Python练习

时间:2024-03-30 17:08:32

1. 求一个十进制的数值的二进制的0、1的个数

def er(x):
  a = bin(x)
  b = str(a).count("1")
  c = str(a).count("0") - 1
  print(f"{a},count 1:{b},count 0:{c}")
 
  
x = int(input("enter a number:"))
er(x)

 

2. 实现一个用户管理系统(要求使用容器保存数据)

     [{name: xxx, pass: xxx, ……},{},{}]

users = []
 
while True:
    print("\t\t欢迎登录用户管理系统")
    print("\t\t 1、用户注册")
    print("\t\t 2、用户登录")
    print("\t\t 3、退出系统")
    choice = input("请输入您的选择:")
 
    if choice == "1":
        while True:
            
            username = input("请输入用户名:")
            password = input("请输入密码:")
            password_again = input("请再次确认密码:")
 
            
            if username == None or len(username) == 0:
                print("对不起,用户为空")
                continue
            
            flag = False
            for u in users:
                if u.get("username") == username:
                    print("对不起,该用户已经存在,请重新注册")
                    flag = True
                    break
            if flag:
                continue
            
            
            if password !=  password_again:
                print("两次密码不一致")
                continue
 
           
            user = {"username": username, "password": password}
            users.append(user)
            print("用户注册成功,请登录")
            print("\n")
            break
    elif choice == "2":
        print("\n")
        username = input("请输入用户名:")
        password = input("请输入密码:")
 
        
        for user in users:
            if user["username"] == username and user.get("password") == password:
                print(f"尊敬的用户{username},欢迎回来")
            else:
                print("对不起,登录失败,请重新登录")
        
    else:
       sys.exit()	

 3. 求1~100之间不能被3整除的数之和 

a = 0
for x in range(1,101):
	if x % 3 == 0:
		continue
	else:
		a = x + a
print(f"百位以内不可被三整除的数和为:{a}")

 4. 给定一个正整数N,找出1到N(含)之间所有质数的总和

def sum(x):
 
    if x == 1:
        return False
    for i in range(2,x//2 + 1):
        if x % i == 0:
            return False
    return True
 
N = int(input('请输入一个正整数N:'))
sum_N = 0
for x in range(1,N+1):
    if sum(x):
        sum_N += x
print(f'1到N(含)之间所有质数的总和为:{sum_N}')

 5. 计算PI(公式如下:PI=4(1-1/3+1/5-1/7+1/9-1.......)

def PI():
	a = 0
	b = 0
	for i in range(1,99999999,4):
		a += 4 * (1 / i)
	for i in range(3,99999999,4):
		b -= 4 * (1 / i)
	print(a + b)
 
PI()

6. 给定一个10个元素的列表,请完成排序(注意,不要使用系统api)

l = eval(input("请输入一个列表:"))
 
ls=list(l)
ls.sort()
s=set(ls)
 
print(s)

7. 求  a+aa+aaa+.......+aaaaaaaaa=?其中a为1至9之中的一个数,项数也要可以指定。

def sum(a,n):
    sum_a = 0
    for i in range(1,n+1):
        num = int(f'{a}'*i)
        sum_a += num
    return sum_a
 
a = int(input('请输入一个在区间[1,9]的正整数:'))
n = int(input('请输入指定的项数:'))
 
print(f'所求多项式的和为:{sum(a,n)}')

8. 合并两个有序数组,合并后还是有序列表

def selection_sort(ls=[]):
    
    for i in range(len(ls)):
        index_min = i
        for j in range(i+1,len(ls)):
            if ls[index_min] > ls[j]:
                index_min = j
        ls[index_min],ls[i] = ls[i],ls[index_min]
    return ls
 
l1 = eval(input("请输入一个列表:"))
l2 = eval(input("请输入一个列表:"))
l1.extend(l2)
print(f'合并后的有序列表为:{selection_sort(l1)}')

9. 给定一个非负整数数组A,将该数组中的所有偶数都放在奇数元素之前 

def even_before_odd(ls = []):
  
    for i in range(len(ls)):
        if ls[i] % 2 != 0:
            for j in range(i+1,len(ls)):
                if ls[j] % 2 == 0:
                    ls[i],ls[j] = ls[j],ls[i]
                    break
    return ls
 
ls = eval(input("请输入一个列表:"))
 
print(f'将偶数置于奇数前,列表变为:{even_before_odd(ls)}')

10. 给定一个包含n+1个整数的数组nums,其数字在1到n之间(包含1和n),可知至少存在一个重复的整数,假设只有一个重复的整数,请找出这个重复的数

nums = [1,3,4,2,2]
repeatDict = {}
for num in nums:
    if num not in repeatDict:
        repeatDict[num] = 1
    else:
        print(num)
nums.sort()
pre = nums[0]
n = len(nums)
for index in range(1, n):
    if pre == nums[index]:
        print (pre)
    pre = nums[index]
nums.sort()
for i in range(len(nums)-1):
    if nums[i] == nums[i+1]:
        print(nums[i])

 11. 找出10000以内能被5或6整除,但不能被两者同时整除的数(函数)

def math_5_6(x):
    count=[]
    for i in range(1,x+1):
        if (i % 5 == 0 or i % 6 ==0 ):
            if i % 5 == 0 and i % 6 ==0:
                continue  
            else:
                count.append(i)
    print(count)
 
    
math_5_6(10000)

 

 3. 写一个方法,计算列表所有偶数下标元素的和(注意返回值)

def list_sum(ls):
    sum = 0
    i = 0
    while i < len(ls):
        sum += ls[i]
        i += 2
    return sum
    
ls = [1,2,3,4,5,6,7,8,9]
print(f"列表是:{ls},偶数下标元素和是:{list_sum(ls)}")

 

 4. 【选做】某个人进入如下一个棋盘中,要求从左上角开始走, 最后从右下角出来(要求只能前进,不能后退),问题:共有多少种走法?

0 0 0 0 0 0 0 0
     	
0 0 0 0 0 0 0 0
     	
0 0 0 0 0 0 0 0
     	
0 0 0 0 0 0 0 0
     	
0 0 0 0 0 0 0 0
def uniquePaths(m, n):
    dp = [[0] * n for _ in range(m)]
    for i in range(m):
        dp[i][0] = 1
    for j in range(n):
        dp[0][j] = 1
    for i in range(1, m):
        for j in range(1, n):
            dp[i][j] = dp[i-1][j] + dp[i][j-1]
    return dp[m-1][n-1]

m = 8
n = 5
result = uniquePaths(m, n)
print("共有", result, "种走法")

5. 【选做】汉诺塔

def rabbit(month):
    if month <= 2:
        return 2
    else:
        return rabbit(month - 1) + rabbit(month - 2)
 
 
if __name__ == "__main__":
    month = int(input("请输入month:"))
    for i in range(0, month):
        print(rabbit(i))