Long year
You are encouraged to solve this task according to the task description, using any language you may know.
Most years have 52 weeks, some have 53, according to ISO8601.
- Task
Write a function which determines if a given year is long (53 weeks) or not, and demonstrate it.
11l
<lang 11l>F is_long_year(year)
F p(year) R (year + (year I/ 4) - (year I/ 100) + (year I/ 400)) % 7 R p(year) == 4 | p(year - 1) == 3
L(year) 2000..2100
I is_long_year(year) print(year, end' ‘ ’)</lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Action!
<lang Action!>BYTE FUNC P(CARD y) RETURN ((y+(y/4)-(y/100)+(y/400)) MOD 7)
BYTE FUNC IsLongYear(CARD y)
IF P(y)=4 OR P(y-1)=3 THEN RETURN (1) FI
RETURN (0)
PROC Main()
CARD y BYTE LMARGIN=$52,oldLMARGIN
oldLMARGIN=LMARGIN LMARGIN=0 ;remove left margin on the screen Put(125) PutE() ;clear the screen
FOR y=1900 TO 2400 DO IF IsLongYear(y) THEN PrintC(y) Put(32) FI OD
LMARGIN=oldLMARGIN ;restore left margin on the screen
RETURN</lang>
- Output:
Screenshot from Atari 8-bit computer
1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195 2201 2207 2212 2218 2224 2229 2235 2240 2246 2252 2257 2263 2268 2274 2280 2285 2291 2296 2303 2308 2314 2320 2325 2331 2336 2342 2348 2353 2359 2364 2370 2376 2381 2387 2392 2398
Ada
The Ada calendar package handles dates for years 1901 through 2399. This program outputs all the long years within that range. <lang Ada>------------------------------------------------------------- -- Calculate long years -- Reference: https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year
with Ada.Text_IO; use Ada.Text_IO; with Ada.Calendar; use Ada.Calendar; with Ada.Calendar.Formatting; use Ada.Calendar.Formatting;
procedure Main is
First_Day : Time; Last_Day : Time; package AC renames Ada.Calendar; type Counter is mod 10; Count : Counter := 0;
begin
for Yr in Year_Number loop
First_Day := AC.Time_Of (Year => Yr, Month => 1, Day => 1); Last_Day := AC.Time_Of (Year => Yr, Month => 12, Day => 31); -- If Jan 1 is Thursday or Dec 31 is Thursday then -- the year is a long year if Day_Of_Week (First_Day) = Thursday or else Day_Of_Week (Last_Day) = Thursday then if Count = 0 then New_Line; end if; Put (Yr'Image); Count := Count + 1; end if; end loop;
end Main; </lang>
- Output:
1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195 2201 2207 2212 2218 2224 2229 2235 2240 2246 2252 2257 2263 2268 2274 2280 2285 2291 2296 2303 2308 2314 2320 2325 2331 2336 2342 2348 2353 2359 2364 2370 2376 2381 2387 2392 2398
ALGOL 68
<lang algol68>BEGIN # find "long years" - years which have 53 weeks this is equivalent to #
# finding years where 1st Jan or 31st Dec are Thursdays # # returns the day of the week of the specified date (d/m/y), Sunday = 1 # PROC day of week = ( INT d, m, y )INT: BEGIN INT mm := m; INT yy := y; IF mm <= 2 THEN mm := mm + 12; yy := yy - 1 FI; INT j = yy OVER 100; INT k = yy MOD 100; (d + ( ( mm + 1 ) * 26 ) OVER 10 + k + k OVER 4 + j OVER 4 + 5 * j ) MOD 7 END # day of week # ; # returns TRUE if year is a long year, FALSE otherwise # PROC is long year = ( INT year )BOOL: day of week( 1, 1, year ) = 5 OR day of week( 31, 12, year ) = 5; # show long years from 2000-2099 # print( ( "long years 2000-2099:" ) ); FOR year FROM 2000 TO 2099 DO IF is long year( year ) THEN print( ( " ", whole( year, 0 ) ) ) FI OD
END</lang>
- Output:
long years 2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
ALGOL-M
<lang ALGOL>BEGIN
COMMENT
FIND ISO CALENDAR YEARS HAVING 53 WEEKS. THE SIMPLEST TEST IS THAT A GIVEN YEAR WILL BE "LONG" IF EITHER THE FIRST OR LAST DAY IS A THURSDAY;
% CALCULATE P MOD Q % INTEGER FUNCTION MOD(P, Q); INTEGER P, Q; BEGIN
MOD := P - Q * (P / Q);
END;
COMMENT
RETURN DAY OF WEEK (SUN=0, MON=1, ETC.) FOR A GIVEN GREGORIAN CALENDAR DATE USING ZELLER'S CONGRUENCE;
INTEGER FUNCTION DAYOFWEEK(MO, DA, YR); INTEGER MO, DA, YR; BEGIN
INTEGER Y, C, Z; IF MO < 3 THEN BEGIN MO := MO + 10; YR := YR - 1; END ELSE MO := MO - 2; Y := MOD(YR, 100); C := YR / 100; Z := (26 * MO - 2) / 10; Z := Z + DA + Y + (Y / 4) + (C /4) - 2 * C + 777; DAYOFWEEK := MOD(Z, 7);
END;
% RETURN 1 IF YEAR IS LONG, OTHERWISE 0 % INTEGER FUNCTION ISLONGYEAR(YR); INTEGER YR; BEGIN
INTEGER THURSDAY; THURSDAY := 4; IF (DAYOFWEEK(1,1,YR) = THURSDAY) OR (DAYOFWEEK(12,31,YR) = THURSDAY) THEN ISLONGYEAR := 1 ELSE ISLONGYEAR := 0;
END;
% MAIN PROGRAM STARTS HERE % INTEGER YEAR; WRITE("ISO YEARS THAT WILL BE LONG IN THIS CENTURY:"); WRITE(""); FOR YEAR := 2000 STEP 1 UNTIL 2099 DO
BEGIN IF ISLONGYEAR(YEAR) = 1 THEN WRITEON(YEAR); END;
END </lang>
- Output:
ISO YEARS THAT WILL BE LONG IN THIS CENTURY: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
