age=18
print(' adult ')
print(' A minor ')
Problem solved ： Perform an operation when conditions are met , When the addition is not satisfied, it will not be executed
explain ：
grammar ：
if Conditional statements :
Code segment ( Code that executes only when conditions are met
age = 28
if age >= 18:
print(' adult ')
print('=====')
Execution process ：
First, judge the conditional statements , The result is True when , Execute code snippets 1; The result is False when , Execute code snippets 2; The code snippet must be executed 3.
# grammar
Applicable if .... Just .... otherwise ...
if Conditional statements :
Code segment 1
else：
Code segment 2
Code segment 3
# grammar
Method 1 ：
# Do different things according to different conditions , This applies to situations where conditions are mutually exclusive .
if Conditional statements 1:
Code segment 1
elif Conditional statements 2：
Code segment 2
elif Conditional statements 3：
Code segment 3
else:
Code segment 4
Method 2 ：
# Do different things according to different conditions , There is no relationship between multiple conditions. If one of the other conditions is true, the other conditions will not be true .
if Conditional statements 1:
Code segment 1
if Conditional statements 2:
Code segment 2
if Conditional statements 3:
Code segment 3
if Conditional statements 4:
Code segment 4
Be careful ：elif It could be any number ;else There can be or not
for Principle of circulation ： Variables are taken from the sequence in turn , Until it's done ; Each order , Perform a loop body .
for The number of cycles of a loop is only related to the number of elements in the sequence .
grammar ：
for Variable in Sequence ：
The loop body
explain - keyword ; Fixed writing
Variable - Valid variable name （ Whether it has been defined does not affect the implementation ）
in - keyword ; Fixed writing
Sequence - Data of container data type （ Dictionaries 、 Tuples 、 character string 、 list 、 aggregate 、 iterator 、 Generator, etc ）
： - Fixed writing
The loop body - and for One or more statements that hold an indent ; Code that needs to be executed repeatedly
example :
for x in 'abc':
print('hello world!')
print('======')
""" Execution process ： The first 1 Time ：x = 'a' -> print('hello world!') The first 2 Time ：x = 'b' -> print('hello world!') The first 3 Time ：x = 'c' -> print('hello world!') The loop ends ! hello world! hello world! hello world! """
for Two basic application scenarios of the loop ：
# seek 100 To 200 All can be 3 The sum of even numbers divided by integers
result = 0
for x in range(102, 201, 6):
result += x
print(result)
# Statistics 1000 Number of odd numbers within
count = 0
for x in range(1, 1000, 2):
count += 1
print(count)
>>>range(10) # from 0 Start to 9
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> range(1, 11) # from 1 Start to 10
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
>>> range(0, 30, 5) # In steps of 5
[0, 5, 10, 15, 20, 25]
>>> range(0, 10, 3) # In steps of 3
[0, 3, 6, 9]
>>> range(0, -10, -1) # negative
[0, -1, -2, -3, -4, -5, -6, -7, -8, -9]
>>> range(0)
[]
>>> range(1, 0)
[]
1. Print according to the range of grades entered pass perhaps fail, .
grade=int(input(' Please enter the score ( Percentage system ）：'))
if 0<=grade<60:
print(' unfortunately , You failed QAQ')
else:
print(" Pass ")
2. Print according to the entered age range adult perhaps A minor , If the age is not within the normal range (0~150) Print This is not a person !.
age=int(input(' Please enter age ：'))
if 0<age<18:
print(' A minor ')
elif 18<=age<150:
print(' adult ')
else:
print(' Not human ')
3. Enter two integers a and b, if a-b The result is an odd number , The result is output , Otherwise, the prompt message will be output a-b The result is not an odd number
a=int(input(' please enter an integer a:'))
b=int(input(' please enter an integer b:'))
result=a-b
if result%2==1:
print(result)
else:
print('a-b The result is not an odd number ')
4. Use for Cyclic output 0~100 All in 3 Multiple .
# Method 1
for x in range(0,101,3):
print(x)
# Method 2
for x in range(0,101):
if x%3==0:
print(x)
Use for Cyclic output 100~200 The inner single digit or ten digit can be 3 Divisible number .
for x in range(100,201):
if x%10%3==0 or x//10%10%3==0:
print(x)
Use for Cycle statistics 100~200 The median ten is 5 The number of
# Method 1
for x in range(100, 201):
if x // 10 % 10 == 5:
print(x)
# Method 2
for x in range(150,160):
print(x)
Use for Loop printing 50~150 All can be 3 Divisible but not by 5 Divisible number
for x in range(51,151,3):
if x%5!=0:
print(x)
Use for Cycle calculation 50~150 All can be 3 Divisible but not by 5 The sum of divisible numbers
count=0
for x in range(51,151,3):
if x%5!=0:
count+=x
print(count)
Statistics 100 The inner single digits are 2 And can be 3 The number of integers .
count=0
for x in range(0,101,3):
if x%10==2:
count+=1
print(count)
Enter any positive integer , Ask him how many digits ？
Be careful : You can't use strings here , Only loop
# Method 1
num=int(input(" Please enter a positive integer :"))
for x in range(0,10):
if num//(10**x)>=1:
continue
print(x)
break
# Method 2
num=int(input(" Please enter a positive integer :"))
n=0
x=0
while n<1:
if num//(10**x)>=1:
x+=1
else:
break
print(x)
Print out all the daffodils , The so-called narcissus number refers to a three digit number , Its figures ⽴ The sum of the squares is equal to the number itself .例 Such as :153 yes
⼀ individual ⽔ Fairy flower number , because 1³ + 5³ + 3³ be equal to 153.
for x in range(100,1000):
a=x//100
b=x//10%10
c=x%10
if a**3+b**3+c**3==x:
print(x)
Judge whether the specified number is a prime number （ Prime numbers are prime numbers , In addition to 1 A number other than itself that cannot be divided by other numbers )
k=0
num=int(input(" Please enter a positive integer :"))
for x in range (2,num):
if num%x==0:
k+=1
if k==0:
print(' prime number ')
else:
print(' Not primes ')
Output 9*9 formula . Program analysis ： Branch and column considerations , common 9 That's ok 9 Column ,i The control line ,j Control the column
for i in range(1,10):
for j in range(1,i+1):
print(i,'*',j,'=',i*j,end=" ")
print('')
This is the classic " A hundred horses and a hundred burdens " problem , There are a hundred horses , Carry a hundred loads , Big horse, big horse 3 Dan , On the back of a horse 2 Dan , Two ponies carry 1 Dan , How big is it , in , How many ponies each ？（ You can directly use the exhaustive method ）
for small in range(0,101):
for mid in range(0,101):
for big in range(0,101):
if big+mid+small==100 and big*3+mid*2+small/2==100:
print(' Malaysia ',big,' horse '','' Zhongma ',mid,' horse '','' The pony ',small,' horse ','.')
