Perfect number in C
This C program checks whether a number is a perfect number or not.
Perfect number #
A perfect number is a number that is equal to the sum of its proper divisors. For example, the sum of the proper divisors of 6 is 1 + 2 + 3 = 6, which is a perfect number. The sum of the proper divisors of 28 is 1 + 2 + 4 + 7 + 14 = 28, which is also a perfect number.
Algorithm #
- Take the number as input.
- Loop through all the numbers from 1 to the number and check whether the number is a divisor of the number.
- If the number is a divisor of the number, add it to the sum.
- If the sum is equal to the number, print "The number is a perfect number".
- If the sum is not equal to the number, print "The number is not a perfect number".
Problem description #
This program will ask the user for a number and then check whether the number is a perfect number or not.
Examples #
Input: 6 Output: The number is a perfect number
Input: 34 Output: The number is not a perfect number
Input: 100 Output: The number is not a perfect number
Naive Approach #
In this approach we'll not make use of any functions. We'll just loop through all the numbers from 1 to the number and check whether the number is a divisor of the number.
Approach #
- Take the number as input.
- Store the number in a variable.
- Inside for loop, take the number and divide it by i and store the result in a variable.
- If the result is equal to the number, add the result to a variable.
- Repeat the process until i is equal to the number.
- If the sum of the variable is equal to the number, print the number is a perfect number.
- Otherwise, print the number is not a perfect number.
- Repeat the process until the number is equal to 0.
- Print the variable.
Program/Source code #
/* Program to check whether a number is a perfect number or not */
#include <stdio.h>
int main(void)
{
	int number, i, sum = 0;
	printf("Enter the number: ");
	scanf("%d", &number);
	for (i = 1; i < number; i++) {
		if (number % i == 0)
			sum = sum + i;
	}
	if (sum == number)
		printf("The number is a perfect number");
	else
		printf("The number is not a perfect number");
}
Explanation #
The program will ask the user for a number. Then it will loop through all the numbers from 1 to the number and check whether the number is a divisor of the number. If the number is a divisor of the number, it will add it to the sum. If the sum is equal to the number, it will print the number is a perfect number. Otherwise, it will print the number is not a perfect number.
Time complexity #
The time complexity of this program is O(n).
Space Complexity #
The space complexity of this program is O(1).
Output #
> ./perfect-number
Enter the number: 28
The number is a perfect number
> ./perfect-number
Enter the number: 34
The number is not a perfect number
Approach using function #
In this approach we will make use of a function. We'll take the number as input and then call the function.
Methods used #
- int is_perfect_number(int number)- This function will check whether the number is a perfect number or not.
- int is_divisor(int number, int divisor)- This function will check whether the number is a divisor of the number.
- int sum_of_divisors(int number)- This function will return the sum of the divisors of the number.
Approach #
- Take the number as input.
- Call the function is_perfect_numberand pass the number as input.
- If the function returns 1, print the number is a perfect number.
- Otherwise, print the number is not a perfect number.
Program/Source code #
/* Program to check whether a number is a perfect number or not */
#include <stdio.h>
int is_divisor(int number, int divisor)
{
    if (number % divisor == 0)
        return 1;
    else
        return 0;
}
int sum_of_divisors(int number)
{
    int i, sum = 0;
    for (i = 1; i < number; i++) {
        if (is_divisor(number, i))
            sum = sum + i;
    }
    return sum;
}
int is_perfect_number(int number)
{
    if (sum_of_divisors(number) == number)
        return 1;
    else
        return 0;
}
int main(void)
{
    int number;
    printf("Enter the number: ");
    scanf("%d", &number);
    if (is_perfect_number(number))
        printf("The number is a perfect number");
    else
        printf("The number is not a perfect number");
}
Explanation #
The program asks the user for a number. Then it calls the function is_perfect_number and passes the number as input. Inside the function another function sum_of_divisors is called and the number is passed as input. Inside this function, we'll loop through all the numbers from 1 to the number and check whether the number is a divisor of the number. If the number is a divisor of the number, it will add it to the sum. If the sum is equal to the number, it will print the number is a perfect number. Otherwise, it will print the number is not a perfect number.
Time complexity #
The time complexity of this program is O(n).
Space Complexity #
The space complexity of this program is O(1).
Output #
> ./perfect-number-fun
Enter the number: 99
The number is not a perfect number
> ./perfect-number-fun
Enter the number: 33550336
The number is a perfect number