ALGOL W
Uses the Day_of_week procedure from the Day_of_the_week task. <lang algolw>begin % find "long years" - years which have 53 weeks %
% this is equivalent to finding years where % % 1st Jan or 31st Dec are Thursdays % % finds the day of the week - Sunday = 1 % integer procedure Day_of_week ( integer value d, m, y ); begin integer j, k, mm, yy; mm := m; yy := y; if mm <= 2 then begin mm := mm + 12; yy := yy - 1; end if_m_le_2; j := yy div 100; k := yy rem 100; (d + ( ( mm + 1 ) * 26 ) div 10 + k + k div 4 + j div 4 + 5 * j ) rem 7 end Day_of_week; % returns true if year is a long year, false otherwise % logical procedure isLongYear ( integer value year ); Day_of_week( 1, 1, year ) = 5 or Day_of_week( 31, 12, year ) = 5; % show long years from 2000-2099 % write( "long years 2000-2099:" ); for year := 2000 until 2099 do begin if isLongYear( year ) then writeon( I_W := 5, S_W := 0, year ) end for_year
end.</lang>
- Output:
long years 2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
APL
<lang apl>dec31weekday ← {7|⍵+⌊(⍵÷4)+⌊(⍵÷400)-⌊⍵÷100} isolongyear ← {(4 = dec31weekday ⍵) ∨ 3 = dec31weekday ⍵ - 1}</lang>
- Output:
{⍵/⍨isolongyear ⍵}1800+⍳300 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
AppleScript
<lang applescript>on isLongYear(y)
-- ISO8601 weeks begin on Mondays and belong to the year in which they have the most days. -- A year which begins on a Thursday, or which begins on a Wednesday and is a leap year, -- has majority stakes in the weeks it overlaps at *both* ends and so has 53 weeks instead of 52. -- Leap years divisible by 400 begin on Saturdays and so don't so need to be considered in the leap year check. tell (current date) to set {Jan1, its day, its month, its year} to {it, 1, January, y} set startWeekday to Jan1's weekday return ((startWeekday is Thursday) or ((startWeekday is Wednesday) and (y mod 4 is 0) and (y mod 100 > 0)))
end isLongYear
set longYears to {} repeat with y from 2001 to 2100
if (isLongYear(y)) then set end of longYears to y
end repeat
return longYears</lang>
- Output:
{2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099}
On the other hand, since the cycle repeats every 400 years, it's possible to cheat with a precalculated look-up list:
<lang applescript>on isLongYear(y)
return (y mod 400 is in {4, 9, 15, 20, 26, 32, 37, 43, 48, 54, 60, 65, 71, 76, 82, 88, 93, 99, 105, 111, 116, 122, 128, 133, 139, 144, 150, 156, 161, 167, 172, 178, 184, 189, 195, 201, 207, 212, 218, 224, 229, 235, 240, 246, 252, 257, 263, 268, 274, 280, 285, 291, 296, 303, 308, 314, 320, 325, 331, 336, 342, 348, 353, 359, 364, 370, 376, 381, 387, 392, 398})
end isLongYear
set longYears to {} repeat with y from 2001 to 2100
if (isLongYear(y)) then set end of longYears to y
end repeat
return longYears</lang>
Arturo
<lang rebol>longYear?: function [year][
date: to :date .format: "dd/MM/yyyy" ~"01/01/|year|"
or? date\Day = "Thursday" and? leap? year date\Day = "Wednesday"
]
print "Years with 53 weeks between 2000 and 2100:" print select 2000..2100 => longYear?</lang>
- Output:
Years with 53 weeks between 2000 and 2100: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
AutoHotkey
<lang AutoHotkey>Long_year(y) { A := Mod(y + floor(y/4) - floor(y/100) + floor(y/400), 7) y--, B := Mod(y + floor(y/4) - floor(y/100) + floor(y/400), 7) return A=4 || B=3 }</lang> Examples:<lang AutoHotkey>loop, 100{ y := 1999+A_Index res .= Long_year(y) ? Y " ": "" } MsgBox % "Long Years 2000-2100 : " res return</lang>
- Output:
Long Years 2000-2100 : 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
AWK
<lang AWK>
- syntax: GAWK -f LONG_YEAR.AWK
BEGIN {
for (cc=19; cc<=21; cc++) { printf("%2d00-%2d99: ",cc,cc) for (yy=0; yy<=99; yy++) { ccyy = sprintf("%02d%02d",cc,yy) if (is_long_year(ccyy)) { printf("%4d ",ccyy) } } printf("\n") }
printf("\n%4d-%4d: ",by=1970,ey=2037) for (y=by; y<=ey; y++) { if (strftime("%V",mktime(sprintf("%d 12 28 0 0 0",y))) == 53) { printf("%4d ",y) } } printf("\n") exit(0)
} function is_long_year(year, i) {
for (i=0; i<=1; i++) { year -= i if ((year + int(year/4) - int(year/100) + int(year/400)) % 7 == 4-i) { return(1) } } return(0)
} </lang>
- Output:
1900-1999: 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 2100-2199: 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195 1970-2037: 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037
BASIC
Applesoft BASIC
<lang gwbasic> 10 DEF FN M7(N) = N - 7 * INT (N / 7)
20 DEF FN WD(Y) = FN M7(Y + INT (Y / 4) - INT (Y / 100) + INT (Y / 400)) 30 DEF FN LY(Y) = (4 = FN WD(Y)) OR (3 = FN WD(Y - 1)) 40 HOME : INVERSE : PRINT "**** LIST OF ISO LONG YEARS ****": NORMAL 50 INPUT "START YEAR? ";S 60 INPUT "END YEAR? ";E 70 PRINT : FOR Y = S TO E 80 IF FN LY(Y) THEN PRINT S$Y;:S$ = " " 90 NEXT Y</lang>
