用Python编写简单的发红包程序:
第一种解法:数轴方法解决
import random
def red_packet(money,num):
money = money * 100 #将钱数转换成分为单位
ret = random.sample(range(1,money),num-1) #在最低钱数1分与总钱数之间生成人数减1个数作为数轴的节点
ret.sort() #对列表进行排序
ret.insert(0,0)
ret.append(money)
for i in range(len(ret)-1):
yield (ret[i+1]-ret[i])/100
ret_g = red_packet(100,10)
for money in ret_g:
print(money)
第二种解法:用概率解决
def red_packet(money,person):
dic_person_money = {}
for i in range(person):
num = random.randint(1,100)
dic_person_money['Person%s' % (i+1)] = num
num_sum = 0
for i in dic_person_money:
num_sum += dic_person_money[i]
for i in dic_person_money:
x = round(dic_person_money[i]/num_sum*money,2)
dic_person_money[i] = '$%s' % x
return dic_person_money
result = red_packet(1,10)
print(result)
用Python设计一个简单的计算器:
import re
def atom_cal(exp):
if '*' in exp:
a,b = exp.split('*')
return str(float(a)*float(b))
elif '/' in exp:
a, b = exp.split('/')
return str(float(a) / float(b)) def format_exp(exp):
exp = exp.replace('--','+')
exp = exp.replace('+-', '-')
exp = exp.replace('-+', '-')
exp = exp.replace('++', '+')
return exp def mul_div(exp):
while 1:
ret = re.search('\d+(\.\d+)?[*/]-?\d+(\.\d+)?',exp)
if ret:
atom_exp = ret.group()
res = atom_cal(atom_exp)
exp = exp.replace(atom_exp,res)
else:return exp def add_sub(exp):
ret = re.findall('[+-]?\d+(?:\.\d+)?',exp)
sum = 0
for i in ret:
sum += float(i)
return sum def cal(exp):
exp = mul_div(exp)
exp = format_exp(exp)
return add_sub(exp) def main(exp):
exp = exp.replace(' ','')
while 1:
ret = re.search('\([^()]+\)',exp)
if ret:
cal_exp = ret.group()
res = str(cal(cal_exp))
exp = exp.replace(cal_exp,res)
else:break
return cal(exp)
s = '1 - 2 * ( (60-30 +(-40/5) * (9-2*5/3 + 7 /3*99/4*2998 +10 * 568/14 )) - (-4*3)/ (16-3*2) )'
print(main(s))