1 day ago

ANIRUDH • 0

Delhi

Problem 1: Binary Cafeteria

Solution: Mathematical Approach

This is a combinatorial counting problem with a unique property that simplifies it significantly. The key insight is that since all dessert costs are distinct powers of two, any total amount of money corresponds to a unique subset of desserts due to the uniqueness of binary representation. This transforms the problem from a complex subset sum variant into a straightforward counting exercise.

The logic is as follows:

• Unique Representation: Because costs are {20,21,dots,2k−1}, any total cost C can be formed in at most one way. For example, a cost of 5 (101_2) can only be achieved by buying desserts costing 20 and 22.

• Counting Affordable Costs: The problem reduces to counting how many distinct total costs are possible within the budget n. The available desserts can form any integer sum from 0 (buying nothing) up to 2k−1 (buying all desserts).

• Finding the Limit: The number of affordable options is limited by two factors: the total number of possible subsets (2k) and the budget (n+1 possible costs, from 0 to n). The answer is the smaller of these two values.

• Handling Large Numbers: Since n and k can be very large, 2k can easily overflow standard integer types. However, since n is at most 109, if k is 30 or greater, 2k will always exceed n + 1. This observation allows us to avoid calculating large powers of two.

#include <iostream>
#include <algorithm>

using namespace std;

long long solveBinaryCafeteria(long long n, long long k) {
    if (k >= 60) return n + 1;
    long long totalCosts = 1LL << k;
    return min(n + 1, totalCosts);
}

Complexity Analysis:

• Time Complexity: O(1). The solution involves a single comparison and a bitwise shift, which are constant-time operations.

• Space Complexity: O(1). No additional space proportional to the input size is required.