定是序列中最大的数，所以最大的因子也就是a【n】）

Ac代码：

#include<stdio.h>#include<string.h>#include<vector>#include<math.h>using namespace std;vector<int >yinzi[100040];int dp[100040];int a[100040];int main(){    int n;    while(~scanf("%d",&n))    {        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++)yinzi[i].clear();        for(int i=1;i<=n;i++)        {            scanf("%d",&a[i]);            yinzi[i].push_back(a[i]);            for(int j=2;j<=(int)(sqrt(a[i]));j++)            {                if(a[i]%j==0)                {                    if(j*j==a[i])                    {                        yinzi[i].push_back(j);                        continue;                    }                    yinzi[i].push_back(j);                    yinzi[i].push_back(a[i]/j);                }            }        }        for(int i=1;i<=n;i++)        {            int maxn=0;            for(int j=0;j<yinzi[i].size();j++)            {                dp[yinzi[i][j]]++;                maxn=max(maxn,dp[yinzi[i][j]]);            }            for(int j=0;j<yinzi[i].size();j++)            {                dp[yinzi[i][j]]=maxn;            }        }        int output=0;        for(int i=1;i<=a[n];i++)        {            output=max(dp[i],output);        }        printf("%d\n",output);    }}