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Coding workflows · Largest-remainder allocation
03
A concrete starting point
Allocate up to 10000 integer seats across at most 1000 nonnegative integer weights using largest remainders. Check zero weights and tied remainders with input-order tie-breaking, rejecting a positive allocation with zero total weight before division.
Define the change
Describe the input, expected output and boundaries that must stay intact.
Inspect the work
Review the proposed edits and any commands before relying on their result.
Check the outcome
Compare actual behavior with the expected result. If a check fails, investigate it before applying more changes.
Largest-remainder allocation
A concrete starting point
def allocate(weights, seats):
if type(weights) not in (list, tuple) or len(weights) > 1000:
raise ValueError("weights.container")
weights = tuple(weights)
if type(seats) is not int or not 0 <= seats <= 10000:
raise ValueError("seats.range")
if len(weights) > 1000 or any(type(w) is not int or w < 0 for w in weights):
raise ValueError("weights.invalid")
total = sum(weights)
if total == 0:
if seats:
raise ValueError("weight.zero")
return [0] * len(weights)
result = [w * seats // total for w in weights]
remaining = seats - sum(result)
order = sorted(range(len(weights)), key=lambda i: (-(weights[i] * seats % total), i))
for i in order[:remaining]:
result[i] += 1
return resultThe example is a starting point for a real project, not a claim that generated code is already tested. Adjust paths, dependencies and expected behavior to match your application.