Нет. Я спросила организаторов, могут ли они его сделать, но это маловероятно.

Список зарегистрированных — пока 43 человека.

В онсайт-раунде Algorithm competition — кто угодно, если присутствует на онсайте. В марафоне — вообще кто угодно.

Думаю, да — список не обновляется в реальном времени :-(

No. Last time I've checked was well after the deadline, and we still had to reach 100 registrants :-(

Interesting that I gave absolutely same problem (as today Hard) on VK Cup Finals 695A - LRU but with n <= 20 (because we haven't managed to solve problem with bigger constraints).

Considering individual bits, note that an AND with x sets bits that are 0 in x to 0, and the rest are unchanged. Similarly, an OR with x sets bits that are 1 in x to 1, and the rest are unchanged.

Therefore, the final state of each bit relates to the last operation that set that bit (if it was an AND, the bit is 0, if it was an OR, the bit is 1). If no operation set that bit, then it's equal to its initial state of 0.

The trick here is to realize that the problem becomes easier if you consider it backwards, because now, once you decide to set a bit using one of the operations, it can never be unset by any earlier operation, so you know immediately its final state. Noticing this, it's much easier, because setting bits to their correct value is obviously always advantageous to you, and setting bits to their incorrect value is obviously forbidden (because they will never change later). So you should do any operation as long as it doesn't set any bit to an incorrect value.

Maybe the code with bitmasks is easier to understand?

string EllysAndor::canGet(vector <int> numbers, int goal) {
	reverse(numbers.begin(), numbers.end());
	
	int fixed_bits = 0;
	for (int x : numbers) {
		// if all 1 bits in x are also in goal or already set, we can use OR
		if ((x & (goal | fixed_bits)) == x) {
			fixed_bits = fixed_bits | x;
		}
		// if all 0 bits in x are also in goal or already set, we can use AND
		else if (((~x) & (~goal | fixed_bits)) == ~x) {
			fixed_bits = fixed_bits | (~x);
		}
	}

	if ((fixed_bits & goal) == goal) return "Possible";
	return "Impossible";
}