2018-2019 9th BSUIR Open Programming Championship. Semifinal

12 problems from 2018-2019 9th BSUIR Open Programming Championship. Semifinal (contest 102439), difficulty -. 12/12 solutions verified against sample I/O.

12 items

2018-2019 9th BSUIR Open Programming Championship. Semifinal

Special | 12 problems | 12/12 verified | Difficulty - | 42m 38s

#	Problem	Time	✓
A	Four minutes until BSUIR Open	2m 22s	✓
B	Varvara and matrix	6m 4s	✓
C	Cockroach Racing	3m 55s	✓
D	Light show	7m 55s	✓
E	Small business	1m 46s	✓
F	Prime or number	1m 6s	✓
G	Sequence exploration	2m 20s	✓
H	Nonfibonacci numbers	3m 21s	✓
I	Equal Mod Segments	6m 26s	✓
J	Boedium	2m 30s	✓
K	Innovations	2m 35s	✓
L	The only winner	2m 18s	✓

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439A - Four minutes until BSUIR Open

We have a straight road from position 0 to the university at position (xn). Camera (i) is at position (xi), and when the car passes that point its speed must be at most (vi). The car starts at speed zero.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439E - Small business

We have a bag of digit blocks, represented by a string s. Every block must be used exactly once to build two decimal integers. The two integers may be equal, zero is allowed, but neither number may contain leading zeroes. Both numbers must be at most (10^{18}).

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439K - Innovations

We have a weighted tree of cities. Because the graph is a tree, between every pair of cities there is exactly one path, so that path is automatically the shortest path.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439J - Boedium

Josya is the first participant in the input. Every athlete runs five identical laps and shoots at twenty targets in total. Ten targets are shot from the prone position and ten from the standing position. A hit costs no extra time, while every miss adds exactly 60 seconds.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439F - Prime or number

We are given one non-negative integer n, with 1 <= n <= 10^18. Instead of ordinary multiplication, we are asked to use bitwise OR as the operation that combines two numbers.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439C - Cockroach Racing

We have n cockroaches, and each cockroach has an m digit number written on its back. Some digits are known, while every ? can independently be replaced by any digit from 0 to 9. Leading zeroes are allowed.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439L - The only winner

We have 2n distinct cards, numbered from 1 through 2n. They are randomly divided into n pairs, with each pair going to one guest. A guest's score is the sum of the two cards in their pair. We need the probability that exactly one guest has the maximum score.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439I - Equal Mod Segments

For every contiguous segment [L, R], there are two ways to evaluate it. Starting at L, repeatedly take the remainder by the next array element: a[L] % a[L+1] % ... % a[R]. Starting at R, do the same thing in the opposite direction: a[R] % a[R-1] % ... % a[L].

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439H - Nonfibonacci numbers

For an integer (x), crossing out digits means deleting some positions from its decimal representation while keeping all remaining digits in their original order. A number is disliked if some positive Fibonacci number can be obtained this way.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439G - Sequence exploration

We start with the string 1. Each next term is obtained by reading the previous term from left to right, grouping equal consecutive digits, and replacing every group by two digits: its length followed by the digit itself.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439D - Light show

There are (n) lightbulbs, so a configuration of the club is a binary vector of length (n). A switch is another binary vector. Pressing a switch means XORing its vector with the current configuration, because every bulb whose corresponding bit is one gets toggled.

Practice › Codeforces Solutions › 2018-2019 9th BSUIR Open Programming Championship. Semifinal ›

CF 102439B - Varvara and matrix

We have an (n times m) matrix whose entries are integers from (0) to (k). A zero marks an unknown cell, and the special condition is that every row and every column contains at most one zero. We must replace every zero independently by either (A) or (B).