ASIC
<lang basic> REM Long year CLS PRINT "**** List of ISO long years ****" PRINT "Start year"; INPUT S PRINT "End year"; INPUT E PRINT FOR Y = S TO E
GOSUB CALCLY: IF LY <> 0 THEN PRINT Y; ENDIF
NEXT Y PRINT END
CALCLY:
REM Nonzero if Y is long LY = 0 AY = Y GOSUB CALCWD: IF WD = 4 THEN LY = -1 ENDIF AY = Y - 1 GOSUB CALCWD: IF WD = 3 THEN LY = -1 ENDIF
RETURN
CALCWD:
REM Weekday of AY-12-31, 0 = Sunday WD = AY TMP = AY / 4 WD = WD + TMP TMP = AY / 100 WD = WD - TMP TMP = AY / 400 WD = WD + TMP WD = WD MOD 7
RETURN </lang>
- Output:
**** List of ISO long years **** Start year?1995 End year?2045 1998 2004 2009 2015 2020 2026 2032 2037 2043
BASIC256
<lang BASIC256>function p(y) return (y + int(y/4) - int(y/100) + int(y/400)) mod 7 end function
function isLongYear(y) return (p(y) = 4) or (p(y - 1) = 3) end function
for y = 2000 to 2100 if isLongYear(y) then print y next y end</lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
BBC BASIC
<lang bbcbasic> INSTALL @lib$ + "DATELIB"
REM The function as per specification. DEF FNLongYear(year%)=FN_dow(FN_mjd(1, 1, year%)) == 4 OR FN_dow(FN_mjd(31, 12, year%)) == 4
REM Demonstrating its use. PROCPrintLongYearsInCentury(20) PROCPrintLongYearsInCentury(21) PROCPrintLongYearsInCentury(22) END
DEF PROCPrintLongYearsInCentury(century%) LOCAL year%, start% start%=century% * 100 - 100 PRINT "The long years between ";start% " and ";start% + 100 " are "; FOR year%=start% TO start% + 99 IF FNLongYear(year%) PRINT STR$year% + " "; NEXT PRINT ENDPROC</lang>
- Output:
The long years between 1900 and 2000 are 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 The long years between 2000 and 2100 are 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 The long years between 2100 and 2200 are 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195
Commodore BASIC
<lang basic>100 REM M7(N) = N MOD 7 110 DEF FNM7(N) = N - 7*INT(N / 7) 120 : 130 REM WD(Y) = WEEKDAY OF Y-12-31, 0 = SUNDAY 140 DEF FNWD(Y) = FNM7(Y + INT(Y / 4) - INT(Y / 100) + INT(Y / 400)) 150 : 160 REM LY(Y) = NONZERO IF Y IS LONG 170 DEF FNLY(Y) = (4 = FNWD(Y)) OR (3 = FNWD(Y-1)) 180 : 190 PRINT CHR$(147); CHR$(18); "**** LIST OF ISO LONG YEARS ****" 200 INPUT "START YEAR"; S 210 INPUT "END YEAR"; E 220 PRINT 230 : 240 FOR Y = S TO E 250 : IF FNLY(Y) THEN PRINT Y, 260 NEXT Y 270 PRINT</lang>
- Output:
**** LIST OF ISO LONG YEARS **** START YEAR? 1995 END YEAR? 2045 1998 2004 2009 2015 2020 2026 2032 2037 2043 READY.
FreeBASIC
<lang freebasic>function p(y as unsigned integer) as unsigned integer
return ( y + int(y/4) - int(y/100) + int(y/400) ) mod 7
end function
function islongyear( y as uinteger ) as boolean
if p(y) = 4 then return true if p(y-1) = 3 then return true return false
end function
print islongyear(1998) print islongyear(2020) print islongyear(2021)</lang>
- Output:
true true false
GW-BASIC
<lang gwbasic>10 INPUT "Enter a year: ", Y 20 X = Y 30 GOSUB 100 40 IF P = 4 THEN L = 1 50 X = Y - 1 60 GOSUB 100 70 IF P = 3 THEN L = 1 80 IF L = 1 THEN PRINT Y; " is a long year." ELSE PRINT Y;" is not a long year." 90 END 100 P = X + INT(X/4) - INT(X/100) + INT(X/400) 110 P = P MOD 7 120 RETURN</lang>
IS-BASIC
<lang IS-BASIC>100 PROGRAM "Longyear.bas" 110 DEF RD(Y)=Y*365+INT(Y/4)-INT(Y/100)+INT(Y/400) 120 DEF LONGYEAR(Y)=(4=MOD(RD(Y),7)) OR(4=MOD((RD(Y-1)+1),7)) 130 INPUT PROMPT "Start year: ":S 140 INPUT PROMPT "End year: ":E 150 FOR Y=S TO E 160 IF LONGYEAR(Y) THEN PRINT Y, 170 NEXT 180 PRINT</lang>
Nascom BASIC
<lang basic> 10 REM Long year 20 REM FNM7(N)=MOD(N,7) 30 DEF FNM7(N)=N-7*INT(N/7) 40 REM FNWD(Y)=Weekday of Y-12-31, 0 Sunday 50 DEF FND(Y)=Y+INT(Y/4)-INT(Y/100)+INT(Y/400) 60 DEF FNWD(Y)=FNM7(FND(Y)) 70 REM FNLY(Y)=Nonzero if Y is long 80 DEF FNLY(Y)=(4=FNWD(Y))OR(3=FNWD(Y-1)) 90 CLS 100 PRINT "**** "; 110 PRINT "List of ISO long years"; 120 PRINT " ****" 130 INPUT "Start year";S 140 INPUT "End year";E 150 PRINT 160 FOR Y=S TO E 170 IF FNLY(Y) THEN PRINT Y; 180 NEXT Y 190 PRINT 200 END </lang>
- Output:
**** List of ISO long years **** Start year? 1995 End year? 2045 1998 2004 2009 2015 2020 2026 2032 2037 2043
PureBasic
<lang PureBasic>Procedure.b p(y)
ProcedureReturn (y + Int(y/4) - Int(y/100) + Int(y/400)) % 7
EndProcedure
Procedure.b isLongYear(y)
ProcedureReturn Bool((p(y) = 4) Or (p(y - 1) = 3))
EndProcedure
If OpenConsole()
For y = 2000 To 2100 If isLongYear(y) PrintN(Str(y)) EndIf Next y Print(""): Input() CloseConsole()
EndIf</lang>
- Output:
Igual que la entrada de BASIC256.
