Thanks to Sergey Kopeliovich, try this

http://acm.math.spbu.ru/~sk1/algo/input-output/fread_write_export.cpp.html

Hi, have you already managed to find out how to make big tests?

http://hsin.hr/coci/archive/2015_2016/contest4_tasks.pdf

Problem "GALAKSIJA" has the same idea in it.

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

I guess ios_base with cin.tie would be enough

On techocup site, where you have registered for this olympiad

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There are some facts I can't prove because of my bad Englsh)) They are not very hard, but I can try to write proofs for them in Russian if you would want)

It's enough. If we can put a point in a_i + 1 and a_i - 1, the result will be the same. So I can be wrong in a situation when we can put a point in a_i - 1 but can't in a_i + 1. Let's take a look at this. The points must be like ...a_i - 1, a_i, a_i + 1(because we can't make a point at here)... Let's increase i until we find a free a_i + 1. What do we get? A sequence of adjacent points. I want to say that the result will be the same if we put a point in the left of this sequence or in the right.

If we don't find this i, it means there is a sequence with the end at M. in that case I say that the result will be the same if we put a point in the left or don't put at all.

No, I meant another thing. It's correct that making a point in a_i-1 and a_i+1 would give the same result. But sometimes it's impossible to put a point in a_i+1(like if there are points with coordinates x and x+1)

hmm

Am I wrong in the situaton when there are two adjacent points?

I didn't think about it because my solution had AC

In B that's enough to check a_i + 1 only