SPOJ NSUBSTR

You are given a string S which consists of 250000 lowercase latin letters at most. We define F(x) as the maximal number of times that some string with length x appears in S. For example for string 'ababa' F(3) will be 2 because there is a string 'aba' that occurs twice. Your task is to output F(i) for every i so that 1<=i<=|S|.

Input

String S consists of at most 250000 lowercase latin letters.

Output

Output |S| lines. On the i-th line output F(i).

Example

Input:
ababa
 
Output:
3
2
2
1
1
 
题解:
这..比上题还简单啊,根据parent树上父节点为子节点的最大子集,直接统计size即可,
英语不好看成了求和........3WA

 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstring>
 #include <cstdio>
 #include <cmath>
 using namespace std;
 const int N=,M=;
 char s[N];int n,last=,cnt=,cur=,dis[M],ch[M][],fa[M],size[M];
 void build(int j){
     int c=s[j]-'a';
     last=cur;cur=++cnt;
     int p=last;dis[cur]=j;
     for(;p && !ch[p][c];p=fa[p])ch[p][c]=cur;
     if(!p)fa[cur]=;
     else{
         int q=ch[p][c];
         if(dis[q]==dis[p]+)fa[cur]=q;
         else{
             int nt=++cnt;
             dis[nt]=dis[p]+;
             memcpy(ch[nt],ch[q],sizeof(ch[q]));
             fa[nt]=fa[q];fa[q]=fa[cur]=nt;
             for(;ch[p][c]==q;p=fa[p])ch[p][c]=nt;
         }
     }
     size[cur]=;
 }
 int sa[M];long long ans[N],c[N];
 void Flr(){
     int p;
     for(int i=;i<=cnt;i++)c[dis[i]]++;
     for(int i=;i<=n;i++)c[i]+=c[i-];
     for(int i=cnt;i>=;i--)sa[c[dis[i]]--]=i;
     for(int i=cnt;i>=;i--){
         p=sa[i];
         if(size[p]>ans[dis[p]])ans[dis[p]]=size[p];
         size[fa[p]]+=size[p];
     }
 }
 void work(){
     scanf("%s",s+);
     n=strlen(s+);
     for(int i=;i<=n;i++)build(i);
     Flr();
     for(int i=;i<=n;i++)printf("%lld\n",ans[i]);
 }
 int main()
 {
     //freopen("pp.in","r",stdin);
     work();
     return ;
 }