Codeforces 427Div2 D、835D Palindromic characteristics

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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:

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： 对于输入的一个字符串判断其中1~|s|级回文串各有几个，对于k级回文串的定义：如果一个长度字符串是k级字符串，则前一半的字符串是k - 1级回文串，后一半也是k - 1 级回文串。

    思路： DP[l][r]表示从l~r之间是DP[l][r]级回文串，当满足str[l] == str[r] && DP[l + 1][r - 1] != 0 时 DP[l][r] = DP[l][r - (r - l)/2 - 1] + 1，否则DP[l][r] = 0。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>

#define MAX 5005
using namespace std;
int DP[MAX][MAX];

int main( )
{
	char str[MAX];
	cin >> str;
	int len = strlen(str);
	for(int j = 0; j < len; j++)
		if(str[j] == str[j + 1]) DP[j][j + 1] = 2,DP[j][j] = 1;
		else DP[j][j] = 1;
	int dif;
	for(int j = 2; j < len; j++){
		for(int k = j - 2; k >= 0; k--){
			if(str[j] == str[k] && DP[k + 1][j - 1] != 0){
				dif = (j - k)/2 + 1;
				DP[k][j] = DP[k][j - dif] + 1;
			}
			//原来写法： 
			/*if(str[j] == str[k]){
				dif = (j - k)/2 + 1;
				if(DP[k][j - dif] != 0 && DP[k + dif][j] != 0) DP[k][j] = DP[k][j - dif] + 1;
				else if(DP[k + 1][j - 1] != 0) DP[k][j] = 1;
			}*/
			//hack date: abaaca
		}
	}
/*	for(int j = 0; j < len; j++){
		for(int k = 0; k < len; k++)
			cout << DP[j][k] << " ";
		cout << endl;
	}*/
	int ans[MAX] = {0};
	for(int j = 0; j < len; j++)
		for(int k = j; k < len; k++)
			while(DP[j][k] > 0) ans[DP[j][k]--]++;
	for(int j = 1; j <= len; j++)
		cout << ans[j] << " ";
	cout << endl;
}

        在晚上打的时候都WA6，在补题中看了眼题解的后马上就明白了。但是，也不明白自己原来是怎么WA的，最后还是想出了一组数据成功证明我原来写的是错的。