Quick BASIC
Translated from Delphi <lang vb> DEFINT A-Z
DECLARE FUNCTION p% (Yr AS INTEGER) DECLARE FUNCTION LongYear% (Yr AS INTEGER)
DIM iYi, iYf, i
CLS PRINT "This program calculates which are 53-week years in a range." PRINT INPUT "Initial year"; iYi INPUT "Final year (could be the same)"; iYf IF iYf >= iYi THEN
FOR i = iYi TO iYf IF LongYear(i) THEN PRINT i; " "; END IF NEXT i
END IF PRINT PRINT PRINT "End of program." END
FUNCTION LongYear% (Yr AS INTEGER)
LongYear% = (p%(Yr) = 4) OR (p%(Yr - 1) = 3)
END FUNCTION
FUNCTION p% (Yr AS INTEGER)
p% = (Yr + INT(Yr / 4) - INT(Yr / 100) + INT(Yr / 400)) MOD 7
END FUNCTION </lang>
- Output:
This program calculates which are 53-week years in a range. Initial year? 1900 Final year (can be the same)? 1999 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 End of program.
S-Basic
<lang BASIC> $lines
rem - compute p mod q function mod(p, q = integer) = integer end = p - q * (p/q)
comment
return day of week (Sun = 0, Mon = 1, etc.) for a given Gregorian calendar date using Zeller's congruence
end function dayofweek (mo, da, yr = integer) = integer
var y, c, z = integer if mo < 3 then begin mo = mo + 10 yr = yr - 1 end else mo = mo - 2 y = mod(yr,100) c = int(yr / 100) z = int((26 * mo - 2) / 10) z = z + da + y + int(y/4) + int(c/4) - 2 * c + 777 z = mod(z,7)
end = z
comment
The simplest of several possible tests is that any calendar year starting or ending on a Thursday is "long", i.e., has 53 ISO weeks
end function islongyear(yr = integer) = integer
var thursday, result = integer thursday = 4 if (dayofweek(1,1,yr) = thursday) or \ (dayofweek(12,31,yr) = thursday) then result = -1 rem "true" else result = 0 rem "false"
end = result
rem - main program begins here
var year = integer print "ISO years that will be long in this century:" for year = 2000 to 2099
if islongyear(year) then print year;
next year
end</lang>
- Output:
ISO years that will be long in this century: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Tiny BASIC
<lang tiny basic> PRINT "What year would you like?"
INPUT Y LET X = Y GOSUB 100 IF P = 4 THEN LET L = 1 LET X = Y - 1 GOSUB 100 IF P = 3 THEN LET L = 1 IF L = 1 THEN PRINT Y," is a long year." IF L = 0 THEN PRINT Y," is not a long year." END
100 LET P = X + X/4 - X/100 + X/400 110 IF P < 7 THEN RETURN
LET P = P - 7 GOTO 110
</lang>
- Output:
What year would you like? 2020 2020 is a long year. What year would you like? 2021 2021 is not a long year.
True BASIC
<lang qbasic>FUNCTION p(y) = REMAINDER((y + INT(y/4) - INT(y/100) + INT(y/400)), 7)
FUNCTION isLongYear(y)
IF p(y) = 4 THEN LET isLongYear = 1 ELSEIF p(y-1) = 3 THEN LET isLongYear = 1 ELSE LET isLongYear = 0 END IF
END FUNCTION
FOR y = 2000 TO 2100
IF isLongYear(y) > 0 THEN PRINT y
NEXT y END</lang>
- Output:
Igual que la entrada de BASIC256.
Yabasic
<lang yabasic>sub p(y) return mod((y + int(y/4) - int(y/100) + int(y/400)), 7) end sub
sub isLongYear(y) return (p(y) = 4) or (p(y - 1) = 3) end sub
for y = 2000 to 2100 if isLongYear(y) then print y : fi next y end</lang>
- Output:
Igual que la entrada de BASIC256.
