How to round floats to integers while preserving their sum?

Let's say I have an array of floating point numbers, in sorted (let's say ascending) order, whose sum is known to be an integer N. I want to "round" these numbers to integers while leaving their sum unchanged. In other words, I'm looking for an algorithm that converts the array of floating-point numbers (call it fn) to an array of integers (call it in) such that:

1. the two arrays have the same length
2. the sum of the array of integers is N
3. the difference between each floating-point number fn[i] and its corresponding integer in[i] is less than 1 (or equal to 1 if you really must)
4. given that the floats are in sorted order (fn[i] <= fn[i+1]), the integers will also be in sorted order (in[i] <= in[i+1])

Given that those four conditions are satisfied, an algorithm that minimizes the rounding variance (sum((in[i] - fn[i])^2)) is preferable, but it's not a big deal.

Examples:

[0.02, 0.03, 0.05, 0.06, 0.07, 0.08, 0.09, 0.1, 0.11, 0.12, 0.13, 0.14]
    => [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
[0.1, 0.3, 0.4, 0.4, 0.8]
    => [0, 0, 0, 1, 1]
[0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1]
    => [0, 0, 0, 0, 0, 0, 0, 0, 0, 1]
[0.4, 0.4, 0.4, 0.4, 9.2, 9.2]
    => [0, 0, 1, 1, 9, 9] is preferable
    => [0, 0, 0, 0, 10, 10] is acceptable
[0.5, 0.5, 11]
    => [0, 1, 11] is fine
    => [0, 0, 12] is technically not allowed but I'd take it in a pinch

To answer some excellent questions raised in the comments:

• Repeated elements are allowed in both arrays (although I would also be interested to hear about algorithms that work only if the array of floats does not include repeats)
• There is no single correct answer - for a given input array of floats, there are generally multiple arrays of ints that satisfy the four conditions.
• The application I had in mind was - and this is kind of odd - distributing points to the top finishers in a game of MarioKart ;-) Never actually played the game myself, but while watching someone else I noticed that there were 24 points distributed among the top 4 finishers, and I wondered how it might be possible to distribute the points according to finishing time (so if someone finishes with a large lead they get a larger share of the points). The game tracks point totals as integers, hence the need for this kind of rounding.

For the curious, here is the test script I used to identify which algorithms worked.

One option you could try is "cascade rounding".

For this algorithm you keep track of two running totals: one of floating point numbers so far, and one of the integers so far. To get the next integer you add the next fp number to your running total, round the running total, then subtract the integer running total from the rounded running total:-

number  running total   integer integer running total
   1.3       1.3          1           1
   1.7       3.0          2           3
   1.9       4.9          2           5
   2.2       8.1          3           8
   2.8      10.9          3          11
   3.1      14.0          3          14

Here is one algorithm which should accomplish the task. The main difference to other algorithms is that this one rounds the numbers in correct order always. Minimizing roundoff error.

The language is some pseudo language which probably derived from JavaScript or Lua. Should explain the point. Note the one based indexing (which is nicer with x to y for loops. :p)

// Temp array with same length as fn.
tempArr = Array(fn.length)

// Calculate the expected sum.
arraySum = sum(fn)

lowerSum = 0
-- Populate temp array.
for i = 1 to fn.lengthf
    tempArr[i] = { result: floor(fn[i]),              // Lower bound
                   difference: fn[i] - floor(fn[i]),  // Roundoff error
                   index: i }                         // Original index

    // Calculate the lower sum
    lowerSum = lowerSum + tempArr[i].result
end for

// Sort the temp array on the roundoff error
sort(tempArr, "difference")

// Now arraySum - lowerSum gives us the difference between sums of these
// arrays. tempArr is ordered in such a way that the numbers closest to the
// next one are at the top.
difference = arraySum - lowerSum

// Add 1 to those most likely to round up to the next number so that
// the difference is nullified.
for i = (tempArr.length - difference + 1) to tempArr.length
    tempArr.result = tempArr.result + 1
end for

// Optionally sort the array based on the original index.
array(sort, "index")