Codeforces Round 272 (Div. 2)

Solutions for Codeforces Round 272 (Div. 2) (contest 476). 5/5 problems verified against sample I/O. Difficulty range: 1000-2200.

5 items

Codeforces Round 272 (Div. 2)

Type: Div. 2 | Problems: 5 | Verified: 5/5 | Rating range: 1000-2200 | Time: 12m 19s

Problem	Name	Rating	Tags	Solve Time	Verified
A	Dreamoon and Stairs	1000	implementation, math	1m 32s	✓
B	Dreamoon and WiFi	1300	bitmasks, brute-force, combinatorics	1m 35s	✓
C	Dreamoon and Sums	1600	math	1m 19s	✓
D	Dreamoon and Sets	1900	constructive-algorithms, greedy, math	2m 8s	✓
E	Dreamoon and Strings	2200	dp, strings	5m 45s	✓

Practice › Codeforces Solutions › Codeforces Round 272 (Div. 2) ›

CF 476A - Dreamoon and Stairs

Dreamoon wants to reach the top of a staircase containing n steps. Every move can cover either 1 step or 2 steps. Among all possible ways to reach exactly step n, we need the number of moves used to be divisible by m. The task is not asking for the number of different sequences.

Practice › Codeforces Solutions › Codeforces Round 272 (Div. 2) ›

CF 476B - Dreamoon and WiFi

Drazil sends a sequence of movement commands. Each '+' moves Dreamoon one step to the right and each '-' moves one step to the left. We know the original command string and also the string that Dreamoon receives. The received string may contain '?

Practice › Codeforces Solutions › Codeforces Round 272 (Div. 2) ›

CF 476C - Dreamoon and Sums

We are asked to calculate the sum of all integers $x$ that satisfy a pair of modular conditions with respect to two given integers $a$ and $b$.

Practice › Codeforces Solutions › Codeforces Round 272 (Div. 2) ›

CF 476D - Dreamoon and Sets

The task is to construct n sets of four distinct positive integers, where each set has a certain "rank" k. The rank condition requires that the greatest common divisor (GCD) of any two numbers within the set equals exactly k.

Practice › Codeforces Solutions › Codeforces Round 272 (Div. 2) ›

CF 476E - Dreamoon and Strings

We start with a string s. We may delete exactly x characters, keeping the remaining characters in their original order. The resulting string is a subsequence of s. Inside that resulting string, we look for as many non-overlapping occurrences of a pattern p as possible.