生成1-1000的随机数，怎么做到如果随机出 非1000的在前面补0（比如 789=0789； 53=0053）

Question

import random
alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
word1 = random.randrange(0,len(alphabet))
word2 = random.randrange(0,len(alphabet))
word3 = random.randrange(0,len(alphabet))
word4 = random.randrange(0,len(alphabet))
number = random.randrange(1,1000)
title = "Password Gernerator"
password = (alphabet[word1] + alphabet[word2] + alphabet[word3] +
alphabet[word4] )
zero_extend = "0" * (len(password)-len(str(number)))
print((len(title)+2) * "*")
print("*",title,"*",sep="")
print((len(title)+2) * "*")
print()

print("Your password is ",password,zero_extend,number,sep="")

Answer 1

线先判断数字能否被1000整除后判断会不会被100整除商大于一说明是个三位数降数字拉入数组后两位前两位置零这个好理解