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Process control
1) Sequential structure: code is executed from top to bottom, and each statement is executed in sequence (default)
2) Branch structure: choose to execute or not execute part of the code according to the conditions (use if)
3) Loop structure: make the code execute repeatedly (for, while)
Branching structure
age = 10
if age >= 18:
print('adult')
else:
print('under age')
if branch structure
if single branch structure
Syntax:
if conditional statement:
Code snippet
explain:
if - keyword; Fixed writing
Conditional statement - can make any expression with results, including: specific data, operation expression (except assignment operation), variables that have been paid, function call expression, etc
: - Fixed writing
Code snippet - structurally, it is one or more statements (at least one) that maintain an indent with if; Logically: code is the code that will be executed when the condition is established
if (): # Null = None=0=False
print('a')
If it is a leap year, print the leap year
a = float(input('Please enter the year'))
if a % 4 == 0 and a % 100 != 0 or a % 400 == 0:
print('leap year')
else:
print('Missing intercalation')
If the variable is saved as an integer, print 'Integer'
num = 34
if type(num) == int:
print('integer')
if double branch structure
Syntax:
if conditional statement:
Code snippet 1 (code executed when conditions are met)
else:
Code snippet 2 (code executed without conditions)
Code snippet 3 (code executed regardless of whether the condition is satisfied or not)
Judge the parity of a number
a = float(input('please enter a number'))
if a % 2 == 0: #if a % 2 :
print('even numbers')
else:
print('Odd number')
if multi branch structure
Syntax:
if condition 1:
Code snippet 1
elif condition 2:
Code snippet 2
elif condition 3:
Code snippet 3
...
else:
Code snippet N
Note: elif can make any number of; else can have or not
a = float(input('please enter a number'))
if a > 90:
print('2000')
elif 85 < a <= 90 : #You can not write < = 90, because the case of a > 90 has been excluded in the previous step, so it is already in
print('1000')
elif 75 < a <= 85 :
print('500')
elif 60 <= a <= 75:
print('Little safflower')
else:
print('What reward do you want for failing?')
for loop
for loop
Syntax:
for variable in sequence:
Loop body (code that needs to be executed repeatedly)
explain:
for - keyword; Fixed writing
Variable - valid variable name (can be defined or undefined)
in - keyword; Fixed writing
Sequence - data of container data type. Container data types include string, list, dictionary, set, tuple, iterator, generator, range, etc
: - Fixed writing
Loop body - one or more statements that maintain an indentation with for; The loop body is the code that needs to be executed repeatedly
Execution process
Let variables take values from the sequence, one by one, until they are finished; Take a value and execute the loop body once.
The number of cycles of the for loop is related to the number of elements in the sequence
for x in 'abc':
print('qwer')
Execution process:
First time: x = 'a' - > Print ('qwer ')
Second time: x = 'b' - > Print ('qwer ')
Third time: x = 'c' - > Print ('qwer ')
range function
range(N) - produces an equal difference sequence of [0,N), and the difference is 1
range(3) -> 0，1，2
range(5) -> 0,1,2,3,4
range(M,N) - produces an equal difference sequence of [M,N), and the difference is 1
range(10,21) -> 10,11,12,13,14,15,16,17,18,19,20
range(M,N,STEP) - generate an equal difference sequence of [M,N), and the difference is step
range(1,10,2) -> 1,3,5,7,9
range(20,2,-3) -> 20,17,14,11,8,5
Print all even numbers within 1000
for x in range(0,1001,2):
print(x)
Two basic application scenarios of for loop
Cumulative summation
Exercise 1 Write code to find the sum of 1 + 2 + 3 + 4... + 100
# Step 1: define the variable and save the result. The initial value of the variable is generally 0 (summation) or 1 (quadrature)
a = 0
# Step 2: obtain cumulative data one by one with a loop
for b in range(1,101):
# Step 3: the recycle volume combines each data obtained into the variable corresponding to the result
a += b
print(a)
Exercise 2 Find the factorial of 10: 123... * 10
c = 1
for d in range(1,11):
c *= d
print(c)
Exercise 3 Find the sum of all even numbers in 100 to 200 that can be divided by 3
e = 0
for f in range(102,201,6):
e += f
print(e)
Number of Statistics
Exercise 1: count the number of odd numbers within 1000
Step 1: define the last number of variables to save. The default value of variables is 0
count = 0
# Step 2: use the loop to obtain the statistical object
for x in range(1000):
# Step 3: add one to the number of data that meet the statistical conditions
if x % 2 == 0:
count += 1
print(count)
Or directly
for x in range(1,1000,2):
count += 1
print(count)
Exercise 2: count the number of numbers within 1000 that can be divided by 3 but cannot be divided by 7
s = 0
for i in range(1000):
if i % 3 == 0 and i % 7 != 0:
s += 1
print(s)
Homework after class
Basic questions
-
Print pass or fail according to the range of grades entered.