BCPL
<lang bcpl>get "libhdr"
let p(y) = (y + y/4 - y/100 + y/400) rem 7 let longyear(y) = p(y)=4 | p(y-1)=3
let start() be
for y = 2000 to 2100 if longyear(y) do writef("%N*N", y)</lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
C#
<lang csharp>using static System.Console; using System.Collections.Generic; using System.Linq; using System.Globalization;
public static class Program {
public static void Main() { WriteLine("Long years in the 21st century:"); WriteLine(string.Join(" ", 2000.To(2100).Where(y => ISOWeek.GetWeeksInYear(y) == 53))); } public static IEnumerable<int> To(this int start, int end) { for (int i = start; i < end; i++) yield return i; }
}</lang>
- Output:
Long years in the 21st century: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
C
<lang cpp>#include <stdio.h>
- include <math.h>
// https://webspace.science.uu.nl/~gent0113/calendar/isocalendar.htm
int p(int year) { return (int)((double)year + floor(year/4) - floor(year/100) + floor(year/400)) % 7; }
int is_long_year(int year) { return p(year) == 4 || p(year - 1) == 3; }
void print_long_years(int from, int to) { for (int year = from; year <= to; ++year) { if (is_long_year(year)) { printf("%d ", year); } } }
int main() {
printf("Long (53 week) years between 1800 and 2100\n\n"); print_long_years(1800, 2100); printf("\n"); return 0; }</lang>
- Output:
Long (53 week) years between 1800 and 2100 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
C++
<lang cpp>// Reference: // https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year
- include <iostream>
inline int p(int year) {
return (year + (year/4) - (year/100) + (year/400)) % 7;
}
bool is_long_year(int year) {
return p(year) == 4 || p(year - 1) == 3;
}
void print_long_years(int from, int to) {
for (int year = from, count = 0; year <= to; ++year) { if (is_long_year(year)) { if (count > 0) std::cout << ((count % 10 == 0) ? '\n' : ' '); std::cout << year; ++count; } }
}
int main() {
std::cout << "Long years between 1800 and 2100:\n"; print_long_years(1800, 2100); std::cout << '\n'; return 0;
}</lang>
- Output:
Long years between 1800 and 2100: 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Clojure
<lang clojure>(defn long-year? [year]
(-> (java.time.LocalDate/of year 12 28) (.get (.weekOfYear (java.time.temporal.WeekFields/ISO))) (= 53)))
(filter long-year? (range 2000 2100))</lang>
- Output:
(2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099)
CLU
<lang clu>% We can't hide one procedure inside another, but % we can hide the helper `p' in a cluster
longyear = cluster is test
rep = null p = proc (n: int) returns (int) return ((n + n/4 - n/100 + n/400) // 7) end p
test = proc (y: int) returns (bool) return (p(y)=4 | p(y-1)=3) end test
end longyear
start_up = proc ()
po: stream := stream$primary_output() for i: int in int$from_to(2000, 2100) do if longyear$test(i) then stream$putl(po, int$unparse(i)) end end
end start_up</lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Common Lisp
<lang lisp>(defun december-31-weekday (year)
(mod (+ year (floor year 4) (- (floor year 100)) (floor year 400)) 7))
(defun iso-long-year-p (year)
(or (= 4 (december-31-weekday year)) (= 3 (december-31-weekday (1- year)))))
(format t "Long years between 1800 and 2100:~&~a~%"
(loop for y from 1800 to 2100 if (iso-long-year-p y) collect y))</lang>
- Output:
Long years between 1800 and 2100: (1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099)
Dc
<lang dc>[0q]s0 [1q]s1 [1r- r 1r- * 1r-]sO # O = logical OR
- .............................................................................
- C: for( initcode ; condcode ; incrcode ) {body}
- .[q] [1] [2] [3] [4]
- # [initcode] [condcode] [incrcode] [body] (for)
[ [q]S. 4:.3:.2:.x [2;.x 0=. 4;.x 3;.x 0;.x]d0:.x
Os.L.o
]sF # F = for
- .............................................................................
- [1] [0]
- (.) [cond_code] [then_code] [else_code] (if_CTE)
[ []S. 0:. 1:. x [0=0 1]x ;. s.L. x]sI # I = if
- -----------------------------------------------------------------------------
[S. l. l.4/+ l.100/- l.400/+ 7% s.L.]sp # p
- .............................................................................
[S. [l. lpx 4=1 0]x
[l. 1- lpx 3=1 0]x lOx s.L.
]si # i = is_long_year
- .............................................................................
[
# f = from # t = to # y = year # c = count st sf # fetch args from stack [lfsy 0sc] [ly lt <0 1] # cond [ly 1+ sy] # incr y [ [ly lix] # is_long_year(y) [ [lc 0 <1 0] # 0<c [ [ lc 10% 0=1 0] # (c % 10) == 0 [ AP ] [ [ ]P ] lIx # if ] [] lIx # if ly n lc 1+ sc ] [] lIx # if ] lFx # for
]sD # D = doit = print_long_years
- .............................................................................
[Long years between 1800 and 2100:]P AP 1800 2100 lDx AP</lang>
- Output:
Long years between 1800 and 2100: 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Delphi
Note: The Library System.DateUtils implement a WeeksInYear,but not working, return 52 always.
<lang Delphi> program Long_year;
{$APPTYPE CONSOLE}
{$R *.res}
uses
System.SysUtils;
function p(const Year: Integer): Integer; begin
Result := (Year + (Year div 4) - (Year div 100) + (Year div 400)) mod 7;
end;
function IsLongYear(const Year: Integer): Boolean; begin
Result := (p(Year) = 4) or (p(Year - 1) = 3);
end;
procedure PrintLongYears(const StartYear: Integer; const EndYear: Integer); var
Year, Count: Integer;
begin
Count := 0; for Year := 1800 to 2100 do if IsLongYear(Year) then begin if Count mod 10 = 0 then Writeln; Write(Year, ' '); inc(Count); end;
end;
var
Year: Integer;
begin
Writeln('Long years between 1800 and 2100:'); PrintLongYears(1800, 2100); Readln;
end. </lang>
Elixir
<lang elixir>defmodule ISO do
def long_year?(y) do {:ok, jan1} = Date.new(y,1,1) {:ok, dec31} = Date.new(y,12,31) Date.day_of_week(jan1) == 4 or Date.day_of_week(dec31) == 4 end
end
IO.inspect(Enum.filter(1990..2050, &ISO.long_year?/1))</lang>
- Output:
[1992, 1998, 2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048]
Factor
<lang factor>USING: calendar formatting io kernel math.ranges sequences ;
- long-year? ( n -- ? ) 12 28 <date> week-number 53 = ;
"Year Long?\n-----------" print 1990 2021 [a,b] [ dup long-year? "yes" "no" ? "%d %s\n" printf ] each</lang>
- Output:
Year Long? ----------- 1990 no 1991 no 1992 yes 1993 no 1994 no 1995 no 1996 no 1997 no 1998 yes 1999 no 2000 no 2001 no 2002 no 2003 no 2004 yes 2005 no 2006 no 2007 no 2008 no 2009 yes 2010 no 2011 no 2012 no 2013 no 2014 no 2015 yes 2016 no 2017 no 2018 no 2019 no 2020 yes 2021 no
Forth
<lang forth>: dec31wd ( year -- weekday ) dup dup 4 / swap dup 100 / swap 400 / swap - + + 7 mod ;
- long? ( year -- flag ) dup dec31wd 4 = if drop 1 else 1 - dec31wd 3 = if 1 else 0 then then ;
- demo ( startyear endyear -- ) cr swap do i long? if i . then loop cr ;</lang>
- Output:
1995 2045 demo 1998 2004 2009 2015 2020 2026 2032 2037 2043 ok
Fortran
<lang fortran> program longyear
use iso_fortran_env, only: output_unit, input_unit implicit none
integer :: start, ende, i, counter integer, parameter :: line_break=10
write(output_unit,*) "Enter beginning of interval" read(input_unit,*) start write(output_unit,*) "Enter end of interval" read(input_unit,*) ende
if (start>=ende) error stop "Last year must be after first year!"
