Codeforces Round #427 (Div. 2) Problem D Palindromic characteristics (Codeforces 835D) - 记忆化搜索

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 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.


  题目大意 一个k阶回文串是左右两半的串(长度为这个串的长度除以2向下取整),且这两个串都是k - 1阶回文串。统计一个串内各阶回文串的个数。

  表示这道题很简单。。然而比赛时我又把题读错。。。(我这文科渣到了某种境界)。我也不知道我怎么想的,明明看到了half,强行理解成一段。。中间那一段解释half的长度的一段,竟然成功归避。。跑去看样例猜题意了。。。绝望。。。然后不说废话了。

  根据数学的直觉和人生的哲理(瞎扯ing),可以知道,你需要用O(n^2)的时间预处理出任意一段子串是否是回文串。

  对于一个回文串是否是k阶回文串,因为它自带二分效果(因为一个回文串是对称的,所以如果它的左半边是k - 1阶回文串,那么右半边也一定是),所以可以考虑递归求解。

  据说直接求解也可以过。但是个人更喜欢把它写成记忆化搜索吧,可以省掉一个log。

  由于一个k阶回文串一定是k - 1阶回文串,所以开始就统计每个子串最高的阶数(如果不是回文串,阶数就当成0吧)的数量,最后再求个后缀和,输答案,完事。

  如果还有什么不清楚可以看代码。

Code

 1 /**
 2  * Codeforces
 3  * Problem#835D
 4  * Accepted
 5  * Time: 171ms
 6  * Memory: 124500k
 7  */
 8 #include <bits/stdc++.h>
 9 using namespace std;
10 typedef bool boolean;
11 
12 int n;
13 char s[5005];
14 int lvls[5002][5002];
15 boolean vis[5002][5002];
16 int ans[5005];
17 
18 inline void init() {
19     scanf("%s", s + 1);
20     n = strlen(s + 1);
21 }
22 
23 inline int getLvl(int l, int r) {
24     if(l == r)    return 1;
25     if(lvls[l][r] != 1 || vis[l][r])    return lvls[l][r];
26     vis[l][r] = true;
27     int len = (r - l + 1) / 2;
28     return lvls[l][r] = (getLvl(l, l + len - 1) + 1);
29 }
30 
31 inline void solve() {
32     s[0] = '+', s[n + 1] = '-', s[n + 2] = 0;
33     for(int i = 1; i <= n; i++) {
34         lvls[i][i] = 1;
35         int l = i - 1, r = i + 1;
36         while(s[l] == s[r])    lvls[l][r] = 1, l--, r++;
37         l = i, r = i + 1;
38         while(s[l] == s[r])    lvls[l][r] = 1, l--, r++;
39     }
40     for(int i = 1; i <= n; i++)
41         for(int j = i; j <= n; j++) {
42 //            int c = getLvl(i, j);
43 //            cout << i << " " << j << " " << c << endl;
44             ans[getLvl(i, j)]++;
45         }
46     for(int i = n - 1; i; i--)
47         ans[i] += ans[i + 1];
48     for(int i = 1; i <= n; i++)
49         printf("%d ", ans[i]);
50 }
51 
52 int main() {
53     init();
54     solve();
55     return 0;
56 } 

 

  

  

posted @ 2017-08-01 10:58 阿波罗2003 阅读(...) 评论(...) 编辑 收藏