Codeforces 264b Good Sequences dp

Codeforces 264b Good Sequences dp

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Good Sequencestime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard output

Squirrel Liss is interested in sequences. She also has preferences of integers. She thinksNIntegersA1 ,?A2 ,?...,?ANAre good.

Now she is interested in good sequences. A sequenceX1 ,?X2 ,?...,?XKIs called good if it satisfies the following three conditions:

• The sequence is strictly increasing, I. e.XI? XI? +? 1 for eachI(1? ≤?I? ≤?K? -? 1 ).
• No two adjacent elements are coprime, I. e.Gcd(XI,?XI? +? 1)?>? 1 for eachI(1? ≤?I? ≤?K? -? 1) (whereGcd(P,?Q) Denotes the greatest common divisor of the integersPAndQ).
• 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 integerN(1? ≤?N? ≤? 105)-the number of good integers. The second line contains a single-space separated list of good integersA1 ,?A2 ,?...,?ANIn 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.

  
  

  An extremely bad question !!

  The question was wrong for two days !!

  Nima was originally input data a (I)

  I thought it was a sub-sequence requirement !!

  Why are there so many ac users !!

  Well, it's my English scum ..

  Okay, I have not reviewed the question ..

  Well, even so, I didn't expect it at the beginning ..

  Status ..

  Code:

  #include
     
      #include
      
       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;}