Five weekends: Difference between revisions
(Modified task description to make it easier to compare entries for correctness.) |
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1-2100 |
1-2100 |
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10-2100</pre> |
10-2100</pre> |
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=={{header|Python}}== |
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<lang python>from datetime import timedelta, date |
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DAY = timedelta(days=1) |
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WEEKEND = {6, 5, 4} # Sunday is day 6 |
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FMT = '%Y %m(%B)' |
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def fiveweekendspermonth(start=date(1900, 1, 1), stop=date(2101, 1, 1)): |
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'Compute months with five weekends between dates' |
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when = start |
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lastmonth = weekenddays = 0 |
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fiveweekends = [] |
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while when < stop: |
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year, mon, _mday, _h, _m, _s, wday, _yday, _isdst = when.timetuple() |
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if mon != lastmonth: |
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if weekenddays >= 15: |
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fiveweekends.append(when - DAY) |
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weekenddays = 0 |
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lastmonth = mon |
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if wday in WEEKEND: |
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weekenddays += 1 |
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when += DAY |
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return fiveweekends |
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dates = fiveweekendspermonth() |
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indent = ' ' |
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print('There are %s months of which the first and last five are:' % len(dates)) |
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print(indent +('\n'+indent).join(d.strftime(FMT) for d in dates[:5])) |
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print(indent +'...') |
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print(indent +('\n'+indent).join(d.strftime(FMT) for d in dates[-5:]))</lang> |
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'''Sample Output''' |
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<pre>There are 201 months of which the first and last five are: |
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1901 03(March) |
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1902 08(August) |
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1903 05(May) |
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1904 01(January) |
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1904 07(July) |
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... |
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2097 03(March) |
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2098 08(August) |
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2099 05(May) |
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2100 01(January) |
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2100 10(October)</pre> |
Revision as of 07:32, 23 October 2010
You are encouraged to solve this task according to the task description, using any language you may know.
The month of October in 2010 has five Fridays, five Saturdays, and five Sundays.
The task
- Write a program to show all months that have this same characteristic of five full weekends from the year 1900 through 2100 (Gregorian calendar).
- Show the number of months with this property.
- Show at least the first and last five dates, in order.
Java
<lang java>import java.util.Calendar; import java.util.GregorianCalendar;
public class FiveFSS {
//dreizig tage habt september... private static int[] month31 = {Calendar.JANUARY, Calendar.MARCH, Calendar.MAY, Calendar.JULY, Calendar.AUGUST, Calendar.OCTOBER, Calendar.DECEMBER}; public static void main(String[] args){ for(int year = 1900; year <= 2100; year++){ for(int month:month31){ Calendar date = new GregorianCalendar(year, month, 1); if(date.get(Calendar.DAY_OF_WEEK) == Calendar.FRIDAY){ //months are 0-indexed in Calendar System.out.println((date.get(Calendar.MONTH) + 1) + "-" + year); } } } }
}</lang> Output (middle results cut out):
3-1901 8-1902 5-1903 1-1904 7-1904 12-1905 3-1907 5-1908 1-1909 10-1909 7-1910 ... 12-2090 8-2092 5-2093 1-2094 10-2094 7-2095 3-2097 8-2098 5-2099 1-2100 10-2100
Python
<lang python>from datetime import timedelta, date
DAY = timedelta(days=1) WEEKEND = {6, 5, 4} # Sunday is day 6 FMT = '%Y %m(%B)'
def fiveweekendspermonth(start=date(1900, 1, 1), stop=date(2101, 1, 1)):
'Compute months with five weekends between dates' when = start lastmonth = weekenddays = 0 fiveweekends = [] while when < stop: year, mon, _mday, _h, _m, _s, wday, _yday, _isdst = when.timetuple() if mon != lastmonth: if weekenddays >= 15: fiveweekends.append(when - DAY) weekenddays = 0 lastmonth = mon if wday in WEEKEND: weekenddays += 1 when += DAY return fiveweekends
dates = fiveweekendspermonth() indent = ' ' print('There are %s months of which the first and last five are:' % len(dates)) print(indent +('\n'+indent).join(d.strftime(FMT) for d in dates[:5])) print(indent +'...') print(indent +('\n'+indent).join(d.strftime(FMT) for d in dates[-5:]))</lang>
Sample Output
There are 201 months of which the first and last five are: 1901 03(March) 1902 08(August) 1903 05(May) 1904 01(January) 1904 07(July) ... 2097 03(March) 2098 08(August) 2099 05(May) 2100 01(January) 2100 10(October)