a = float(input('Please enter')) if a >= 60: print('pass') else: print('fail,') -
Print adults or minors according to the entered age range. If the age is not within the normal range (0 ~ 150), it is not a person!.
a = float(input('Please enter')) if 150 > a >= 18: print('adult') elif a < 18 : print('under age') else: print('This is not a person') -
Enter two integers a and b. if the result of a-b is odd, the result will be output. Otherwise, the result of output prompt information a-b is not odd
a = float(input('Please enter a')) b = float(input('Please enter b')) c = a - b if c % 2 != 0: print(c) else: print('a-b The result is not odd') -
Use the for loop to output multiples of all 3 in 0 ~ 100.
for x in range(0,101,3): print(x) -
Use the for loop to output the number of single digits or tens within 100 ~ 200 that can be divided by 3.
for x in range(100,201): if (x % 10) % 3 == 0: print(x) elif ((x % 100)// 10) % 3 == 0: print(x) -
Use the for loop to count the number of 10 digits in 100 ~ 200 that are 5
a = 0 for x in range(100,201): if ((x % 100) // 10) / 5 == 1: a += 1 print(x) print(a) -
Use the for loop to print all numbers in 50 ~ 150 that can be divided by 3 but cannot be divided by 5
for x in range(50,151): if x % 3 == 0 and x % 5 != 0: print(x) -
Use the for loop to calculate the sum of all numbers in 50 ~ 150 that can be divided by 3 but cannot be divided by 5
a = 0 for x in range(50,151): if x % 3 == 0 and x % 5 != 0: a += x print(a) -
Count the number of digits within 100 that are 2 and can be divided by 3.
a = 0 for x in range(2,100,10): if x % 3 == 0: a += 1 print(a)
Advanced questions
-
Enter any positive integer. How many digits is it?
Note: you can't use strings here. You can only use loops
b = 1 a = int(input('Please enter a')) while a // 10 > 0: a //= 10 b += 1 print(b) perhaps a = int(input('Please enter a')); b = 0 for x in range(a): a //= 10 b += 1 if a == 0: print(b) break -
Print out all the numbers of daffodils. The so-called daffodils number refers to a three digit number, and the square sum of each digit is equal to the number itself. For example: 153 yes
The number of fairy flowers is 1 ³ + five ³ + three ³ Equal to 153.
for x in range(100,1000): if (x // 100) ** 3 + ((x % 100) // 10) ** 3 + (x % 10) ** 3 == x: print(x) -
Judge whether the specified number is a prime number (prime number is a prime number, that is, a number that cannot be divided by other numbers except 1 and itself)
a = int(input('Please enter a')) for x in range(2,a): if a % x == 0: print('Not prime') break else: print('Is a prime number') break -
Output 9 * 9 formula. Procedure analysis: Considering branches and columns, there are 9 rows and 9 columns in total, i control row and j control column.
# i = 0 ; j = 0 for i in range(9): print() # i += 1 for j in range(9): # j += 1 s = i * j if j <= i: print(i, '*', j, '=', s,end = ' ') After simplification for i in range(1,10): print() for j in range(1,i+1): s = i * j print(j, '*', i, '=', s,end = ' ') -
This is the classic question of "100 horses and 100 loads". There are 100 horses, carrying 100 loads of goods, 3 loads of horses, 2 loads of medium horses and 1 load of two ponies. How many are the big, medium and small horses? (exhaustive method can be used directly)
# a = 0 ; b = 0 ; c = 0 for a in range(34): # a += 1 for b in range(50): # b += 1 for c in range(200): # c += 2 if a * 3 + b * 2 + c * 0.5 == 100 and a + b + c == 100: print(a,b,c)