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 |
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1 | tourist | 3690 |
2 | jiangly | 3647 |
3 | Benq | 3581 |
4 | orzdevinwang | 3570 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | Radewoosh | 3509 |
8 | ecnerwala | 3486 |
9 | jqdai0815 | 3474 |
10 | gyh20 | 3447 |
Страны | Города | Организации | Всё → |
№ | Пользователь | Вклад |
---|---|---|
1 | maomao90 | 174 |
2 | awoo | 164 |
3 | adamant | 162 |
4 | TheScrasse | 159 |
5 | nor | 157 |
6 | maroonrk | 156 |
7 | -is-this-fft- | 152 |
8 | orz | 146 |
9 | pajenegod | 145 |
9 | Petr | 145 |
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 $$$\sum_{i = 1} ^ {n} [n / i] = \Theta(n log n).$$$