Given a string. Check if the string is a concatenation of any prefix of this string (any number of times concatenation is possible). If present then print the prefix, otherwise -1.
№ | Пользователь | Рейтинг |
---|---|---|
1 | tourist | 3993 |
2 | jiangly | 3743 |
3 | orzdevinwang | 3707 |
4 | Radewoosh | 3627 |
5 | jqdai0815 | 3620 |
6 | Benq | 3564 |
7 | Kevin114514 | 3443 |
8 | ksun48 | 3434 |
9 | Rewinding | 3397 |
10 | Um_nik | 3396 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | cry | 167 |
2 | Um_nik | 163 |
3 | maomao90 | 162 |
3 | atcoder_official | 162 |
5 | adamant | 159 |
6 | -is-this-fft- | 158 |
7 | awoo | 155 |
8 | TheScrasse | 154 |
9 | Dominater069 | 153 |
10 | nor | 152 |
Given a string. Check if the string is a concatenation of any prefix of this string (any number of times concatenation is possible). If present then print the prefix, otherwise -1.
Название |
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Let's prove there are always only two prefixes required if it's possible. Let's say we get the answer with more than two strings: $$$s_1, s_2, ..., s_k$$$. Then we can get a string $$$s_1 + s_2 + ... + s_{k - 1}$$$ using only one prefix.
So all we have to do is to find the suffix, such that it is a prefix either. You can do it using Z-function
UPD: I'm quite stupid I guess
We can solve it using simple polynomial hash method.
For each prefix, we can calculate the hash value of it. And then, we can try to copy it a lot of times, calculating the hash value as the same time, until the length exceed n.
Time complexity is $$$\sum_{i = 1} ^ {n} [n / i] = \Theta(n log n).$$$