Pet Store Dogs

Calculate the number of ways to arrange N dogs in cages where aggressive dogs must be alone and passive dogs can share cages in pairs

Understand the Problem

Problem Statement

There are N dogs in a pet store. The pet store wants to keep the N dogs in cages. There can be two dogs in each cage. A dog can be either passive or aggressive.

If the dog is passive, it can be clubbed with another passive dog. If the dog is aggressive, it has to be alone in a cage. The program must print the number of ways W to put the N dogs in the cages as the output.

Note: The number of cages in the pet store is sufficient to keep N dogs according to the given condition.

Constraints

1 ≤ N ≤ 1000 (for practical implementation with large number handling)
The answer can grow very large, so appropriate data types (BigInteger, etc.) should be used for larger values
Dogs are distinguishable (the arrangement matters)

Examples

Example 1

Input

Output

Explanation

For 2 dogs, there are 2 ways: (1) Keep both dogs in separate cages, (2) Keep both dogs together in one cage (since they can be passive).

Example 2

Input

Output

Explanation

For 3 dogs (A, B, C), the arrangements are: (1) All three separate, (2) A with B, C separate, (3) A with C, B separate, (4) B with C, A separate.

Example 3

Input

Output

Explanation

This demonstrates that the number grows rapidly. Using the recurrence relation W(N) = W(N-1) + (N-1) × W(N-2), we get W(10) = 9496.

Solution

#include <stdio.h>
#include <math.h>

long double f(long double a) {
    if (a <= 1)
        return 1;
    return a * f(a - 1);
}

long double calculate_ways(long double a) {
    long double sum = 1;
    long double val = f(a);

    for (int i = 1; i <= a / 2; i++) {
        long double k = val;
        k /= pow(2, i);
        k /= f(a - (2 * i));
        k /= f(i);
        sum += k;
    }

    return sum;
}

int main() {
    long double a;
    scanf("%Lf", &a);
    long double result = calculate_ways(a);
    printf("%.Lf", result);
    return 0;
}

Time:O(N²) - Due to factorial calculations and the main loop

Space:O(N) - For recursive factorial calls

Approach:

The C solution uses a mathematical approach based on combinations. The function f() calculates factorial recursively. The main logic calculates the sum of ways to arrange dogs by considering all possible pairings.

For each possible number of pairs i (from 1 to N/2), it calculates the combinations using the formula: N! / (2^i × (N-2i)! × i!), which represents choosing i pairs from N dogs.

Visual Explanation

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All Problems