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

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 time
def cal(sum,count,value):
	if sum == 200:
		count += 1
		return
	if sum > 200 :
		return
	for 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=0
cal(0,count,num)
end = time.time()
print count
print 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 1
	else:
		coins = sorted(coins)
		largest = coins[-1]
		uses = value / largest
		total = 0
		for i in range(uses+1):
			total += cal(value-largest*i, coins[:-1])
		return total

print cal(200, [1,2,5,10,20,50,100,200])

第二种 贪心凑极值 AC

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