How to divide number into integer pieces that are each a multiple of n?

Question

Had a hard time coming up with a concise title for this. I'm sure there are terms for what I want to accomplish and there is no doubt a common algorithm to accomplish what I'm after - I just don't know about them yet.

I need to break up a number into n pieces that are each a multiple of 50. The number is itself a multiple of 50. Here is an example: Divide 5,000 by 3 and end up with three numbers that are each multiples of 50:

• 1,650
• 1,700
• 1,650

I also would like to have the numbers distributed so that they flip back and forth, here is an example with more numbers to illustrate this: Divide 5,000 by 7 and end up with 7 numbers that are each multiples of 50:

• 700
• 750
• 700
• 750
• 700
• 700
• 700

Note that in the above example I'm not worried that the extra 50 is not centered in the series, that is I don't need to have something like this:

• 700
• 700
• 750 <--- note the '50s' are centered
• 700
• 750 <--- note the '50s' are centered
• 700
• 700

Hopefully I've asked this clearly enough that you understand what I want to accomplish.

Update: Here is the function I'll be using.

var number = 5000;
var n = 7;
var multiple = 50;

var values = getIntDividedIntoMultiple(number, n, multiple)

function getIntDividedIntoMultiple(dividend, divisor, multiple)
{
    var values = [];
    while (dividend> 0 && divisor > 0)
    {
        var a = Math.round(dividend/ divisor / multiple) * multiple;
        dividend -= a;
        divisor--;
        values.push(a);
    }

    return values;
}

Solution 1

var number = 5000;
var n = 7;

var values = [];
while (number > 0 && n > 0) {
    var a = Math.floor(number / n / 50) * 50;
    number -= a;
    n--;
    values.push(a);
}  // 700 700 700 700 700 750 750

Edit

You can alternate Math.floor and Math.ceil to obtain the desired result:

while (number > 0 && n > 0) {
    if (a%2 == 0)
        a = Math.floor(number / n / 50) * 50;
    else
        a = Math.ceil(number / n / 50) * 50;
    number -= a;
    n--;
    values.push(a);
}  // 700 750 700 750 700 700 700

Solution 2

// i - an integer multiple of k
// k - an integer
// n - a valid array length
// returns an array of length n containing integer multiples of k
// such that the elements sum to i and the array is sorted,
// contains the minimum number of unique elements necessary to
// satisfy the first condition, the elements chosen are the
// closest together that satisfy the first condition.
function f(i, k, n) {
  var minNumber = (((i / k) / n) | 0) * k;
  var maxNumber = minNumber + k;
  var numMax = (i - (minNumber * n)) / k;
  var nums = [];
  for (var i = 0; i < n - numMax; ++i) {
    nums[i] = minNumber;
  }
  for (var i = n - numMax; i < n; ++i) {
    nums[i] = maxNumber;
  }
  return nums;
}

So your second example would be

f(5000, 50, 7)

which yields

[700,700,700,700,700,750,750]

Solution 3

Let a be your starting number, k - number of parts you want to divide to.
Suppose, that b = a/n.
Now you want to divide b into k close integer parts.

• Take k numbers, each equal to b/k (integer division).
• Add 1 to first b%k numbers.
• Multiply each number by n.

Example: a = 5000, n = 50, k = 7.
b = 100
Starting series {14, 14, 14, 14, 14, 14, 14}
Add 1 to first 2 integers {15, 15, 14, 14, 14, 14, 14}.
Multiply by 50 {750, 750, 700, 700, 700, 700, 700}.

Solution 4

I see your problem as basically trying to divide a sum of money into near-equal bundles of bills of a certain denomination.

For example, dividing 10,000 dollars into 7 near-equal bundles of 50-dollar bills.

function getBundles(sum, denomination, count, shuffle)
{
  var p = Math.floor(sum / denomination);
  var q = Math.floor(p / count);
  var r = p - q * count;

  console.log(r + " of " + ((q + 1) * denomination)
      + " and " + (count - r) + " of " + (q * denomination));

  var b = new Array(count);
  for (var i = 0; i < count; i++) {
    b[i] = (r > 0 && (!shuffle || Math.random() < .5 || count - i == r)
        ? (--r, q + 1) : q)
      * denomination;
  }

  return b;
}

// Divide 10,000 dollars into 7 near-equal bundles of 50-dollar bills
var bundles = getBundles(10000, 50, 7, true);

console.log("bundles: " + bundles);

Output:

4 of 1450 and 3 of 1400
bundles: 1400,1450,1450,1400,1450,1400,1450

If the last argument shuffle is true, it distributes the extra amount randomly between the bundles.

Solution 5

Your problem is the same as dividing a number X into N integer pieces that are all within 1 of each other (just multiply everything by 50 after you've found the result). Doing this is easy - set all N numbers to Floor(X/N), then add 1 to X mod N of them.

View more solutions

Comments

[scubasteve]

Nice examples, how would you extend it to toggle the round up and round down so that you have 700, 750, 700, 750, etc. ?

[Tadeck]

Wrong language. At least it shows how much code you need to type in Java to do a simple thing ;) (no offense)

[scubasteve]

Yes, that works well. I was just stepping through the code in the debugger and watching how it worked and I get it.

[scubasteve]

Nice and clean, provided a JS example which wasn't needed but is appreciated.

[scubasteve]

Nice to see a different approach, not exactly what I wanted as the 750 values are back loaded consecutively.

[scubasteve]

At first it "wooosh!" over my head w/ the terminology but after playing with some code I see what you're saying now ;)

[scubasteve]

I wasn't saying it as a negative, was just commenting that the extra effort for a JS example was appreciated considering you first supplied a c example.

[Azaz Khan]

multiple is undefined