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
没什么可说的
1 #include<iostream> 2 #include<cstring> 3 #include<cstdlib> 4 #include<cstdio> 5 #include<cmath> 6 #include<algorithm> 7 #define maxn 500005 8 using namespace std; 9 char s[maxn]; 10 int len; 11 struct data { 12 int son[maxn][26],link[maxn],step[maxn],last,cnt,size[maxn]; 13 int v[maxn],pos[maxn],f[maxn]; 14 data() {last=cnt=1;} 15 void extend(int c) { 16 int p=last,np=last=++cnt;step[np]=step[p]+1;size[np]=1; 17 while(p&&!son[p][c]) son[p][c]=np,p=link[p]; 18 if(!p) link[np]=1; 19 else { 20 int q=son[p][c]; 21 if(step[q]==step[p]+1) link[np]=q; 22 else { 23 int nq=++cnt; 24 memcpy(son[nq],son[q],sizeof(son[q])); 25 link[nq]=link[q]; 26 link[q]=link[np]=nq; 27 step[nq]=step[p]+1; 28 while(p&&son[p][c]==q) son[p][c]=nq,p=link[p]; 29 } 30 } 31 } 32 void pre() { 33 for(int i=1;i<=cnt;i++) v[step[i]]++; 34 for(int i=1;i<=cnt;i++) v[i]+=v[i-1]; 35 for(int i=cnt;i;i--) pos[v[step[i]]--]=i; 36 for(int i=cnt;i;i--) size[link[pos[i]]]+=size[pos[i]]; 37 for(int i=1;i<=cnt;i++) f[step[i]]=max(f[step[i]],size[i]); 38 for(int i=len;i;i--) f[i]=max(f[i],f[i+1]); 39 for(int i=1;i<=len;i++) printf("%d\n",f[i]); 40 } 41 }sam; 42 int main() { 43 scanf("%s",s+1); 44 len=strlen(s+1); 45 for(int i=1;i<=len;i++) sam.extend(s[i]-‘a‘); 46 sam.pre(); 47 return 0; 48 }
