Codeforces835D-Palindromic characteristics(DP)

Palindromic characteristics

time limit per test

 3 seconds

memory limit per test

 256 megabytes

input

 standard input

output

 standard output

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:

1. Its left half equals to its right half.
2. 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.

题意：

给出一个字符串，在其中找到1..k阶的回文子串并统计它们的数量。如果一个字符串是一个回文串,则它可以是1阶子串（k阶字符串,要求它的左边和右边都是k-1阶子串）

解题思路：

dp[i][j]用来记录这段字符串是几阶子串即可

#include<bits/stdc++.h>using namespace std;const int maxn = 5005;int dp[maxn][maxn], ans[maxn];char s[maxn];int main(){    while(scanf("%s", s+1) != EOF)    {        memset(dp, 0, sizeof(dp));        memset(ans, 0, sizeof(ans));        int len = strlen(s+1);        for(int i = 1; i <= len; i++)        {            dp[i][i] = 1;            ans[1]++;            if(i < len && s[i] == s[i+1])            {                dp[i][i+1] = 2;                ans[2]++;ans[1]++;            }        }        for(int j = 3; j <= len; j++)            for(int i = 1; i+j-1 <= len; i++)                if(dp[i+1][i+j-2] && s[i] == s[i+j-1])                {                    dp[i][i+j-1] = dp[i][i+j/2-1] + 1;                    for(int k = 1; k <= dp[i][i+j-1]; k++)                        ans[k]++;                }        for(int i = 1; i <= len; i++)            printf("%d%c", ans[i], i == len ? '\n' : ' ');    }    return 0;}