Integer-cents plus divmod remainder makes expense splits exact
When splitting money, convert to integer cents with round(amount*100), then use divmod(total, n_sharers): each sharer pays the quotient and the first remainder sharers pay one extra cent. Shares then sum exactly to the total, so group balances always net to zero. Verified with a 5000-case fuzz over random groups/expenses: greedy settlement (biggest debtor pays biggest creditor) then zeroes everyone out in at most n-1 transfers.
Checked the round(amount*100) step: for amounts that really have <=2 decimals it is exact. round(float(s)*100) == cents for all 0..2e6 cents plus 2e6 random cents up to 1e15 (i.e. up to 1e13 currency units), 0 mismatches; the product error is ~|x|*100*2^-53, far below 0.5. It breaks once inputs carry sub-cent digits (computed prices, tax, FX): round(1.005*100)=100, round(2.675*100)=268, round(0.125*100)=12 (float representation plus Python's round-half-even). If that can happen, convert with Decimal(str_amount).quantize(Decimal('0.01'), ROUND_HALF_UP) and pick the rounding rule on purpose. divmod is also fine for refunds: divmod(-1000, 3) = (-334, 2), so -333,-333,-334 still sums to -1000.