Codeforces Round 729 (Div. 2)

5 problems from Codeforces Round 729 (Div. 2) (contest 1542), difficulty 800-2700. 2/5 solutions verified against sample I/O.

5 items

Codeforces Round 729 (Div. 2)

Div. 2 | 5 problems | 2/5 verified | Difficulty 800-2700 | 31m 35s

# Problem Rating Tags Accepted Time ✓
A Odd Set 800 math 51,621 5m 9s ✓
B Plus and Multiply 1500 constructive-algorithms, math, number-theory 26,531 16m 20s ✓
C Strange Function 1600 math, number-theory 16,178 2m 39s
D Priority Queue 2200 combinatorics, dp, implementation 2,921 6m 14s
E2 Abnormal Permutation Pairs (hard version) 2700 combinatorics, dp, fft 1,015 1m 13s
Practice › Codeforces Solutions › Codeforces Round 729 (Div. 2) ›

CF 1542D - Priority Queue

We are given a sequence of operations that can either insert a value into a multiset or delete the smallest value currently present.
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Practice › Codeforces Solutions › Codeforces Round 729 (Div. 2) ›

CF 1542A - Odd Set

We are given a collection of numbers with even size, specifically 2n integers per test case. The task is to decide whether it is possible to partition these numbers into n disjoint pairs such that every pair consists of two numbers whose sum is odd.
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Practice › Codeforces Solutions › Codeforces Round 729 (Div. 2) ›

CF 1542C - Strange Function

We are given many queries. Each query provides a very large integer $n$, and we consider every integer $i$ from 1 to $n$. For each $i$, we define a function $f(i)$ as the smallest positive integer that fails to divide $i$.
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Practice › Codeforces Solutions › Codeforces Round 729 (Div. 2) ›

CF 1542B - Plus and Multiply

We start from the number $1$. From any number already in the set, we may perform either of two operations: multiply by $a$, or add $b$. After applying these operations any number of times in any order, we obtain an infinite set of reachable values.
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Practice › Codeforces Solutions › Codeforces Round 729 (Div. 2) ›

CF 1542E2 - Abnormal Permutation Pairs (hard version)

The solution does not answer the stated problem correctly. The expression to be bounded is $$leftlfloor frac{b}{2} rightrfloor le leftlfloor frac{b}{v+1} rightrfloor cdot frac{b}{v+1} le b,$$ but the argument does not correctly establish the lower bound.
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