Freaks City

A straightforward Rust implementation for finding all factors of a number is shown below. This approach iterates up to the square root of the input value. fn compute_factors_simple(num: u64) -> Vec<u64> { let sqrt_val = (num as f64).sqrt().floor() as u64; let mut factors = Vec::new(); for divisor in 1..=sqrt_val { i ...

Problem A: Counting Digit Holes The task requires calculating the total number of closed loops (holes) in a sequence of digits. Digits '0', '6', and '9' contain one loop each, while '8' contains two loops. The solution involves iterating through the string and accumulating the count based on the digit encountered. #include <iostream> #inc ...

Fast Exponentiation Recursive Approach function fastExponentiation(base, exponent, modulus) { if (exponent === 0) return 1; const halfExp = fastExponentiation(base, Math.floor(exponent / 2), modulus); let result = (halfExp * halfExp) % modulus; if (exponent % 2 === 1) { result = (result * base) % modulus; ...

Introduction Consider the following geometric problem: mark all vertical lines x = c and horizontal lines y = c where c ∈ ℤ on a plane. Now consider a line y = (px + r)/q where p, r ∈ ℕ and q ∈ ℕ₊. Since the residue class of r modulo q determines the behavior, we can assume r < q without loss of generality. Imagine a moving point traveling a ...

Problem A: Strategic Allocation This challenge involves selecting a subset of items to meet a weight capacity requirement with the minimum count. The optimal approach utilizes a greedy strategy. By prioritizing larger weights first, we minimize the number of items required to reach the target threshold. void processAllocation() { int itemCo ...

To solve this problem, we scan every possible 2x2 submatrix in a n × m grid and check whether the four characters within it collectively contain at least one 'y', one 'o', and one 'u'. The order does not matter—only the presence of all three required characters. The algorithm iterates over all valid top-left corners of 2x2 blocks (from row 1 to ...

A. Ribbon Coloring Strategy Alice's optimal strategy is to color the ribbon in a repeating pattern like "123123...". Bob's optimal counter-strategy is to recolor the ribbon to the most frequent color. The most frequent color appears at least ⌈n/m⌉ times, leaving Bob with at most (n - ⌈n/m⌉) recoloring operations. Compare this value wi ...

Problem D: DiviDuelo Approach This problem requires number theory analysis and prime factorization techniques. The solution involves categorizing cases based on the prime factorization of the input number. The implementation uses advanced primality testing and factorization algorithms including Miller-Rabin and Pollard's rho method. Implementat ...

Number Theory Primality Testing via Trial Division #include <iostream> using namespace std; bool isPrime(int num) { if (num < 2) return false; for (int d = 2; d <= num / d; ++d) if (num % d == 0) return false; return true; } int main() { int queries; cin >> queries; while (queries--) { ...

Resource Redistribution with Asymmetric Operations Given a sequence (a) of length (n) and two positive integers (x, y) where (y \le x). You may repeatedly pick two elements and transfer resources: subtract (x) from one and add (y) to the other, provided the first remains non‑negative. Determine the maximum possible value of any element after op ...