Вот слегка протестированный код для решения вашей проблемы точно за время, полиномиальное по количеству доступных денег, количеству акций, которые у вас есть, и максимальному количеству акций, которые вы можете купить.
#! /usr/bin/env python
from collections import namedtuple
Stock = namedtuple('Stock', ['id', 'price', 'profit'])
def optimize (stocks, money=10000, max_stocks=8, max_per_stock=2500):
Investment = namedtuple('investment', ['profit', 'stock', 'quantity', 'previous_investment'])
investment_transitions = []
last_investments = {money: Investment(0, None, None, None)}
for _ in range(max_stocks):
next_investments = {}
investment_transitions.append([last_investments, next_investments])
last_investments = next_investments
def prioritize(stock):
# This puts the best profit/price, as a ratio, first.
val = [-(stock.profit + 0.0)/stock.price, stock.price, stock.id]
return val
for stock in sorted(stocks, key=prioritize):
# We reverse transitions so we have not yet added the stock to the
# old investments when we add it to the new investments.
for transition in reversed(investment_transitions):
old_t = transition[0]
new_t = transition[1]
for avail, invest in old_t.iteritems():
for i in range(int(min(avail, max_per_stock)/stock.price)):
quantity = i+1
new_avail = avail - quantity*stock.price
new_profit = invest.profit + quantity*stock.profit
if new_avail not in new_t or new_t[new_avail].profit < new_profit:
new_t[new_avail] = Investment(new_profit, stock, quantity, invest)
best_investment = investment_transitions[0][0][money]
for transition in investment_transitions:
for invest in transition[1].values():
if best_investment.profit < invest.profit:
best_investment = invest
purchase = {}
while best_investment.stock is not None:
purchase[best_investment.stock] = best_investment.quantity
best_investment = best_investment.previous_investment
return purchase
optimize([Stock('A', 100, 10), Stock('B', 1040, 160)])
И вот с крошечной оптимизацией удаления инвестиций, как только мы видим, что дальнейшее добавление акций к нему не может улучшиться. Это, вероятно, будет работать на несколько порядков быстрее, чем старый код с вашими данными.
#! /usr/bin/env python
from collections import namedtuple
Stock = namedtuple('Stock', ['id', 'price', 'profit'])
def optimize (stocks, money=10000, max_stocks=8, max_per_stock=2500):
Investment = namedtuple('investment', ['profit', 'stock', 'quantity', 'previous_investment'])
investment_transitions = []
last_investments = {money: Investment(0, None, None, None)}
for _ in range(max_stocks):
next_investments = {}
investment_transitions.append([last_investments, next_investments])
last_investments = next_investments
def prioritize(stock):
# This puts the best profit/price, as a ratio, first.
val = [-(stock.profit + 0.0)/stock.price, stock.price, stock.id]
return val
best_investment = investment_transitions[0][0][money]
for stock in sorted(stocks, key=prioritize):
profit_ratio = (stock.profit + 0.0) / stock.price
# We reverse transitions so we have not yet added the stock to the
# old investments when we add it to the new investments.
for transition in reversed(investment_transitions):
old_t = transition[0]
new_t = transition[1]
for avail, invest in old_t.items():
if avail * profit_ratio + invest.profit <= best_investment.profit:
# We cannot possibly improve with this or any other stock.
del old_t[avail]
continue
for i in range(int(min(avail, max_per_stock)/stock.price)):
quantity = i+1
new_avail = avail - quantity*stock.price
new_profit = invest.profit + quantity*stock.profit
if new_avail not in new_t or new_t[new_avail].profit < new_profit:
new_invest = Investment(new_profit, stock, quantity, invest)
new_t[new_avail] = new_invest
if best_investment.profit < new_invest.profit:
best_investment = new_invest
purchase = {}
while best_investment.stock is not None:
purchase[best_investment.stock] = best_investment.quantity
best_investment = best_investment.previous_investment
return purchase
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21.02.2018