MEX Foundation Contest (supported by AIM Tech)

10 problems from MEX Foundation Contest (supported by AIM Tech) (contest 102412), difficulty -. 10/10 solutions verified against sample I/O.

10 items

MEX Foundation Contest (supported by AIM Tech)

Special | 10 problems | 10/10 verified | Difficulty - | 51m 55s

#	Problem	Time	✓
A	The One Polynomial Man	8m 20s	✓
B	Alexey the Sage of The Six Paths	5m 52s	✓
C	Steel Ball Run	17m 10s	✓
D	The Jump from Height of Self-importance to Height of IQ Level	3m 13s	✓
E	Minimums on the Edges	2m 19s	✓
F	IQ Test	1m 33s	✓
G	AtCoder Quality Problem	4m 5s	✓
H	Mex on DAG	2m 21s	✓
I	Find the Vertex	1m 12s	✓
J	Yet Another Mex Problem	5m 50s	✓

Practice › Codeforces Solutions › MEX Foundation Contest (supported by AIM Tech) ›

CF 102412D - The Jump from Height of Self-importance to Height of IQ Level

We have a row of (n) skyscrapers, and their heights form a permutation of (1,2,ldots,n). A valid jump uses three skyscrapers in increasing position order whose heights are also strictly increasing.

Practice › Codeforces Solutions › MEX Foundation Contest (supported by AIM Tech) ›

CF 102412E - Minimums on the Edges

We have an undirected multigraph with (n) vertices and (m) edges, together with exactly (s) tokens. We choose a nonnegative integer (av) for every vertex (v), where (av) is the number of tokens placed there and the total is exactly (s). An edge ((u,v)) has capacity (min(au,av)).

Practice › Codeforces Solutions › MEX Foundation Contest (supported by AIM Tech) ›

CF 102412F - IQ Test

We start with a set containing 0, 1, and 2. One operation chooses any two numbers x and y already in the set and inserts x 2 −y. The inserted value must stay between 0 and 10 18, and we may perform at most 43 operations.

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CF 102412A - The One Polynomial Man

We have a prime modulus (p), a subset (S) of residues modulo (p), and another subset (V). For every ordered pair ((a,b)) with both values in (S), we evaluate [ F(a,b)= frac{(2a+3b)^2+5a^2}{(3a+b)^2} + frac{(2a+5b)^2+3b^2}{(3a+2b)^2} pmod p.

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CF 102412J - Yet Another Mex Problem

The problem is from the MEX Foundation Contest, Gym 102412, Problem J. The official limits are (2le nle 2cdot10^5), (1le kle n), (0le aile n), with a 4 second time limit and 512 MiB of memory. We have a nonnegative integer array (a).

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CF 102412G - AtCoder Quality Problem

We have a set (S) with (n) elements, and every subset of (S) must be painted either red or blue. For a subset (T), painting it red costs (RT), while painting it blue costs (BT). The costs may be negative, so the goal is not simply to choose the cheaper color independently.

Practice › Codeforces Solutions › MEX Foundation Contest (supported by AIM Tech) ›

CF 102412I - Find the Vertex

We have a connected undirected graph. One vertex, called the source, was chosen secretly. For every vertex (v), we are given the shortest-path distance from the source to (v), but only its remainder modulo (3). The task is to recover the source vertex.

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CF 102412H - Mex on DAG

We have a directed acyclic graph with (n) vertices and exactly (2n) edges. The edges are numbered from (0) to (2n-1), and edge (i) has value (lfloor i/2rfloor). Thus every value from (0) through (n-1) occurs on exactly two edges.

Practice › Codeforces Solutions › MEX Foundation Contest (supported by AIM Tech) ›

CF 102412C - Steel Ball Run

We have a tree whose vertices may or may not currently contain a chip. A query toggles one vertex between these two states. After every toggle, we need the minimum total number of edge traversals required to gather all currently present chips at one common vertex.

Practice › Codeforces Solutions › MEX Foundation Contest (supported by AIM Tech) ›

CF 102412B - Alexey the Sage of The Six Paths

Think of every party member as a vertex. There are (n) vertices on the left, numbered (1) through (n), and (n) vertices on the right, numbered (n+1) through (2n). Every problem creates one edge between one left vertex and one right vertex.