codeforces.contest/835/problem/D Palindromic characteristics （记忆化搜索）

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Palindromic characteristics of string s with length |s| is a sequence of |s| integers, where k-th number is the total number of non-empty substrings of s which are k-palindromes.

A string is 1-palindrome if and only if it reads the same backward as forward.

A string is k-palindrome (k > 1) if and only if:

Its left half equals to its right half.
Its left and right halfs are non-empty (k - 1)-palindromes.

The left half of string t is its prefix of length ⌊|t| / 2⌋, and right half — the suffix of the same length. ⌊|t| / 2⌋ denotes the length of string tdivided by 2, rounded down.

Note that each substring is counted as many times as it appears in the string. For example, in the string "aaa" the substring "a" appears 3 times.

Input

The first line contains the string s (1 ≤ |s| ≤ 5000) consisting of lowercase English letters.

Output

Print |s| integers — palindromic characteristics of string s.

Examples
input
abba
output
6 1 0 0 
input
abacaba
output
12 4 1 0 0 0 0 

Note

In the first example 1-palindromes are substring «a», «b», «b», «a», «bb», «abba», the substring «bb» is 2-palindrome. There are no 3- and 4-palindromes here.

一开始用STL T了，这题带有结构性质，所以可以用记忆化搜索，预处理任意两个区间是否是合法（双向搜索），想不到这个预处理就很伤

#include<iostream>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<string>#include<vector>#include<cmath>#include<queue>#include <bits/stdc++.h>using namespace std;const int N = 5000+7;typedef  long long LL;const LL mod = 1e9+7;char str[N];int a[N][N], dp[N][N], ans[N];int dfs(int l,int r){    if(l>r) return 0;    int len=(r-l+1)/2;    if(dp[l][r]!=-1) return dp[l][r];    if(!a[l][r]) return dp[l][r]=0;    return  dp[l][r]=dfs(l,l+len-1)+1;}int main(){    scanf("%s",str);    int len=strlen(str);    memset(dp,-1,sizeof(dp));    memset(a,0,sizeof(a));    memset(ans,0,sizeof(ans));    for(int i=0;i<len;i++)    {        for(int j=i, k=i;j>=0&&k<len;k++,j--)        {            if(str[j]==str[k]) a[j][k]=1;            else break;        }        for(int j=i, k=i+1;j>=0&&k<len;k++,j--)        {            if(str[j]==str[k]) a[j][k]=1;            else break;        }    }    for(int i=0;i<len;i++)    {        for(int j=i;j<len;j++)        {            dp[i][j]=dfs(i,j);            ans[dp[i][j]]++;        }    }    for(int i=len;i>=1;i--)   ans[i]+=ans[i+1];    for(int i=1;i<=len;i++)  printf("%d%c",ans[i],i==len?'\n':' ');    return 0;}

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