Codeforces835D Palindromic characteristics_codeforces 835 d. palindromic characteristics-优快云博客

本文介绍了一种算法，用于解决给定字符串中不同长度的K回文子串的数量计算问题，通过区间动态规划的方法高效地实现了这一目标。

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.

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题目的意思是给出一个字符串，求各个k回文的个数。k回文的定义，本身是回文且左一半是k-1回文

思路：区间dp，dp[i][j]表示区间[i,j]的回文等级，具体看代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;
#define mod 10000007
#define mem(a,b) memset(a,b,sizeof a)

int dp[5005][5005];
char s[5005];
int ans[50005];
int main()
{
    while(~scanf("%s",s))
    {
        int n=strlen(s);
        mem(dp,0);
        mem(ans,0);
        for(int i=0; i<n; i++)
        {
            ans[1]++;
            dp[i][i]=1;
        }
        for(int i=1; i<n; i++)
        {
            if(s[i-1]==s[i])
                dp[i-1][i]=2,ans[1]++,ans[2]++;
        }
       for(int j=2; j<n; j++)
            for(int i=0; i<n; i++)
            {
                if(i+j>=n)
                    break;
                if(s[i]==s[i+j]&&dp[i+1][i+j-1])
                {
                    dp[i][i+j]=dp[i][i+(j+1)/2-1]+1;
                    for(int q=1; q<=dp[i][i+j]; q++)
                    {
                        ans[q]++;
                    }
                }
            }
        for(int i=1; i<n; i++)
            printf("%d ",ans[i]);
            printf("%d\n",ans[n]);

    }

    return 0;
}