Meaning of the questions:
To give you a length \ (n-\) string length is now ask you \ (1,2 \ dots n \) of the substring, the number of substrings \ (STR \) is a palindromic sequence, and it to shorten the length of $ \ lceil \ frac {| str |} {2} \ $ rceil the palindromic sequence also.
analysis:
Obviously, we can use the tree are palindromic \ (CNT \) arrays and \ (Fail \) array, an arbitrary length of the \ (len \) palindromic substrings and its longest string suffix palindromic \ (\ mathcal {O} ( n) \) time out process. Know how to find things after the above, we only need to consider how to eliminate some of the illegal palindrome substring.
For a palindromic sequence \ (STR \) , he wants him corresponds to the first half palindrome half palindrome, the palindrome then if his longest suffix length less than \ (\ frac {| str | } {2 } \) , it is clearly not, therefore be considered equal to a length greater than the longest palindromic suffix \ (\ frac {|} { 2} \) | str case. We found that if you want it after half palindrome, it is to shorten delete a number of original series and the longest palindrome suffix poor, so we only need to determine $ \ lfloor \ frac {| str |} {2} \ rfloor $ whether divisible \ ((| str | - | str_ {maxsuffix} |) \) to
Code:
#include <bits/stdc++.h>
#define maxn 300005
using namespace std;
char str[maxn];
typedef long long ll;
struct PAM{//回文树
int next[maxn][26],fail[maxn],len[maxn],cnt[maxn],S[maxn];
int id,n,last;
int newnode(int x){
for(int i=0;i<26;i++){
next[id][i]=0;
}
cnt[id]=0;
len[id]=x;
return id++;
}
void init(){
id=0;
newnode(0);
newnode(-1);
fail[0]=1;
S[0]=-1;
last=n=0;
}
int getfail(int x){
while(S[n-len[x]-1]!=S[n]) x=fail[x];
return x;
}
void Insert(int c){
c-='a';
S[++n]=c;
int cur=getfail(last);
if(!next[cur][c]){
int now=newnode(len[cur]+2);
fail[now]=next[getfail(fail[cur])][c];
next[cur][c]=now;
}
last=next[cur][c];
cnt[last]++;
}
void getsum(){//自下向上更新
for(int i=id-1;i>=0;i--){
cnt[fail[i]]+=cnt[i];
}
}
}pam;
int Ans[maxn];
bool judge(int x,int Len){
int tmp1=pam.len[pam.fail[x]];
int tmp2=pam.len[x]-tmp1;
if((pam.len[x]/2)%tmp2==0) return true;
else return false;
}
int main()
{
while(~scanf("%s",str)){
pam.init();
int len=strlen(str);
for(int i=0;i<=len;i++){
Ans[i]=0;
}
for(int i=0;i<len;i++){
pam.Insert(str[i]);
}
pam.getsum();
ll res=0;
for(int i=2;i<pam.id;i++){
if(judge(i,(pam.len[i]+1)/2)){
Ans[pam.len[i]]+=pam.cnt[i];
}
}
for(int i=1;i<=len;i++){
if(i==1) printf("%d",Ans[i]);
else printf(" %d",Ans[i]);
}
puts("");
}
return 0;
}