# [Tutor] Help!

David Hutto smokefloat at gmail.com
Thu Mar 3 23:08:12 CET 2011

```On Thu, Mar 3, 2011 at 4:56 PM, Knacktus <knacktus at googlemail.com> wrote:
> Am 03.03.2011 22:28, schrieb Andrew Bouchot:
>>
>> okay so this is my comp sci lab
>> *
>>
>> Problem:
>>
>> *ProductionTime.py It takes exactly 2 minutes and 7 second to produce an
>> item. Unfortunately, after 143 items are produced, the fabricator must
>> cool off for 5 minutes and 13 seconds before it can continue. Write a
>> program that will calculate the amount of time required to manufacture a
>> given number of items. *
>>
>> Output:
>>
>> *Output the amount of time D days HH:MM:SS *
>>
>> Sample Input :
>>
>> *
>>
>> numItems =1340 Represents the numbers items to be manufactured
>>
>> *
>>
>> Sample Output :
>>
>> *
>>
>> 2 days 00:03:17
>>
>> this is the coding i have written for it!
>>
>> numitems= int(raw_input("Enter the number of items needed to be
>> manufactured:"))
>> seconds=numitems*127
>> m, s = divmod(seconds, 60)
>> h, m = divmod(m, 60)
>> print "%d:%02d:%02d" % (h, m, s)
>>
>> but how would i add the 5 min and 13 seconds after 143 items have been
>> produced???
>
> For any given number of item, how many cool-down periods do you have in
> total?
> What's the prodcution time without cooling?
>
> -----------|++++++++|-----------|++++++++|----------|++++++++|------
> producing  |cooling |producing  |cooling |producing |cooling |producing
>
Maybe I understood it wrong, but they ask "but how would i add the 5
min and 13 seconds after 143 items have been produced???", which is
produce 143, then wait 5:13. Those times are irrelevant, unless the
need is in between the produced 143, which would make the pseudo:

> count = 0
> while count != 144
>    do stuff
time.sleep(5:13/143 interval)
count += 1

> HTH,
>
> Jan
>
>>
>>
>>
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