D. 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:

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 t divided 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.

回文串

递归解法

#include<iostream>

#include<cstdio>

#include<queue>

#include<algorithm>

#include<cstring>

#include<string>

using namespace std;

typedef long long LL;

#define MAXN 5050

#define N 100

/*

1阶回文串就是左右两边读相等

k 回文串就是从中间分开 左右两边都是k-1阶回文串

预处理a[i][j]确定i 到j 是不是回文串

然后递归 找每个固定阶的子串 k阶串肯定是k+1阶串！

*/

int  g[MAXN][MAXN], ans[MAXN];

char str[MAXN];

int solve(int l, int r)

{

    if (!g[l][r]) return ;//不回文

    if (l == r) return ;

    if (l +  == r) return ;

    if ((l - r + ) %  == )

        return solve(l, (l + r) / ) + ;

    else

        return solve(l, (l + r) /  - ) + ;

}

int main()

{

    scanf("%s", str + );

    int len = strlen(str+);

    for (int i = ; i <= len; i++)

    {

        g[i][i] = ;

        if (i +  <= len&&str[i] == str[i + ])

            g[i][i + ] = ;

    }

    for (int k = ; k <= len; k++)

    {

        for (int i = ; i <= len; i++)

        {

            int r = i + k - ;

            if (r > len) break;

            if (g[i + ][r - ] && str[r] == str[i])

                g[i][r] = ;

        }

    }

    for (int i = ; i <= len; i++)

    {

        for (int j = ; j <= len; j++)

            ans[solve(i, j)]++;//solve(i,j)b表示从i到j的连续子串是一个几阶回文串

    }

    for (int i = len - ; i >= ; i--)

        ans[i] += ans[i + ];

    for (int i = ; i <= len; i++)

        printf("%d ", ans[i]);

    return ;

}

巴特西

D. Palindromic characteristics

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