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=250005,M=550005;
char s[N];int n,last=1,cnt=1,cur=1,dis[M],ch[M][27],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]=1;
    else{
        int q=ch[p][c];
        if(dis[q]==dis[p]+1)fa[cur]=q;
        else{
            int nt=++cnt;
            dis[nt]=dis[p]+1;
            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]=1;
}
int sa[M];long long ans[N],c[N];
void Flr(){
    int p;
    for(int i=1;i<=cnt;i++)c[dis[i]]++;
    for(int i=1;i<=n;i++)c[i]+=c[i-1];
    for(int i=cnt;i>=1;i--)sa[c[dis[i]]--]=i;
    for(int i=cnt;i>=1;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+1);
    n=strlen(s+1);
    for(int i=1;i<=n;i++)build(i);
    Flr();
    for(int i=1;i<=n;i++)printf("%lld
",ans[i]);
}
int main()
{
    //freopen("pp.in","r",stdin);
    work();
    return 0;
}