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Mobius Function Definition \[\mu(n) = \begin{cases} 1 & n = 1 \\ 0 & n \text{ has squared prime factors} \\ (-1)^k & \text{where } k \text{ is the count of distinct prime factors of } n \end{cases}\] Properties The Mobius function is multilpicative: \[\sum_{d \mid n} \mu(d) = \varepsilon(n) \\ \mu \ast 1 = \varepsilon\] where \(\var ...

1.1 Basic Sieve Approach The fundamental strategy involves iterating through integers starting from 2 and marking all their multiples as composite. Because a prime number is divisible only by 1 and itself, it remains unmarked. The total operations correspond to the sum of the harmonic series ($\frac{n}{2} + \frac{n}{3} + \dots$), yielding a tim ...

Rewriting the Polynomial Directly solving high-degree polynomials is computationally infeasible due to the lack of general root-finding formulas. Instead, we restructure the expression by factoring out terms: Modular Evaluation Strategy To verify whether a value of x satisfies the equation, we evaluate \(f(x) \bmod p\), where p is a large prim ...

Given a positive integer $ n $, determine whether it can be expressed as a non-negative integer combination of 2020 and 2021 — that is, whether there exist integers $ a \geq 0 $ and $ b \geq 0 $ such that: $$ n = 2020a + 2021b $$ Rewriting the equation reveals a useful structure: $$ n = 2020(a + b) + b = 2020k + b, $$ where $ k = a + b $. Since ...

Linear Sieve for Prime Numbers (O(2N)) bool composite[N]; int primes[N], prime_count; void build_sieve(int limit) { for (int num = 2; num <= limit; num++) { if (!composite[num]) { primes[prime_count++] = num; } for (int j = 0; j < prime_count && num * primes[j] <= limit; j++) { ...

Chino’s Take-Away Game The game ends when only one stone remains. If the initial pile contains an odd count, the first player can always remove exactly one stone, leaving an even number. Any subsequent move by the second player must remove an odd amount (co-prime with the even remainder), again leaving an odd count. By induction the first playe ...

Given integers l, r, and k, determine how many distinct greatest common divisors (GCDs) can be formed by selecting any k numbers from the range [l, r]. The constraint is: 1 ≤ l ≤ r ≤ 10^12, 2 ≤ k ≤ r - l + 1. Rather than computing all possible GCD values directly, we count all potential divisors. Any integer m = i × j that divides two numbers x ...

Problem 1: Counting Valid Pairs with LCM Condition Problem Description Given an integer n where 1 ≤ n ≤ 10^8, determine the number of pairs (x, y) such that 1 ≤ x, y ≤ n and the following inequality holds: lcm(x, y) / gcd(x, y) ≤ 3 Note that lcm(x, y) = (x * y) / gcd(x, y)^2. Solution Approach This is a straightforward number theory problem. L ...

Subtree Budget Allocation and Optimal Path Reconstruction The core challenge revolves around allocating a fixed budget across a binary tree structure to maximize a specific threshold value. The solution begins by analyzing the root's contribution and propagating constraints downward. We define a dynamic programming state dp_max[u][b] representi ...

The Ternary Goldbach Conjecture asserts that any odd integer greater than 7 can be represented as the sum of three prime numbers. While proven for sufficiently large numbers, verifying this for smaller integers requires an effiicent computational approach. Given an odd integer n (9 < n < 20,000), our objective is to find a triplet of prim ...