Basic Python programming code exercises (2)

1. Find the sum of numbers between 1 and 100 that are not divisible by 3

1. Loop condition: i<100
2. Loop operation

 The implementation code is as follows:

def sums():
   sum = 0
   for num in range(1, 101):
      if num % 3 != 0:
         sum += num
   print("1~100之间不能被3整除的数之和为：%s" % (sum))

sums()
print("1~100之间不能被3整除的数之和为：%s" % (sum))

operation result:

 

2. Calculate the sum of odd numbers within 100

1. Calculate the sum of odd numbers within 100 , set a breakpoint to debug the program, and track the execution order of the three expressions and the change of the loop variable

 The implementation code is as follows:

sum = 0
n = 99
while n > 0:
    sum = sum + n
    n = n - 2
print('100以内的奇数之和为：%s' % (sum))

operation result:

 

3. Calculate the proportion of customers

1. Shopping malls conduct surveys on the age level of customers

Calculate the proportion of customers in each age group

 The implementation code is as follows:

num=0
num2=0
for i in range(1,11):
    age = int(input('请输入第%d位顾客的年龄：'%i))
    if age>=30:
        num=num+1
    else:
        num2=num2+1
a=num/10*100
b=num2/10*100
print('30岁以下的比例是%.1f%%'%b)
print('30岁以上的比例是%.1f%%'%a)

operation result:

 

Four, cycle accumulation

1. Add integers between 1 and 10 to get the current number whose accumulated value is greater than 20
2. hint
3. Use a loop to add up from 1 to 10
4. Determine whether the cumulative value is greater than 20
5. If it is greater than 20, break out of the loop and print the current value

 The implementation code is as follows:

sum =0
for i in range(1,11):
   sum+= i
   if sum>20:
    print('1~10之间的整数相加，得到累加值大于20的当前数有：%s' % sum)

operation result:

 

5. Accumulation of even numbers

1. Find the sum of all even numbers between 1 and 10
2. hint
3. Use a loop to accumulate, the range of the loop is from 1 to 10
4. Check if the current number is even

If it is an odd number skip, execute the next cycle. If even, add up

 The implementation code is as follows:

sum_1 = 0

for i in range(1,11):
  if (i % 2==0):
    sum_1=sum_1+i

print ("1~10之间的所有偶数和为%d" % sum_1)

operation result:

 

6. Cyclic entry of member information

1. training points
2. for loop structure
3. continue statement
4. Statement of needs
   1. Circular entry of information for 3 members
   2. If the membership number is valid, the input information will be displayed; otherwise, the input failure will be displayed
5. Implementation ideas
   1. Analysis problem: there are repeated operations and the number of repetitions is determined
   2. Circular entry of 3 member information
   3. The membership number is invalid, use continue to realize the program jump
   4. use continue statement

 The implementation code is as follows:

text1 = input('MyShopping管理系统 》 客户信息管理 》 添加客户信息')
text2 = input('请输入会员号（<4位整数>)：')
a = int(text2)


for i in range(0,1):
  if 1000 <= a <= 9999:

    text3 = input('请输入会员生日（月/日<用两位数表示>）：')
    text4 = input('请输入积分：')
    print("已录入的会员信息是：" + text2, text3, text4)

    continue

for i in range(0,2):
  if 1000 <= a <= 9999:

    text2 = input('请输入会员号（<4位整数>)：')
    text3 = input('请输入会员生日（月/日<用两位数表示>）：')
    text4 = input('请输入积分：')
    print("已录入的会员信息是：" + text2, text3, text4)

    continue

operation result:

 

7. Verify user login information

User login verification, the number of verifications is up to 3 times

 The implementation code is as follows:

Account = 'admin'
Password = 123
test1='MyShopping系统'
for i in range(0,3):
   a = input('输入您的账户：')
   p = eval(input('输入您的密码：'))
   if(a == Account and p == Password):
    i = i + 1
    print('欢迎登录'+test1+'!')
    break
   elif(i < 2):
       i = i + 1
   print('账户或密码有误,您还有%d次机会' %(3-i))
else:
   print('账户或密码有误,3次机会已用完，请明天再登录，退出！')

operation result: