Deficient Numbers in C++

A deficient number is a positive integer where the sum of its proper divisors (excluding the number itself) is less than the number. For example, 8 is deficient because its divisors (1, 2, 4) sum to 7, which is less than 8.

Input: 10

Output: Deficient

Input: 12

Output: Not Deficient

Explanation:

For the number 10, its proper divisors are 1, 2, and 5. Adding them gives a sum of 8, which is less than 10, making 10 a deficient number. For 12, its proper divisors are 1, 2, 3, 4, and 6, with a sum of 16, which is greater than 12, so 12 is not deficient.

Approach 1: Using Divisor Sum Comparison:

Algorithm: If a number is deficient

Step 1: Understand the Problem: Identify the goal: check whether a given number n is deficient. To do this, calculate the sum of its proper divisors and compare it with n. A number is deficient if the sum is less than n.

Step 2: Define Deficient Number: A deficient number is a positive integer where the sum of its proper divisors (excluding the number itself) is less than the number. The proper divisors are all integers less than the number that divide it evenly, without including the number itself. For example, 8 is deficient because the sum of its divisors (1, 2, 4) is 7, which is less than 8.

Step 3: Input the Number: Start with an input number n. This number should be a positive integer, as deficiency is only defined for positive integers.

Step 4: Initialize Variables: Set up a variable sum to store the cumulative total of the divisors. Initialize sum=0 before starting the divisor calculation.

Step 5: Find Proper Divisors: Proper divisors of a number are all integers less than n that divide n evenly (remainder is 0). Loop through all integers from 1 to n-1.

For each number i, check if n mod i=0 (i.e., i divides n without remainder). If true, add i to sum.

Step 6: Calculate Sum of Divisors: At the end of the loop, sum will hold the total value of all proper divisors of n.

Step 7: Compare the Sum with n: If sum<n, then n is a deficient number because the sum of its proper divisors is smaller than n. Otherwise, n is not a deficient number.

Step 8: Output the Result: Print the result based on the comparison:"Deficient" if sum<n. Otherwise, "Not Deficient".

Algorithm: If a number is not deficient

Step 1: Understand the Problem: A number is not deficient if the sum of its proper divisors (excluding the number itself) is greater than or equal to the number. The task is to check if this condition holds for a given number n.

Step 2: Input the Number: Take a positive integer n as the input. Only positive integers are valid for this determination.

Step 3: Initialize Variables: Set up a variable sum to accumulate the total of proper divisors. Initialize it to zero: sum=0

Step 4: Identify Proper Divisors: Proper divisors are all integers smaller than n that divide n without a remainder. Use a loop to iterate through numbers from 1 to n-1.

For each number i, check if n modi=0. If the condition is true, i is a proper divisor. Add i to sum.

Step 5: Calculate the Sum of Divisors: As the loop completes, sum will hold the sum of all proper divisors of n.

Step 6: Compare sum with n: If sum≥n, the number n is not deficient because its divisors are sufficient or exceed n. If sum<n, the number is deficient.

Step 7: Output the Result: Display the result: "Not Deficient" if sum≥n. "Deficient" otherwise.

Program:

Output:

10 is Deficient
12 is Not Deficient   

Complexity Analysis:

Time Complexity:

The time complexity of determining whether a number is deficient is O(n), where n is the input number. This is because we need to check each integer from 1 to n-1 to find the divisors. Thus, the process involves iterating up to n.

Space Complexity:

The space complexity of determining whether a number is deficient is O(1), since the algorithm only requires a fixed amount of extra space. It uses a single variable to sum divisors and does not store any large data structures, regardless of the input size.

Approach: Efficient Divisor Calculation (Using √n)

The above approach loops through all numbers from 111 to n-1n-1n-1 to find proper divisors, which has a time complexity of O(n)O(n)O(n). A more efficient method can reduce this complexity by considering the divisors in pairs and iterating only up to n\sqrt{n}n.