【头歌-Python】Python第五章作业（初级）
第1关：回文素数
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
def is_palindrome(num):
return str(num) == str(num)[::-1]
def generate_palindrome_primes(n):
palindrome_primes = []
num = 2
while len(palindrome_primes) < n:
if is_prime(num) and is_palindrome(num):
palindrome_primes.append(num)
num += 1
return palindrome_primes
if __name__ == "__main__":
n = int(input())
result = generate_palindrome_primes(n)
print(" ".join(map(str, result)))
第2关：反素数
def is_prime(num):
if num < 2:
return False
for i in range(2, int(num**0.5) + 1):
if num % i == 0:
return False
return True
def is_palindrome(num):
if str(num) == str(num)[::-1]:
return False
return True
def generate_palindrome_primes(n):
palindrome_primes = []
num = 2
while len(palindrome_primes) < n:
num1 = int(str(num)[::-1])
if is_prime(num) and is_palindrome(num)and is_prime(num1):
palindrome_primes.append(num)
num += 1
return palindrome_primes
if __name__ == "__main__":
n = int(input())
result = generate_palindrome_primes(n)
print(" ".join(map(str, result)))
第3关：哥德巴赫猜想
def is_prime(n):
"""判断素数的函数,接收一个正整数为参数，参数是素数时返回True，否则返回False"""
if n < 2:
return False
for i in range(2, int(n ** 0.5) + 1):
if n % i == 0:
return False
return True
def goldbach_conjecture(num):
""" 哥德巴赫猜想, 接收一个不小于4的正整数为参数。
当参数为不小于4的偶数时，将其分解为两个素数的加和，按小数+数的格式输出。
有多种组合时全部输出，但不输出重复的组合，例如输出8=3+5，不输出8=5+3。
参数为奇数或小于4时，输出'Data error!'
"""
if num % 2 != 0 or num < 4:
print('Data error!')
a = int(num/2)+1
for i in range(1, a):
if is_prime(i) and is_prime(num - i):
print('{}={}+{}'.format(num, i, num - i))
if __name__ == '__main__':
positive_even = int(input()) # 输入一个正数
goldbach_conjecture(positive_even)第4关：货币转换
a = input()
b = float(input())
str1=''
if a[- 1] != '¥' and a[- 1] != '$' or b < 0.0:
print('Data error!')
elif a[- 1] == '$':
for i in a[:-1]:
if i != ' ':
str1 = str1 + i
str1 = float(str1)
money = float(b * str1)
print('{:.2f}¥'.format(money))
elif a[- 1] == '¥':
for i in a[:-1]:
if i != ' ':
str1 = str1 + i
str1 = float(str1)
money = float(str1 / b)
print('{:.2f}$'.format(money))第5关：个人信息提取
n=input()
t=n.split(" ")
print('姓名：'+t[1])
print('班级：'+t[2])
j = 0
for i in t[4]:
if i=='年':
break
j = j+1
print('出生：'+t[4][j-4:j+1])第6关：身份证号基本信息
a = input()
str1 = a[6:10]
str2 = a[10:12]
str3 = a[12:14]
num = int(a[16])
print('出生：'+str1+'年'+str2+'月'+str3+'日')
if num%2==0:
print('性别：女')
else:
print('性别：男')第7关：各位数字之和为5的数
n = int(input())
for i in range(0, n + 1):
k = i
a = k % 10 + int(k / 10) % 10 + int(k / 100) % 10
if a == 5:
print(i, end=" ")
第8关：字符串长度
print(len(input()))第9关：字符串加密
import string
a = input()
lower = string.ascii_lowercase
upper = string.ascii_uppercase
before = string.ascii_letters
after = lower[3:] + lower[:3] + upper[5:] + upper[:5]
table = "".maketrans(before, after)
print(a.translate(table))
第10关：输出单词
a = input()
t = a.split(" ")
for i in t:
print(i)第11关：字符大小写转换
n =input()
str1 = n.swapcase()
print(str1)第12关：查找指定字符
a = input()
b =input()
index = b.find(a)
if index == -1:
print('Not Found')
else:
print('index = {}'.format(index))第13关：随机密码生成器
import random
import string
n,s = map(int,input().split(','))
random.seed(s)
all_chars = string.digits + string.ascii_letters + '!"#$%&\\()*+,-.'
password = ''.join(random.choice(all_chars) for _ in range(n))
print(password)第14关：模拟生成微软序列号
import random
import string
n = int(input())
s = int(input())
random.seed(s)
all_chars = r'BCEFGHJKMPQRTVWXY2346789'
for i in range(n):
password = ''.join(random.choice(all_chars) for _ in range(5)) + '-' + ''.join(
random.choice(all_chars) for _ in range(5)) + '-' + ''.join(
random.choice(all_chars) for _ in range(5)) + '-' + ''.join(
random.choice(all_chars) for _ in range(5)) + '-' + ''.join(random.choice(all_chars) for _ in range(5))
print(password)
第15关：数列求和
a = int(input())
b = int(input())
if a>=1 and a<=9 and b>=0:
sum = 0
k = a
if b ==0:
print('0')
else:
for i in range(1,b+1):
sum = sum + k
k=k*10+a
print(sum)
else:
print('data error')
第16关：单词个数
a = input()
num=0
for i in a:
if i==" ":
num = num+1
print(num+1)原文地址：https://blog.csdn.net/2301_78289455/article/details/138089521
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