Absolutely correct, thanks, I half-forgot that this problem was there, probably because it is not that hard :)

You can see the statement by clicking on the problem.

Or do you want to print it out? I can send you some version, since we changed limits in 2 problems.

For D the answer is 4*a-1/3. With the power of WolframAlpha.

/*input
2
*/
#include <bits/stdc++.h>
typedef long double ld;
using namespace std;
int main()
{
	ios_base::sync_with_stdio(false);
	ld a;
	cin >> a;
	ld x1 = (sqrtl(4 * a + 1) - 1);
	x1 *= x1;
	x1 /= 4;
	ld x2 = (sqrtl(4 * a - 3) + 1);
	x2 *= x2;
	x2 /= 4;
	ld x3 = (sqrtl(4 * a + 1) + 1);
	x3 *= x3;
	x3 /= 4;
	ld S = ((1 + 2 * a) * sqrtl(1 + 4 * a)) / 2;
	S += (1 - sqrtl(1 + 4 * a) + a * (-6 + sqrtl(-3 + 4 * a) - sqrtl(1 + 4 * a))) / 6;
	S += 2.0 * powl(x2, 1.5) / 3.0 - a * x2;
	S -= 2.0 * powl(x1, 1.5) / 3.0 - a * x1;
	S *= 2;
	cout << fixed << setprecision(10) << S << "\n";
}

$$$ans=4a-\frac 1 3$$$ for $$$a \ge 2$$$

Did you get accepted on G? My solution does not involve computing discrete logarithms.

You know degree for all not_-1 vertices, and all other vertices are the same, so they have equal $$$E[deg(v)]$$$. Sum of all deg is $$$2n-2$$$, so you know $$$E[deg(v)]$$$ for all $$$v$$$.

Maybe G is hard because people like me don't know what group means in mathematics?

let $$$A = {a_1,a_2,\cdots,a_k} $$$ (sorted by dfn)

$$$f(x)$$$ = the sum of the weight from root to x.

$$$ans = f(a_1) + f(a_2) + \cdots + f(a_k) - f(lca(a_1,a_2)) - f(lca(a_2,a_3)) - \cdots - f(lca(a_k,a_1))$$$

A: First of all, we consider that whether a single 0 or 1 occur in the rule. Obviously, only one of 0 or 1 occur in the rule, Otherwise the answer is N or -1. Suppose there exists 0 but not 1, because it is a prefix-free system, a fact is that if W can be decoded, only if 0^k+W can be decoded. For any S, if we consider restrciting a suffix cannot be decoded, we only have to think about how to divide the prefix part. Because 0 can be decoded, and each part we devided at least contains a 1, so we can only divide into p(the number of 1s) parts. We can only create this way by connecting every 1 with the continuous 0s. Now the problem becomes finding the shortest prefix that cannot be decoded and the answer should be added by the number of 1s. We can easily find it by reversing those strings and building an AC-Automaton.

Well, I think it is a quite good problem and also a nice contest. Many thanks to Umnik~

PS: I actually translate this solution from my solution in Chinese written in August,2018. Maybe I miss some important details or have some mistakes.

For E, $$$O(n^3)$$$ is obvious but got TLE. I guess some methods like FFT/NTT would be used to optimize it (since 40961 is a modulus of NTT). But I have no clue.

E: We need to compute $$$\sum((s[i] == 0) \times C(i, x)\times C(n - i - 1, y)$$$ for each $$$x, y$$$. Fix $$$y$$$, it turns out to use FFT/NTT.

I`ve passed F with just bruteforce from a highest weight (but I have nothing to prove it).

I: Lucas's theorem. We need to compute $$$\sum{C(n, i)\times C(n - i, i)}$$$. Represent $$$n$$$ as $$$(a_1a_2...a_k)_p$$$. So we concern only $$$i$$$ such that its $$$p-base$$$ representation $$$(b_1b_2...b_k)_p$$$, $$$2b_i \le a_i$$$.