《计算机科学导论》学习笔记（17） - 课程 17 编程练习 2

1. 练习：字数统计

2. 练习：网络延迟

3. 练习：秒数换算

题目：

# Write a procedure, convert_seconds, which takes as input a non-negative 
# number of seconds and returns a string of the form 
# '<integer> hours, <integer> minutes, <number> seconds' but
# where if <integer> is 1 for the number of hours or minutes, 
# then it should be hour/minute. Further, <number> may be an integer
# or decimal, and if it is 1, then it should be followed by second.
# You might need to use int() to turn a decimal into a float depending
# on how you code this. int(3.0) gives 3
#
# Note that English uses the plural when talking about 0 items, so
# it should be "0 minutes".
#

def convert_seconds():



print convert_seconds(3661)
#>>> 1 hour, 1 minute, 1 second

print convert_seconds(7325)
#>>> 2 hours, 2 minutes, 5 seconds

print convert_seconds(7261.7)
#>>> 2 hours, 1 minute, 1.7 seconds

我的答案：

def convert_seconds(seconds):
    second = seconds % 60
    minute = int(seconds) / 60 % 60
    hour   = int(seconds) / 60 / 60

    if hour == 1:
        hour = str(hour) + ' hour'
    else:
        hour = str(hour) + ' hours'

    if minute == 1:
        minute = str(minute) + ' minute'
    else:
        minute = str(minute) + ' minutes'

    if second == 1:
        second = str(second) + ' second'
    else:
        second = str(second) + ' seconds'

    return hour + ', ' + minute + ', ' + second

(这里的有一点需要注意，跟我的常识有些冲突，那就是，题目的视频提供的测试用例认为， 0 秒应该是 0 seconds ，即应该是复数形式。)

4. 练习：下载时间计算器

题目：

# Write a procedure download_time which takes as inputs a file size, the
# units that file size is given in, bandwidth and the units for
# bandwidth (excluding per second) and returns the time taken to download 
# the file.
# Your answer should be a string in the form
# "<number> hours, <number> minutes, <number> seconds"

# Some information you might find useful is the number of bits
# in kilobits (kb), kilobytes (kB), megabits (Mb), megabytes (MB),
# gigabits (Gb), gigabytes (GB) and terabits (Tb), terabytes (TB).

#print 2 ** 10 # one kilobit, kb
#print 2 ** 10 * 8 # one kilobyte, kB

#print 2 ** 20 # one megabit, Mb
#print 2 ** 20 * 8 # one megabyte, MB

#print 2 ** 30 # one gigabit, Gb
#print 2 ** 30 * 8 # one gigabyte, GB

#print 2 ** 40 # one terabit, Tb
#print 2 ** 40 * 8 # one terabyte, TB

# Often bandwidth is given in megabits (Mb) per second whereas file size 
# is given in megabytes (MB).

def download_time():



print download_time(1024,'kB', 1, 'MB')
#>>> 0 hours, 0 minutes, 1 second

print download_time(1024,'kB', 1, 'Mb')
#>>> 0 hours, 0 minutes, 8 seconds # 8.0 seconds is also acceptable

print download_time(13,'GB', 5.6, 'MB')
#>>> 0 hours, 39 minutes, 37.1428571429 seconds

print download_time(13,'GB', 5.6, 'Mb')
#>>> 5 hours, 16 minutes, 57.1428571429 seconds

print download_time(10,'MB', 2, 'kB')
#>>> 1 hour, 25 minutes, 20 seconds # 20.0 seconds is also acceptable

print download_time(10,'MB', 2, 'kb')
#>>> 11 hours, 22 minutes, 40 seconds # 40.0 seconds is also acceptable

(这道题不难，我的思路是把文件大小和带宽转换成同样的单位，然后就很方便计算了，但是单位转换时，要考虑很多种不同的情况，会有很多重复的代码，合并有时一个小麻烦，这道题我跳过了。。)