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
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.
给定一个字符串，求1-len阶回文串，k阶回文串的左边是有k-1阶回文串组成的，1阶回文串就是字符串的每一个子串回文串。
用dp来做，dp[i][j]表示字符串str[i...j]是否是回文串。O（n^2）的复杂度。
然后求dp[i][j]是第几阶回文串，可以用递归来做，如果str[i..j]不是回文串的话，则返回0，否则返回str[i...m]+1,m表示子串str[i....j]的左串的右边位置。i==j,在递归会出现m<l的因为，返回0；
只好可以把1阶回文串的左边看成由0阶回文串组成的。由于k阶回文串又是k-1,k-2...1阶会文串，所以要ans[i-1] += ans[i]
1 #include <bits/stdc++.h>
2 using namespace std;
3 const int MAX = 5010;
4 char str[MAX];
5 int dp[MAX][MAX], ans[MAX], n;
6 int getK(int l, int r) {
7 if(dp[l][r] == 0) return 0;
8 else {
9 int m = (l+r);
10 if(m&1) return getK(l,m/2)+1;
11 else return getK(l,m/2-1)+1;
12 }
13 }
14 int main() {
15 scanf("%s",str+1);
16 int n = strlen(str+1);
17 for(int i = 1; i <= n; i ++) {
18 dp[i][i] = 1;
19 if(i+1 <= n && str[i] == str[i+1]) dp[i][i+1] = 1;
20 }
21 for(int i = 3; i <= n; i ++) {
22 for(int j = 1; j <= n-i+1; j ++) {
23 int r = j+i-1;
24 if(dp[j+1][r-1] && str[j] == str[r]) dp[j][r] = 1;
25 }
26 }
27 for(int i = 1; i <= n; i ++) {
28 for(int j = i; j <= n; j ++) {
29 ans[getK(i,j)] ++;
30 }
31 }
32 for(int i = n; i > 0; i --)
33 ans[i-1] += ans[i];
34 for(int i = 1; i <= n; i ++)
35 printf("%d%c",ans[i],(i==n?‘\n‘:‘ ‘));
36 return 0;
37 }