Fine-tuning CodeLlama-7B for Fortran Code Generation

Given a positive integer n, return the smallest positive integer that is a multiple of both 2 and n.

Example 1:

Input: n = 5 Output: 10 Explanation: The smallest multiple of both 5 and 2 is 10.

Example 2:

Input: n = 6 Output: 6 Explanation: The smallest multiple of both 6 and 2 is 6. Note that a number is a multiple of itself.

Constraints:

1 <= n <= 150

Note: This problem is from LeetCode.

✅ Compiled

✅ Executed

❌ Correct

program main
    implicit none

    integer, parameter :: n = 5
    integer :: result

    result = smallestMultiple(n)
    print '(I0)', result

contains

    function smallestMultiple(n) result(result_)
        implicit none
        integer, value :: n
        integer :: result_

        result_ = 2 * (n - 1) + 1
    end function smallestMultiple
end program main

stdout

✅ Compiled

✅ Executed

❌ Correct

PROGRAM MULTIPLE_OF_2_AND_N
      IMPLICIT NONE

      INTEGER, PARAMETER :: n = 5
      INTEGER :: i, j, k

      ! Initialize the smallest multiple of 2 and n to 0
      i = 0
      j = 0

      ! Loop through all positive integers
      DO k = 1, 150

         ! Check if k is a multiple of 2
         IF (MODULO(k, 2) == 0) THEN
            i = k
         END IF

         ! Check if k is a multiple of n
         IF (MODULO(k, n) == 0) THEN
            j = k
         END IF

         ! If k is a multiple of both 2 and n, break the loop
         IF (MODULO(k, 2) == 0 .AND. MODULO(k, n) == 0) THEN
            EXIT
         END IF

      END DO

      ! Print the smallest multiple of 2 and n
      WRITE (*, *) "The smallest multiple of 2 and ", n, " is ", i * j

      END PROGRAM MULTIPLE_OF_2_AND_N

stdout

 The smallest multiple of 2 and            5  is          100

🌐 Data from online sources

def smallest_multiple(n):
    return n if n % 2 == 0 else n * 2

The algorithm to solve this problem is quite simple. The smallest positive integer that is a multiple of both 2 and n can be determined by checking if n is even or odd. If n is even, it means that it is already a multiple of 2, so the answer should be n. If n is odd, we need to multiply n by 2 to get the smallest multiple that is also divisible by 2, so the answer should be n * 2.

🌐 Data from online sources

int smallestMultiple(int n) {
    return n % 2 == 0 ? n : n * 2;
}

Smallest Even Multiple