Perfect Number

🏠 ⬅️ ➡️

A perfect number is a positive integer that is equal to the sum of its positive divisors, excluding the number itself. A divisor of an integer x is an integer that can divide x evenly.

Given an integer n, return true if n is a perfect number, otherwise return false.

Example 1:

Input: num = 28 Output: true Explanation: 28 = 1 + 2 + 4 + 7 + 14 1, 2, 4, 7, and 14 are all divisors of 28.

Example 2:

Input: num = 7 Output: false

Constraints:

  • 1 <= num <= 108
Note: This problem is from LeetCode.
❌ Compiled
❌ Executed
❌ Correct 
program main
    implicit none

    integer :: n
    logical :: result

    ! Example 1
    n = 28
    result = is_perfect(n)
    if (result) then
        print '(A, I0, A)', 'Perfect number: ', n, '.'
    else
        print '(A, I0, A)', 'Not a perfect number: ', n, '.'
    end if

    ! Example 2
    n = 7
    result = is_perfect(n)
    if (result) then
        print '(A, I0, A)', 'Perfect number: ', n, '.'
    else
        print '(A, I0, A)', 'Not a perfect number: ', n, '.'
    end if

contains

    function is_perfect(n) result(res)
        implicit none
        integer, intent(in) :: n
        logical :: res
        integer :: i

        res = .false.
        do i = 1, n - 1
            if (mod(n, i) == 0) then
                res = res .or. is_perfect(i)
            end if
        end do
        if (n == sum(1, n)) then
            res = .true.
        end if
    end function is_perfect

end program main
❌ Compiled
❌ Executed
❌ Correct 
```
program perfect_number
    implicit none
    integer :: n, i, sum
    logical :: is_perfect_number

    ! read input
    read *, n

    ! initialize sum
    sum = 0

    ! loop through divisors
    do i = 1, n
        if (mod(n, i) == 0) then
            sum = sum + i
        end if
    end do

    ! check if sum is equal to n
    if (sum == n) then
        is_perfect_number = .true.
    else
        is_perfect_number = .false.
    end if

    ! print output
    if (is_perfect_number) then
        print *, "Yes"
    else
        print *, "No"
    end if
end program perfect_number
```
🌐 Data from online sources 
def checkPerfectNumber(num: int) -> bool:
    if num <= 1:
        return False
    sum = 1
    i = 2
    while i * i <= num:
        if num % i == 0:
            sum += i
            if i * i != num:
                sum += num // i
        i += 1
    return sum == num

The algorithm checks if a given number num is a perfect number by iterating through the positive divisors of the number and adding them to a sum. If the number is less than or equal to 1, it returns false as these are not perfect numbers. The loop starts from 2 and goes until the square root of the number. If i is divisible by num, it's added to the sum. Also, check if i * i is not equal to num, we add num / i to the sum, this is because we don't want to add the same divisor twice. In the end, it checks if the calculated sum is equal to the given number. If it is, return true, else return false.

🌐 Data from online sources 
bool checkPerfectNumber(int num) {
    if (num <= 1) return false;
    int sum = 1;
    for (int i = 2; i * i <= num; ++i) {
        if (num % i == 0) {
            sum += i;
            if (i * i != num) sum += num / i;
        }
    }
    return sum == num;
}

The algorithm checks if a given number num is a perfect number by iterating through the positive divisors of the number and adding them to a sum. If the number is less than or equal to 1, it returns false as these are not perfect numbers. The loop starts from 2 and goes until the square root of the number. If i is divisible by num, it's added to the sum. Also, check if i * i is not equal to num, we add num / i to the sum, this is because we don't want to add the same divisor twice. In the end, it checks if the calculated sum is equal to the given number. If it is, return true, else return false.