题库练习 k-substrings
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A11718 | k-substrings

时间限制1s
内存限制256MB
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题目描述

You are given a string $s$ consisting of $n$ lowercase Latin letters.

Let's denote $k$ -substring of $s$ as a string $subs_{k}=s_{k}s_{k+1}..s_{n+1-k}$ . Obviously, $subs_{1}=s$ , and there are exactly ![](/uploads/acgo/image/8891562983aeb516_23d871660cae.jpeg) such substrings.

Let's call some string $t$ an odd proper suprefix of a string $T$ iff the following conditions are met:

- $|T|>|t|$ ;
- $|t|$ is an odd number;
- $t$ is simultaneously a prefix and a suffix of $T$ .

For evey $k$ -substring (![](/uploads/acgo/image/d0fe2cf0ac574dc1_72a31fdcf600.jpeg)) of $s$ you have to calculate the maximum length of its odd proper suprefix.

输入格式

The first line contains one integer $n$ $(2<=n<=10^{6})$ — the length $s$ .

The second line contains the string $s$ consisting of $n$ lowercase Latin letters.

输出格式

Print ![](/uploads/acgo/image/46d75d0fd3807939_3d68b83272b8.jpeg) integers. $i$ -th of them should be equal to maximum length of an odd proper suprefix of $i$ -substring of $s$ (or $-1$ , if there is no such string that is an odd proper suprefix of $i$ -substring).

输入输出样例

输入 #1
15
bcabcabcabcabca
输出 #1
9 7 5 3 1 -1 -1 -1
输入 #2
24
abaaabaaaabaaabaaaabaaab
输出 #2
15 13 11 9 7 5 3 1 1 -1 -1 1
输入 #3
19
cabcabbcabcabbcabca
输出 #3
5 3 1 -1 -1 1 1 -1 -1 -1
C++ 编辑器
输入
输出