我有以下任务。有24小时的时间表。还有一些固定长度的事件。我有一个预定长度的新事件(例如1小时25分钟)。任务是知道 - 我可以在白天插入此任务多少次(此任务不得与其他事件相交。
答案 0 :(得分:1)
我们的问题可能被认为有点过分了, 所以让答案也简单化了(在Ruby中):
#! /usr/bin/ruby
DAY_START = 0
DAY_END = 24*60
# now, busy periods as array of size-2 arrays:
BUSY_PERIODS = [[9*60, 9*60+30], [18*60+30, 20*60]]
ADDITIONAL_EVENT_LENGTH = 1*60 + 25 # that is, 1 hour and 25 minutes
# in this simple program,
# time will be expressed as a number of minutes of the day
# for more advanced use, try to use "Time" class
# now let define a function to compute the list of free periods
# from a given list of busy periods:
def compute_free_periods( list_of_events )
# at the begining of the calculation, our whole day is assumed free:
free_periods = Array[ [ DAY_START, DAY_END] ]
# now, one by one, let us take away the busy periods:
list_of_events.each_with_object free_periods do | busy_period, free_periods |
# we use 'each_with_object' version of 'each' enumerator
# list of free_periods is passed in as an external object
# so that we can gradually take each busy period away from it
# let us first note the end time for the last free period
# ( [-1] refers to the last element of the list of free periods,
# subsequent [1] to the second element of the size-2 array )
last_free_period_end = free_periods[-1][1]
# and now, let us split this last free period into two by the
# current busy period (busy periods assumed non-overlapping and
# sorted in ascending order)
# first, end time of the last free period is set to the beginning
# of the busy period that we consider in this iteration:
free_periods[-1][1] = busy_period[0]
# then, additional free period is appended to the list of
# free periods usind << operator, starting when the busy period ends:
free_periods << [ busy_period[1], last_free_period_end ]
end
end
# now, let us use the function we just defined:
free_periods = compute_free_periods( BUSY_PERIODS )
# Now, for each free period we will count how many times we can stuff
# in the desired additional event:
free_period_capacities = free_periods.map { |free_period|
# we are using map function for this, which will return
# an array of the same length as free_periods, having
# performed prescribed modifications on each element:
# first, we calculate how many minutes a free period has
period_length = free_period[1] - free_period[0]
# then, we will use whole-number division to get the number
# of additional events that can fit in:
period_length / ADDITIONAL_EVENT_LENGTH
# (in Ruby, whole number division is the default behavior of / operator)
}
# and finally, we sum up the free period capacities for our new event:
total_events_possible = free_period_capacities.reduce( :+ )
# (summing is done elegantly by using reduce function with + operator)
# the last remaining thing we can do to consider program finished is to
puts total_events_possible # print the result on the screen
如果您发表评论并缩短评论,那么该计划将变得非常简短:
#! /usr/bin/ruby
DAY_START, DAY_END = 0, 24*60
BUSY_PERIODS = [[9*60, 9*60+30], [18*60+30, 20*60]]
ADDITIONAL_EVENT_LENGTH = 1*60 + 25
def compute_free_periods( list_of_events )
free_periods = Array[ [ DAY_START, DAY_END] ]
list_of_events.each_with_object free_periods do | busy_period, free_periods |
last_free_period_end = free_periods[-1][1]
free_periods[-1][1] = busy_period[0]
free_periods << [ busy_period[1], last_free_period_end ] end
end
free_periods = compute_free_periods( BUSY_PERIODS )
free_period_capacities = free_periods.map { |free_period|
period_length = free_period[1] - free_period[0]
period_length / ADDITIONAL_EVENT_LENGTH }
puts ( total_events_possible = free_period_capacities.reduce( :+ ) )
你知道,算法并不多。只是简单的编码。拾起 例如。 Ole Foul Zed的教科书Learn Ruby The Hard Way并开始 阅读练习0到结尾。或者,你可以去找其他语言, 例如Python。 Ole Zed拥有教学人才。如果你欣赏的话 经验,不要忘记以电子书的形式购买。