Codeforces Round 235 (Div. 2)

4 problems from Codeforces Round 235 (Div. 2) (contest 401), difficulty 800-2000. 4/4 solutions verified against sample I/O.

4 items

Codeforces Round 235 (Div. 2)

Div. 2 | 4 problems | 4/4 verified | Difficulty 800-2000 | 11m 13s

#	Problem	Rating	Tags	Accepted	Time	✓
A	Vanya and Cards	800	implementation, math	17,417	4m 15s	✓
B	Sereja and Contests	1200	greedy, implementation, math	4,599	1m 24s	✓
C	Team	1400	constructive-algorithms, greedy, implementation	21,696	4m 15s	✓
D	Roman and Numbers	2000	bitmasks, brute-force, combinatorics	4,566	1m 19s	✓

Practice › Codeforces Solutions › Codeforces Round 235 (Div. 2) ›

CF 401A - Vanya and Cards

Vanya has found a subset of his playing cards, each card labeled with an integer between $-x$ and $x$. He wants the total sum of all his cards to be zero. We need to determine the minimum number of additional cards he should find to achieve that zero sum.

Practice › Codeforces Solutions › Codeforces Round 235 (Div. 2) ›

CF 401C - Team

We are asked to arrange a set of cards, some marked with zero and some with one, in a line so that no two zeros are adjacent and no three ones are consecutive. We know the counts of each type of card upfront: n zeros and m ones.

Practice › Codeforces Solutions › Codeforces Round 235 (Div. 2) ›

CF 401B - Sereja and Contests

The contest platform numbers every round consecutively by start time. A round can either be a standalone Div2 round, or a pair of simultaneous rounds where Div2 gets identifier i and Div1 gets identifier i + 1. Sereja only participates in Div2 rounds.

Practice › Codeforces Solutions › Codeforces Round 235 (Div. 2) ›

CF 401D - Roman and Numbers

We are asked to count the number of integers that can be formed by permuting the digits of a given number $n$, do not start with zero, and are divisible by a given modulus $m$. The input $n$ can be as large as $10^{18}$, which means up to 18 digits, and $m$ is at most 100.