projecteuler---->problem=31----Coin sums 无限背包计算可能存在的次数

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Problem 31

In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

It is possible to make £2 in the following way:

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

How many different ways can £2 be made using any number of coins?

import timedef cal(sum,count,value):if sum == 200:count += 1 returnif sum > 200 :returnfor i in range(0,8):sum += value[i]cal(sum,count,value)sum -= value[i]begin = time.time()num={1,2,5,10,20,50,100,200};num=list(num)count=0cal(0,count,num)end = time.time()print countprint end-begin

第一种办法，深搜超时

## 解析：每次从列表中取出最大的coin值，得到取最大值个数的上限#      个数叠加，接下来的用较小的数凑。##def cal(value, coins):if value == 0 or len(coins)==1 :return 1else:coins = sorted(coins)largest = coins[-1]uses = value / largesttotal = 0for i in range(uses+1):total += cal(value-largest*i, coins[:-1])return totalprint cal(200, [1,2,5,10,20,50,100,200])

第二种贪心凑极值 AC

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