counter = 0 do i = start, ende if (is_long_year(i)) then write(output_unit,'(I0,x)', advance="no") i counter = counter + 1 if (modulo(counter,line_break) == 0) write(output_unit,*) end if end do
contains
pure function p(year) integer, intent(in) :: year integer :: p
p = modulo(year + year/4 - year/100 + year/400, 7) end function p
pure function is_long_year(year) integer, intent(in) :: year logical :: is_long_year
is_long_year = p(year) == 4 .or. p(year-1) == 3 end function is_long_year
end program longyear </lang>
- Output:
Enter beginning of interval 1800 Enter end of interval 2100 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 %
Go
<lang go>package main
import (
"fmt" "time"
)
func main() {
centuries := []string{"20th", "21st", "22nd"} starts := []int{1900, 2000, 2100} for i := 0; i < len(centuries); i++ { var longYears []int fmt.Printf("\nLong years in the %s century:\n", centuries[i]) for j := starts[i]; j < starts[i] + 100; j++ { t := time.Date(j, time.December, 28, 0, 0, 0, 0, time.UTC) if _, week := t.ISOWeek(); week == 53 { longYears = append(longYears, j) } } fmt.Println(longYears) }
}</lang>
- Output:
Long years in the 20th century: [1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998] Long years in the 21st century: [2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099] Long years in the 22nd century: [2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195]
Haskell
<lang haskell>import Data.Time.Calendar (fromGregorian) import Data.Time.Calendar.WeekDate (toWeekDate)
longYear :: Integer -> Bool longYear y =
let (_, w, _) = toWeekDate $ fromGregorian y 12 28 in 52 < w
main :: IO () main = mapM_ print $ filter longYear [2000 .. 2100]</lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
J
<lang j> p =: 1 4 _100 400&(7 | [: <. +/ @: %~)"1 0
ily =: (4=p) +. 3=p@:<: ply =: (#~ ily)@:([ + 1+i.@:-~)</lang>
- Output:
ply/ 1800 2100 1801 1807 1812 1818 1823 1829 1835 1840 1846 1852 1857 1863 1869 1874 1880 1885 1891 1897 1902 1908 1914 1919 1925 1930 1936 1942 1947 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2021 2026 2032 2038 2043 2049 2054 2060 2066 2071 2077 2083 2...
Java
<lang java> import java.time.LocalDate; import java.time.temporal.WeekFields;
public class LongYear {
public static void main(String[] args) { System.out.printf("Long years this century:%n"); for (int year = 2000 ; year < 2100 ; year++ ) { if ( longYear(year) ) { System.out.print(year + " "); } } } private static boolean longYear(int year) { return LocalDate.of(year, 12, 28).get(WeekFields.ISO.weekOfYear()) == 53; }
} </lang>
- Output:
Long years this century: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
JavaScript
<lang javascript>const isLongYear = (year) => {
const jan1 = new Date(year, 0, 1); const dec31 = new Date(year, 11, 31); return (4 == jan1.getDay() || 4 == dec31.getDay())
}
for (let y = 1995; y <= 2045; y++) {
if (isLongYear(y)) { console.log(y) }
}</lang>
- Output:
1998 2004 2009 2015 2020 2026 2032 2037 2043
jq
Works with gojq, the Go implementation of jq
Using Zeller's congruence
<lang jq>
- Use Zeller's Congruence to determine the day of the week, given
- year, month and day as integers in the conventional way.
- Emit 0 for Saturday, 1 for Sunday, etc.
def day_of_week($year; $month; $day):
if $month == 1 or $month == 2 then [$month + 12, $year - 1] else [$month, $year] end | $day + (13*(.[0] + 1)/5|floor) + (.[1]%100) + ((.[1]%100)/4|floor) + (.[1]/400|floor) - 2*(.[1]/100|floor) | . % 7 ;
def has53weeks:
day_of_week(.; 1; 1) == 5 or day_of_week(.; 12; 31) == 5;
- To display results neatly:
def nwise($n):
def n: if length <= $n then . else .[0:$n] , (.[$n:] | n) end; n;
"Long years from 1900 to 2100 inclusive:", ([range(1900;2101) | select(has53weeks)] | nwise(10) | join(", "))</lang>
- Output:
Long years from 1900 to 2100 inclusive: 1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953 1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998, 2004, 2009 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065 2071, 2076, 2082, 2088, 2093, 2099
Using mktime and gmtime
<lang jq># Use jq's mktime and gmtime to produce the day of week,
- with 0 for Sunday, 1 for Monday, etc
- $year $month $day are conventional
def day_of_week_per_gmtime($year; $month; $day):
[$year, $month - 1, $day, 0, 0, 1, 0, 0] | mktime | gmtime | .[-2];
- 4 corresponds to Thursday
def has53weeks:
day_of_week_per_gmtime(.; 1; 1) == 4 or day_of_week(.; 12; 31) == 4;
def nwise($n):
def n: if length <= $n then . else .[0:$n] , (.[$n:] | n) end; n;
"Long years from 1900 to 2100 inclusive:", ([range(1900;2101) | select(has53weeks)] | nwise(10) | join(", "))</lang>
- Output:
As above.
