Codeforces Round #427 (Div. 2) D-Palindromic characteristics （回文串，暴力）

D. Palindromic characteristics

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

inputabbaoutput6 1 0 0 inputabacabaoutput12 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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题目大意

 找回文串，一个k阶回文串的定义是他的左边一半是一个k-1阶回文串。输出1-|s|阶回文串的个数。

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注意题目要求，一个k阶回文串也会在k-1阶里面计数，但这问题不大。

首先用通用技巧，在每两个字符之间增加一个字符，来避免讨论奇偶。
然后为了方便的判断阶数，我们可以暴力的开一个数组，来记录以L为起点以R为终点的字符串的阶数，非回文串为0，于是我们只需要枚举中间数，向两边拓展，然后阶数就等于一边的阶数加一即可。注意不要把前面增加的字符算进来。具体可以参见代码。

代码：

#include<stdio.h>#include<iostream>#include<algorithm>#include<cstring>#define ll long longusing namespace std;ll cnt[12345],i,j,k,l,p,x,y,tmp,n;string s;char A[12345];short Q[10005][10005];int main(){    cin>>s;n=s.length();s=" "+s;    for(i=1;i<=n;i++)    {        A[2*i-1]=s[i];        A[2*i]='#';    }    cnt[1]+=n;    n=2*n-1;    for(i=1;i<=n;i++)    if(A[i]!='#')Q[i][i]=1;    for(k=1;k<=n;k++)    {        for(l=1;k-l>=1&&k+l<=n;l++)        {            if(A[k-l]=='#')continue;            i=k-l;j=k+l;            if(A[i]!=A[j])break;            if(A[k-1]=='#')x=k-2;            else x=k-1;            tmp=Q[i][x]+1;            cnt[tmp]++;            Q[i][j]=tmp;        }    }    for(i=s.length()-1;i>0;i--)cnt[i]+=cnt[i+1];    for(i=1;i<s.length();i++)printf("%I64d ",cnt[i]);}