Homework

April 3, 2020

[ I continue working from home. It is interesting that my commute has gone from a daily 75-minute round-trip by car to a 75-foot round-trip by foot, so I think I should have more energy in the evening, but the opposite is true; I am exhausted by the end of the day, just from sitting in front of my laptop. ]

Today’s exercise is somebody’s homework:

Given positive integers C and N, find N numbers that sum up to C and the difference between the highest and the lowest of these number should not be more than one. For example: with C = 26 and N = 7, the desired output is [4 4 4 4 4 3 3].

Your task is to write a program to solve the student’s homework. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Pages: 1 2

Posted by programmingpraxis
Filed in Exercises
9 Comments »

9 Responses to “Homework”

  1. James Curtis-Smith said
    April 3, 2020 at 9:22 AM

    A perl “1-liner”:

    [sourceode lang=”perl”]
    print “@{[ hw(@ARGV) ]}\n”;

    sub hw {
    my($C,$N) = @_;
    return map { int($C/$N) + ($_<$C%$N) } 0..($N-1);
    }
    [/sourcecode]

  2. chaw said
    April 3, 2020 at 10:44 AM

    A simple solution in standard R7RS Scheme and a couple of popular
    helpers.

    (import (scheme base)
        (scheme write)
        (only (srfi 8) receive)
        (only (srfi 1) list-tabulate))

(define (homework c n)
  (receive (q r) (floor/ c n)
    (list-tabulate n (lambda (i) (if (< i r) (+ q 1) q)))))

(display (homework 26 7))
(newline)

    Output:

    (4 4 4 4 4 3 3)

  3. Daniel said
    April 3, 2020 at 5:21 PM

    Here’s a solution in Python.

    def homework(C, N):
    q, r = divmod(C, N)
    return [q + (x < r) for x in range(N)]

print(homework(26, 7))

    Output:

    [4, 4, 4, 4, 4, 3, 3]
  4. Globules said
    April 4, 2020 at 5:23 AM

    Here’s a Haskell version.

    -- A list of n numbers whose sum is c, each of which differs by at most 1 from
-- all the others.  We assume n >= 1.
sumTo :: Int -> Int -> [Int]
sumTo c n = let (q, r) = c `divMod` n in replicate r (q+1) ++ replicate (n-r) q

test :: Int -> Int -> IO ()
test c n = do
  let xs = sumTo c n
  putStrLn $ show c ++ " " ++ show n ++ ": " ++
             show xs ++ " sum to " ++ show (sum xs)

main :: IO ()
main = do
  test 26  7
  test 18  3
  test  5 10
  test  0  2
  -- Negative numbers work, too.
  test (-8) 3

    $ ./homework
26 7: [4,4,4,4,4,3,3] sum to 26
18 3: [6,6,6] sum to 18
5 10: [1,1,1,1,1,0,0,0,0,0] sum to 5
0 2: [0,0] sum to 0
-8 3: [-2,-3,-3] sum to -8
  5. Sochima Biereagu, KodeJuice said
    April 5, 2020 at 9:03 AM 
    def homework(N, C):
    for k in range(1, N+1):
        a = (k-N+C)/N
        if a==int(a):
            return [int(a)]*k + [int(a)+1]*(N-k)

homework(7, 26) #=> [3, 3, 4, 4, 4, 4, 4]
homework(2, 25) #=> [12, 13]
homework(3, -8)  #=> [-3, -3, -2]
  7. Dave said
    April 10, 2020 at 3:34 AM

    def homework(C, N):
    if C % N == 0:
    answer = [C / N] * N
    return answer
    else:
    remainder = C % N
    C -= remainder
    answer = [C / N] * N
    for i in range(remainder):
    answer[i] += 1
    return answer

    print(solver(26,7))

  9. Jan Van lent said
    April 30, 2020 at 2:28 PM

    Using Python with + for concatenation of lists and * for repetition ([1] * 3 == [1, 1, 1]).

    [C//N+1]*(C%N) + [C//N]*(N-C%N)

Leave a comment