Julia
<lang julia>using Dates
has53weeks(year) = week(Date(year, 12, 28)) == 53
println(" Year 53 weeks?\n----------------") for year in 1990:2021
println(year, " ", has53weeks(year) ? "Yes" : "No")
end
</lang>
- Output:
Year 53 weeks? ---------------- 1990 No 1991 No 1992 Yes 1993 No 1994 No 1995 No 1996 No 1997 No 1998 Yes 1999 No 2000 No 2001 No 2002 No 2003 No 2004 Yes 2005 No 2006 No 2007 No 2008 No 2009 Yes 2010 No 2011 No 2012 No 2013 No 2014 No 2015 Yes 2016 No 2017 No 2018 No 2019 No 2020 Yes 2021 No
Kotlin
<lang Kotlin> fun main() {
val has53Weeks = { year: Int -> LocalDate.of(year, 12, 28).get(WeekFields.ISO.weekOfYear()) == 53 } println("Long years this century:") (2000..2100).filter(has53Weeks) .forEach { year -> print("$year ")}
} </lang>
- Output:
Long years this century: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Logo
<lang logo>to div :x :y
output int quotient :x :y
end
to dec31_weekday :year
output remainder (sum :year div :year 4 div :year -100 div :year 400) 7
end
to iso_long_year? :year
output or 4 = dec31_weekday :year 3 = dec31_weekday difference :year 1
end
for [y 1995 2045 1] [if iso_long_year? :y [print :y]]</lang>
- Output:
1998 2004 2009 2015 2020 2026 2032 2037 2043
Mathematica/Wolfram Language
<lang Mathematica>firstyear = 2000; lastyear = 2099; years = Range[firstyear, lastyear]; firstday = Table[DayName[{yearsn, 01, 01}], {n, Length[years]}]; lastday = Table[DayName[{yearsn, 12, 31}], {n, Length[years]}]; Table[If[yearsn >= 1582,
If[firstdayn == Thursday || lastdayn == Thursday, Style[yearsn " long year \n", Bold, Red] , yearsn " short \n"], "error \n"], {n, Length[years]}]</lang>
- Output:
{2000 short , 2001 short , 2002 short , 2003 short , 2004 long year , 2005 short , 2006 short , 2007 short , 2008 short , 2009 long year , 2010 short , 2011 short , 2012 short , 2013 short , 2014 short , 2015 long year , 2016 short , 2017 short , 2018 short , 2019 short , 2020 long year , 2021 short , 2022 short , 2023 short , 2024 short , 2025 short , 2026 long year , 2027 short , 2028 short , 2029 short , 2030 short , 2031 short , 2032 long year , 2033 short , 2034 short , 2035 short , 2036 short , 2037 long year , 2038 short , 2039 short , 2040 short , 2041 short , 2042 short , 2043 long year , 2044 short , 2045 short , 2046 short , 2047 short , 2048 long year , 2049 short , 2050 short , 2051 short , 2052 short , 2053 short , 2054 long year , 2055 short , 2056 short , 2057 short , 2058 short , 2059 short , 2060 long year , 2061 short , 2062 short , 2063 short , 2064 short , 2065 long year , 2066 short , 2067 short , 2068 short , 2069 short , 2070 short , 2071 long year , 2072 short , 2073 short , 2074 short , 2075 short , 2076 long year , 2077 short , 2078 short , 2079 short , 2080 short , 2081 short , 2082 long year , 2083 short , 2084 short , 2085 short , 2086 short , 2087 short , 2088 long year , 2089 short , 2090 short , 2091 short , 2092 short , 2093 long year , 2094 short , 2095 short , 2096 short , 2097 short , 2098 short , 2099 long year }
Modula-2
<lang modula2>MODULE LongYear; FROM InOut IMPORT WriteCard, WriteLn;
VAR year: CARDINAL;
PROCEDURE isLongYear(year: CARDINAL): BOOLEAN;
PROCEDURE p(year: CARDINAL): CARDINAL; BEGIN RETURN (year + year DIV 4 - year DIV 100 + year DIV 400) MOD 7; END p;
BEGIN
RETURN (p(year) = 4) OR (p(year-1) = 3);
END isLongYear;
BEGIN
FOR year := 2000 TO 2100 DO IF isLongYear(year) THEN WriteCard(year, 4); WriteLn; END; END;
END LongYear.</lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Nim
<lang Nim>import times
proc has53weeks(year: Positive): bool =
let dt = initDateTime(monthday = 1, month = mJan, year = year, hour = 0, minute = 0, second= 0) result = dt.weekday == dThu or year.isLeapYear and dt.weekday == dWed
when isMainModule:
echo "Years with 53 weeks between 2000 and 2100:" for year in 2000..2100: if year.has53weeks: echo year</lang>
- Output:
Years with 53 weeks between 2000 and 2100: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
МК-61/52
<lang mk-61>П0 ИП0 4 / [x] + ИП0 1 ВП 2 / [x] - ИП0 4 ВП 2 / [x] + ^ ^ 7 / [x] 7 * - П1 4 - x#0 40 ИП1 3 - x#0 40 0 С/П 1 С/П</lang>
- Output:
Result for 2020-2030 years: 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0.
