codeforces 264b Good Sequences dp

最新推荐文章于 2019-07-06 15:20:52 发布

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最新推荐文章于 2019-07-06 15:20:52 发布

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分类专栏： acm_dp 文章标签： acm dp

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本博客探讨了一个关于序列的数学问题，目标是找到一个严格递增序列的最长长度，其中序列中任意相邻元素均不互为质数且属于一组预定义的好整数。通过输入好整数列表，读者将学会如何解决此问题。

点击打开链接题目链接

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
   
5
2 3 4 6 9

     output
   
4

     input
   
9
1 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.

极其坑爹的题目！！

题意看错想了两天！！

尼玛原来是输入的数据a(i)<a(i+1)啊！！

我以为是子序列的要求啊！！

还在想怎么这么多人ac了啊！！

好吧是我英语渣。。

好吧是我没审题。。

好吧就算是这样其实我一开始也是没想到做法。。

状态需要注意一下。。

贴代码：

#include<cstdio>
#include<cstring>
int dp[100100],mot[100100],a[100100];
int xx;
int maxx(int a,int b)
{
    return a>b?a:b;
}
void getmot(int x)
{
    xx=0;
    int i;
    for(i=2;i*i<=x;i++)
    {
        if(x%i==0)
        {
            while(x%i==0)
                x/=i;
            mot[++xx]=i;
        }
    }
    if(x>1)
        mot[++xx]=x;
}
int main()
{
    int n,i,j;
    int ans;
    int tmp;
    while(scanf("%d",&n)!=EOF)
    {
        ans=1;
        memset(dp,0,sizeof(dp));
        memset(mot,0,sizeof(mot));
        for(i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            getmot(a[i]);
            tmp=1;
            if(a[i]==mot[1])
            {
                dp[mot[1]]=1;
            }
            else
            {
                for(j=1;j<=xx;j++)
                {
                    tmp=maxx(tmp,dp[mot[j]]+1);
                }
            }
            for(j=1;j<=xx;j++)
            {
                dp[mot[j]]=maxx(tmp,dp[mot[j]]);
            }
            ans=maxx(ans,tmp);
        }
        printf("%d\n",ans);
    }
    return 0;
}