随机密码生成

• 以整数17为随机数种子，获取用户输入整数N为长度，产生3个长度为N位的密码，密码的每位是一个数字。每个密码单独一行输出。

import random

def genpwd(length):
    a = 10**(length-1)
    b = 10**length - 1
    return "{}".format(random.randint(a, b))

length = eval(input())
random.seed(17)
for i in range(3):
    print(genpwd(length))

连续质数计算

• ‪‬‪‬‪‬‪‬‪‬‮‬‫‬‮‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‫‬‮‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‮‬‪‬‪‬‪‬‪‬‪‬‮‬‫‬‭‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬获得用户输入数字N，计算并输出从N开始的5个质数，单行输出，质数间用逗号,分割。‪‬‪‬‪‬‪‬‪‬‮‬‫‬‮‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‪‬‪‬‪‬‪‬‪‬‪‬‮‬‫‬‮‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‮‬‪‬‪‬‪‬‪‬‪‬‮‬‫‬‭‬‪‬‪‬‪‬‪‬‪‬‮‬‪‬‫‬

• 注意：需要考虑用户输入的数字N可能是浮点数，应对输入取整数；最后一个输出后不用逗号。

def prime(m):
    for i in range(2,m):
        if m % i == 0:
            return False
    return True

n = eval(input())
a = int(n)
a = a+1 if a < n else a
count = 5
while count > 0:
    if prime(a):
        if count > 1:
            print(a, end=",")
        else:
            print(a, end="")
        count -= 1 
    a += 1

---

七段数码管绘制

import turtle, time

def drawGap(): #绘制数码管间隔
    t.penup()
    t.fd(5)
    
def drawLine(draw):   #绘制单段数码管
    drawGap()
    t.pendown() if draw else t.penup()
    t.fd(40)
    drawGap()
    t.right(90)
    
def drawDigit(d): #根据数字绘制七段数码管
    drawLine(True) if d in [2,3,4,5,6,8,9] else drawLine(False)
    drawLine(True) if d in [0,1,3,4,5,6,7,8,9] else drawLine(False)
    drawLine(True) if d in [0,2,3,5,6,8,9] else drawLine(False)
    drawLine(True) if d in [0,2,6,8] else drawLine(False)
    t.left(90)
    drawLine(True) if d in [0,4,5,6,8,9] else drawLine(False)
    drawLine(True) if d in [0,2,3,5,6,7,8,9] else drawLine(False)
    drawLine(True) if d in [0,1,2,3,4,7,8,9] else drawLine(False)
    t.left(180)
    t.penup()
    t.fd(20)
    
def drawDate(date):
    t.pencolor("red")
    for i in date:
            drawDigit(eval(i))
            
def main():
    t.setup(800, 350, 200, 200)
    t.penup()
    t.fd(-300)
    t.pensize(5)
    drawDate(time.strftime('%Y%m%d',time.gmtime()))
    t.done()
    
main()

科赫雪花小包裹

import turtle
def koch(size, n):
    if n == 0:
        turtle.fd(size)
    else:
        if angle in [0, 60, -120, 60]:
            turtle.left(angle)
            koch(size/3, n-1)

def main(level):
    turtle.setup(600,600)
    turtle.penup()
    turtle.goto(-200, 100)
    turtle.pendown()
    turtle.pensize(2)
    koch(400,level)
    turtle.right(120)
    koch(400, level)
    turtle.right(120)
    koch(400, level)
    turtle.hideturtle()
    turtle.done()    
try:
    level = eval(input("请输入科赫曲线的阶: "))
    main(level)
except:
    print("输入错误")

任意累积

def cmul(a, *b):
    m = a
    for i in b:
        m *= i
    return m

print(eval("cmul({})".format(input())))

# 无限制数量函数定义的方法，其中b在函数cmul中表达除了a之外的所有输入参数

斐波那契数列计算

def fbi(n):
    if n == 1 or n == 2:
        return 1
    else:
        return fbi(n-1)+fbi(n-2)

n = eval(input())
print(fbi(n))

汉诺塔实践

steps = 0
def hanoi(src, des, mid, n):
    global steps
    if n == 1:
        steps += 1
        print("[STEP{:>4}] {}->{}".format(steps, src, des))
    else:
        hanoi(src, mid, des, n-1)
        steps += 1
        print("[STEP{:>4}] {}->{}".format(steps, src, des))
        hanoi(mid, des, src, n-1)
        
N = eval(input())
hanoi("A", "C", "B", N)