10 Row Abacus

输入一个数字，按规定输出符号排列。此题的原型为算盘，用符号排列模拟算盘的显示。具体要求如下

#########################################################################
#                 10-row School abacus
#                         by
#                      Michael H
#########################################################################
#       Description partially extracted from from wikipedia 
#
#  Around the world, abaci have been used in pre-schools and elementary
#
# In Western countries, a bead frame similar to the Russian abacus but
# with straight wires and a vertical frame has been common (see image).
# Helps schools as an aid in teaching the numeral system and arithmetic
#
#         |00000*****   |     row factor 1000000000
#         |00000*****   |     row factor 100000000
#         |00000*****   |     row factor 10000000 
#         |00000*****   |     row factor 1000000
#         |00000*****   |     row factor 100000
#         |00000*****   |     row factor 10000
#         |00000*****   |     row factor 1000
#         |00000****   *|     row factor 100     * 1
#         |00000***   **|     row factor 10      * 2
#         |00000**   ***|     row factor 1       * 3
#                                        -----------    
#                             Sum                123 
#
# Each row represents a different row factor, starting with x1 at the
# bottom, ascending up to x1000000000 at the top row.     
######################################################################

# TASK:
# Define a procedure print_abacus(integer) that takes a positive integer
# and prints a visual representation (image) of an abacus setup for a 
# given positive integer value.
# 
# Ranking
# 1 STAR: solved the problem!
# 2 STARS: 6 < lines <= 9
# 3 STARS: 3 < lines <= 6
# 4 STARS: 0 < lines <= 3

最终的判定是以实现效果所写代码的行数，可见这里是要我们在练习基本语法的同时，去思考最紧凑的方法。

对于算盘来说，在初始状态每一列都是相同的，且输入数字的权重和列数是对应的，从对实际问题的映射和评分要求考虑，下面的代码十分吻合

def print_abacus(value):
    for c in ((10 - len(str(value))) * '0' + str(value)):
        print('|00000*****|'[ : -1 - int(c)] + '   ' + '|00000*****|'[-1 - int(c) : ])

但是一种更为容易想到，使用简单逻辑判断的方法如下

def print_abacus(value):
    num_len = len(str(value))
    for i in range(10 - num_len):
        print('|00000*****   |')
    for c in str(value):
        c_int = int(c)
        if(c_int > 5):
            print('|' + (10 - c_int) * '0' + '   ' + (c_int - 5) * '0' + '*****|')
        else:
            print('|00000' + (5 - c_int) * '*' + '   ' + c_int * '*' + '|')
        
        

题目及解答参考来自Udacity的Intro to Computer Science课程：https://classroom.udacity.com/courses/cs101/lessons/48689154/concepts/6986185940923