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比如 100 元红包 10 个人抢，以下算法：先随机生成 10 个数，之和等于红包金额，然后依次从这个列表随机取一个
，取了后移除，在随机取一个，这样算法是否合理，公平不呢
def hongbao(amount=5, count=3):
s = []
sum = 0
min = 0.01
max = float(amount) - count * 0.01
num = 0
for i in range(count-1):
num = round(random.uniform(0, float(amount) - sum), 2)
while num < min or num > max:
num = round(random.uniform(0, float(amount) - sum), 2)
sum += num
s.append(num)
s.append(round(float(amount) - sum, 2))
# return s
for i in range(count):
temp = random.choice(s)
print("第{0}个人抢的红包是{1}".format(i + 1, temp))
s.remove(temp)
hongbao(100, 10)
测试有问题，问题在哪里呢
Supplement 1 · Oct 15, 2018
def hongbao(amount=5, count=3):
s = []
sum = 0
min = 0.01
max = float(amount) - (count-1) * 0.01
for i in range(count-1):
num = round(random.uniform(0, float(amount) - sum), 2)
while num < min or num > max:
num = round(random.uniform(0, float(amount) - sum), 2)
sum += num
s.append(num)
s.append(round(float(amount) - sum, 2))
# return s
for i in range(count):
temp = random.choice(s)
print("第{0}个人抢的红包是{1}".format(i + 1, temp))
s.remove(temp)
hongbao(100, 10)
5 replies • 2018-10-15 22:26:51 +08:00
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2
crab Oct 15, 2018
二倍均值法
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4
beny2mor Oct 15, 2018
我看了下那些 stackoverflow 的回复，
最简单的是生成 10 个随机数，然后按照比例来分 100 元
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5
ropon Oct 15, 2018
整理思路如下：
#算法一
#################################################################
def hongbao(amount=5, count=3):
s = []
min = 1 #分
amount = amount * 100 #单位分
max = amount - (count-1) #单位分
for i in range(count-1):
# num = random.randint(0, amount)
num = random.uniform(0, int(amount))
while num < min or num > max:
num = random.uniform(0, int(amount))
amount -= num
s.append(round((num / 100), 2))
s.append(round((amount / 100), 2))
random.shuffle(s)
return s
print(hongbao(1, 6))
##############################################################
算法二
#############################################################
# 0--|----|--------|-----------|-------20
# 0 2 4 10 16 20
def hongbao(amount=10, count=5):
ret = random.sample(range(1, amount * 100), count - 1) #分为单位
ret.extend([0, amount * 100]) #追加多个值 0 amount * 100
ret.sort() #排序
# return [((ret[i+1] - ret[i]) / 100 for i in range(num))] # 列表生产式
for i in range(count):
yield (ret[i+1] - ret[i]) / 100
res = hongbao(10, 6)
for i in res:
print(i)
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