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Factorization into Factorial Divisors Given integers (n) and (m) where (1 \le m \le n!) and (n \le 20), decompose (m) into a sum of at most (n) divisors of (n!). A solution is guaranteed to exist. Define a sequence (d_i = \frac{n!}{i!}) for (i) from 1 to (n). By iterating downwards from (i=n) to (1) and greedily subtracting the largest possible ...

Problem 1: Dumpling Purchase Optimization The problem reduces to a daily deciison: buy dumplings at the current price or rely on an earlier purchase plus storage cost. The total expense for day i if we buy on day j ≤ i is price[j] + c*(i - j). This can be rewritten as (price[j] - c*j) + c*i. Thus we can maintain the minimum value of price[j] - ...

Problem A - 4 Question With (n) horizontal lines and (m) vertical lines drawn on a plane, how many axis-aligned rectangles are formed that contain no interior lines? Solution Consider each dimension independent. Along any straight line, (n) distinct points partition the line into (n - 1) segments. These segments serve as the edges of our rectan ...

Consider a sequence $A = (a_1, a_2, \dots, a_n)$. We want to partition $A$ into contiguous subsequences, and then arrange these subsequences to form a new sequence $f_1, f_2, \dots, f_k$. Each contiguous subsequence $f_1, \dots, f_p$ must satisfy either $f_1 \le f_2 \le \dots \le f_p$ or $f_1 \ge f_2 \ge \dots \ge f_p$. The cost associated with ...

A. Word Rearrangement This is a straightforward problem requiring only basic input/output handling. w1, w2 = input().split() print(w2) print(w1) B. Cooking Tangyuan The key idea is to simulate the process of using packages to fulfill cooking rounds. Each package contributes a fixed number of tangyuan, and excess can carry over. n, x, k = map(i ...

Progress: Dynamic Programming - Linear DP, Knapsack, Interval DP Merging Palindromic Substrings Tags: Interval DP Foundation - Longest Palindromic Substring Problem: Given a string (S), find the length of its longest palindromic substring. Approach: A longer palindrome can always be constructed by adding identical characters to both ends of a s ...

Greedy Algorithm: Minimum Cameras to Monitor a Binary Tree Given a binary tree, we need to place cameras on nodes such that every node in the tree is monitored. A camera placed on a node monitors itself, its parenet, and its immediate children. Determine the minimum number of cameras required. Approach We can solve this problem using a greedy a ...

Core Core Data Structures Structure Purpose LiveIntervals Stores live ranges for every virtual register LiveRegMatrix Tracks physical-to-virtual mappings and interference PriorityQueue Heap-based queue ordered by Priority VirtRegMap Final virtual → physical assignment EvictAdvisor Decides whether evicting an existing allocation i ...

A. Primary Task Approach The string is invalid in the following cases: Length ≤ 2. Does not start with "10". The substring after "10" converts to an integer less than 2, or has leading zeros. #include <bits/stdc++.h> using namespace std; using i64 = long long; void solve() { string s; cin >> s; if ...

Problem A: Check multiples of 4 Given three digits a, b, c (most significant first), form the three-digit number n = 100*a + 10*b + c. Determine whether n is divisible by 4. int a, b, c; cin >> a >> b >> c; int n = a * 100 + b * 10 + c; cout << (n % 4 == 0 ? "YES" : "NO") << "\n"; Pro ...