## The Game

Consider a points-based game with the following rules:

- Players spend money within bucketed intervals of time (e.g. 1 interval = 1 day from 00:00 to 23:59)
- Money spent in one interval is split up evenly and spent over N intervals.
- A player is active within an interval if they have money spent during it (including money spread)
- Every interval, 1 point is split among active players, proportional to how much each spent in that interval.
- If no players are active within an interval, no point is distributed.
- If a player is inactive for one or more intervals, their points reset to zero. 
- N is fixed for all players of the game. 

## The challenge
Given the current interval, we should be able to compute the number of active points for each of the previous (completed) N intervals. This algorithm can not have a storage overhead exceeding O(C*N), where N is the number of intervals a single spend is spread across. C is a constant factor.

## Examples

Suppose N=4

### Just Alice
Alice wishes to spend $1. The money is spent as follows:

Interval        1       2       3       4       5       6       7       8
Alice           $0.25   $0.25   $0.25   $0.25   $0      $0      $0      $0

If Alice is the only player, the point totals at the end of each interval are:

Interval        1       2       3       4       5       6       7       8
Alice           1       2       3       4       0       0       0       0

Totals          1       2       3       4       0       0       0       0

### Alice, then Bob

Now, suppose Bob spends $2 in the second interval, as follows:

Interval        1       2       3       4       5       6       7       8
Alice           $0.25   $0.25   $0.25   $0.25   $0      $0      $0      $0
Bob             $0      $0.50   $0.50   $0.50   $0.50   $0      $0      $0

The new point totals are as follows:

Interval        1       2       3       4       5       6       7       8
Alice           1       1.33    1.66    2       0       0       0       0
Bob             0       0.66    1.33    2       3       0       0       0

Totals          1       2       3       4       3       0       0       0

### Alice Renews

Now, suppose Alice renews with an additional $1 starting in the fourth interval:

Interval        1       2       3       4       5       6       7       8
Alice           $0.25   $0.25   $0.25   $0.50   $0.25   $0.25   $0.25   $0
Bob             $0      $0.50   $0.50   $0.50   $0.50   $0      $0      $0

The new point totals are as follows:

Interval        1       2       3       4       5       6       7       8
Alice           1       1.33    1.66    2.16    2.50    3.50    4.50    0
Bob             0       0.66    1.33    1.83    2.50    0       0       0

Totals          1       2       3       4       5       3.50    4.50    0