[first song] rebirth of me in py introductory training (4): Circular program

It's not easy to create , Before reference , Point a praise , Collection , You can't pay too much attention , Family

The first 1 Turn off ： Output diamond character pattern

Programming requirements

According to the prompt , Add code to the editor on the right , And output diamond pattern .

Be careful ：

The last line of diamond pattern does not need to wrap , You can use the following format to remove line breaks ：

1. print("*" ,end = "")

2. #  Output the upper half of the diamond , The lower half is output directly through reverse control 
   #********* Begin *********#
   for n in range( row-1, 0, -1):
       i = 2 * n - 1   # Every line * The number of 
       j = ( row  - n ) * 2    # The number of spaces in each line 
       for jj in range(j): # Controls the space of each line of output 
           print( " ", end="" )
       if i > 1:   # If a line of letters * The number of is greater than 1, Control * and * Space format between and then output 
           for ii in range(i):
               if ii != i - 1:
                   print("* ", end="" )
               else:
                   print("*")
       else:   # If a line of letters * The number of is equal to 1, Direct output *
           print("*")
   #********* End *********#

   The first 2 Turn off ： Output even numbers in the specified range

3. Programming requirements

   Enter two integers , Two numbers as a range , Output all even numbers in the range through the program .

   for example ：
   Input ： 1 12, The output ：2,4,6,8,10,12; Input ： -2 10, The output ：-2,0,2,4,6,8,10; Input ： 100 110, The output ：100,102,104,106,108,110 .

   Be careful ：

   print() Output does not wrap .

# -*- coding:utf-8 -*-
begin, end = map( int, input("").split() )
result = []
#********* Begin *********#
abc = []
for x in range(begin,end+1):
    result.append(x)
for i in result:
    if i % 2 == 0:
        abc.append(i)
print(','.join(map(str,abc)))
#********* End *********#

The first 3 Turn off ： Output the multiplication table

Programming requirements

Demand the use of for Loop out the multiplication table , Through double layer for Cycle to control , Then output the 99 multiplication table in the corresponding format .

• Pay attention to line feed .

# -*- coding:utf-8 -*-
for i in range(1,10):
    for j in range(i):
        #********* Begin *********#
        if i == j+1:
            print('{} * {} = {}'.format(i, j+1, i*(j+1)),end='')
            continue
        print('{} * {} = {} '.format(i, j+1, i*(j+1)),end='')
        #********* End *********#
    if i != 9:
        print()

The first 4 Turn off ： stone 、 scissors 、 Cloth game

Programming requirements

Write traditional stones 、 scissors 、 Cloth's game program , The rule of a single game is , Scissors are better than cloth , Busson stone , Stone is better than scissors . User players compete with computers , use 7 game 4 To win . For simplicity and convenience , The choice of computer is fixed , It's not a random selection sequence , Finally, output the victory or draw information .

for example ：
Input ： stone , scissors , scissors , cloth , cloth , stone , stone , The output ： Computers win !;

Input ： scissors , cloth , scissors , stone , cloth , scissors , stone , The output ： Computers win !;

Input ： cloth , stone , stone , cloth , scissors , stone , scissors , The output ： Game player wins ! .

#********* Begin *********#
for x in range(0,7):
    if [guess_list[x],people_guess[x]] == computer_win_condition[0]:
        computer_win_number += 1
    elif [guess_list[x],people_guess[x]] == computer_win_condition[1]:
        computer_win_number += 1
    elif [guess_list[x],people_guess[x]] == computer_win_condition[2]:
        computer_win_number += 1
    elif people_guess[x] == guess_list[x]:
        computer_win_number += 1
        people_win_number += 1
    else:
        people_win_number += 1
#********* End *********#

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