Codeforces Round 116 (Div. 2, ACM-ICPC Rules)

Solutions for Codeforces Round 116 (Div. 2, ACM-ICPC Rules) (contest 180). 5/5 problems verified against sample I/O. Difficulty range: 1200-2300.

5 items

Codeforces Round 116 (Div. 2, ACM-ICPC Rules)

Type: Div. 2 | Problems: 5 | Verified: 5/5 | Rating range: 1200-2300 | Time: 8m 3s

Problem	Name	Rating	Tags	Solve Time	Verified
A	Defragmentation	1800	implementation	1m 20s	✓
B	Divisibility Rules	2300	math, number-theory	1m 50s	✓
C	Letter	1400	dp	1m 46s	✓
D	Name	1900	greedy, strings	1m 39s	✓
F	Mathematical Analysis Rocks!	1200	constructive-algorithms, implementation, math	1m 28s	✓

Practice › Codeforces Solutions › Codeforces Round 116 (Div. 2, ACM-ICPC Rules) ›

CF 180A - Defragmentation

We are tasked with reorganizing a hard disk so that files occupy contiguous clusters from the beginning of the disk, and free clusters are pushed to the end. The disk has n clusters, numbered from 1 to n, and m files.

Practice › Codeforces Solutions › Codeforces Round 116 (Div. 2, ACM-ICPC Rules) ›

CF 180B - Divisibility Rules

We work in a positional numeral system with base b. For a divisor d, we must determine which kind of divisibility rule exists in this base. The problem defines several categories.

Practice › Codeforces Solutions › Codeforces Round 116 (Div. 2, ACM-ICPC Rules) ›

CF 180C - Letter

We are given a string containing uppercase and lowercase English letters. We want to transform it into a "fancy" string where every uppercase letter appears before every lowercase letter.

Practice › Codeforces Solutions › Codeforces Round 116 (Div. 2, ACM-ICPC Rules) ›

CF 180D - Name

We are given a multiset of characters in string s. We may rearrange these characters in any order, but we must use every character exactly once. Among all such permutations, we want the lexicographically smallest string that is still strictly larger than another string t.

Practice › Codeforces Solutions › Codeforces Round 116 (Div. 2, ACM-ICPC Rules) ›

CF 180F - Mathematical Analysis Rocks!

Every student points to exactly one best friend, and every student is pointed to by exactly one other student. That means the friendship relation forms a permutation p. On day 1, notebook i stays with student i. On every following day, notebooks move according to p.