Pascal
<lang pascal>program long_year(input);
var y: integer;
function rd_dec31(year: integer): integer; begin { Rata Die of Dec 31, year } rd_dec31 := year * 365 + year div 4 - year div 100 + year div 400 end;
function rd_jan1(year: integer): integer; begin rd_jan1 := rd_dec31(year - 1) + 1 end;
function weekday(rd: integer): integer; begin weekday := rd mod 7; end;
function long_year(year: integer): boolean; var jan1: integer; dec31: integer; begin jan1 := rd_jan1(year); dec31 := rd_dec31(year); long_year := (weekday(jan1) = 4) or (weekday(dec31) = 4) end;
begin for y := 1990 to 2050 do if long_year(y) then writeln(y) end.</lang>
- Output:
1993 1999 2004 2010 2016 2021 2027 2032 2038 2044 2049
Free Pascal
Using DateUtils and WeeksInYear to not reinvent this. <lang pascal> program Long_year;
uses
SysUtils, DateUtils;
procedure PrintLongYears(StartYear, EndYear: Uint32); var Year, Count: Uint32; DateSep: char; begin DateSep := FormatSettings.DateSeparator; Writeln('Long years between ', StartYear, ' and ', EndYear); Count := 0; for Year := StartYear to EndYear do if WeeksInYear(StrToDate('01' + DateSep + '01' + DateSep + IntToStr(Year))) = 53 then begin if Count mod 10 = 0 then Writeln; Write(Year, ' '); Inc(Count); end; if Count mod 10 <> 0 then Writeln; writeln('Found ', Count, ' long years between ', StartYear, ' and ', EndYear); end;
begin
PrintLongYears(1800, 2100); {$IFDEF WINDOWS} Readln; {$ENDIF}
end.</lang>
- Output:
Long years between 1800 and 2100 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 Found 54 long years between 1800 and 2100
Perl
<lang perl>use strict; use warnings; use DateTime;
for my $century (19 .. 21) {
for my $year ($century*100 .. ++$century*100 - 1) { print "$year " if DateTime->new(year => $year, month => 12, day => 28)->week_number > 52 } print "\n";
}</lang>
- Output:
1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195
Phix
with javascript_semantics function week_number(integer y,m,d) integer doy = day_of_year(y,m,d), dow = day_of_week(y,m,d), week = floor((doy-dow+10)/7) return week end function for c=20 to 22 do sequence long_years = {} integer century = (c-1)*100 for year=century to century+99 do if week_number(year,12,28)=53 then long_years &= year end if end for printf(1,"Long years in the %d%s century:%v\n", {c,ord(c),long_years}) end for
- Output:
Long years in the 20th century:{1903,1908,1914,1920,1925,1931,1936,1942,1948,1953,1959,1964,1970,1976,1981,1987,1992,1998} Long years in the 21st century:{2004,2009,2015,2020,2026,2032,2037,2043,2048,2054,2060,2065,2071,2076,2082,2088,2093,2099} Long years in the 22nd century:{2105,2111,2116,2122,2128,2133,2139,2144,2150,2156,2161,2167,2172,2178,2184,2189,2195}
PHP
<lang php>function isLongYear($year) {
return (53 == strftime('%V', gmmktime(0,0,0,12,28,$year)));
}
for ($y=1995; $y<=2045; ++$y) {
if (isLongYear($y)) { printf("%s\n", $y); }
}</lang>
- Output:
1998 2004 2009 2015 2020 2026 2032 2037 2043
PowerShell
<lang powershell>Function Is-Long-Year {
param([Int]$year) 53 -eq (Get-Date -Year $year -Month 12 -Day 28 -UFormat %V)
}
For ($y=1995; $y -le 2045; $y++) {
If (Is-Long-Year $y) { Write-Host $y }
}</lang>
- Output:
1998 2004 2009 2015 2020 2026 2032 2037 2043
Prolog
<lang prolog>% See https://en.wikipedia.org/wiki/ISO_week_date#Weeks_per_year
p(Year, P):-
P is (Year + (Year//4) - (Year//100) + (Year//400)) mod 7.
long_year(Year):-
p(Year, 4), !.
long_year(Year):-
Year_before is Year - 1, p(Year_before, 3).
print_long_years(From, To):-
writef("Long years between %w and %w:\n", [From, To]), print_long_years(From, To, 0), nl.
print_long_years(From, To, _):-
From > To, !.
print_long_years(From, To, Count):-
long_year(From), !, (Count > 0 -> (0 is Count mod 10 -> nl ; write(' ')) ; true ), write(From), Count1 is Count + 1, Next is From + 1, print_long_years(Next, To, Count1).
print_long_years(From, To, Count):-
Next is From + 1, print_long_years(Next, To, Count).
main:-
print_long_years(1800, 2100).</lang>
- Output:
Long years between 1800 and 2100: 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Python
<lang python>Long Year ?
from datetime import date
- longYear :: Year Int -> Bool
def longYear(y):
True if the ISO year y has 53 weeks. return 52 < date(y, 12, 28).isocalendar()[1]
- --------------------------TEST---------------------------
- main :: IO ()
def main():
Longer (53 week) years in the range 2000-2100 for year in [ x for x in range(2000, 1 + 2100) if longYear(x) ]: print(year)
- MAIN ---
if __name__ == '__main__':
main()</lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Quackery
dayofweek
is defined at Day of the week#Quackery
<lang Quackery> [ dup dip
[ 1 1 rot dayofweek 4 = ] 31 12 rot dayofweek 4 = or ] is longyear ( n --> b )
say "Long Years in the 21st Century" cr cr 100 times [ 2000 i^ + longyear if [ 2000 i^ + echo sp ] ]</lang>
- Output:
Long Years in the 21st Century 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Raku
(formerly Perl 6)
December 28 is always in the last week of the year. (By ISO8601) <lang perl6>sub is-long ($year) { Date.new("$year-12-28").week[1] == 53 }
- Testing
say "Long years in the 20th century:\n", (1900..^2000).grep: &is-long; say "\nLong years in the 21st century:\n", (2000..^2100).grep: &is-long; say "\nLong years in the 22nd century:\n", (2100..^2200).grep: &is-long;</lang>
- Output:
Long years in the 20th century: (1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998) Long years in the 21st century: (2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099) Long years in the 22nd century: (2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195)
REXX
<lang rexx>/*REXX program determines if a (calendar) year is a SHORT or LONG year (52 or 53 weeks).*/ parse arg LO HI . /*obtain optional args. */ if LO== | LO=="," | LO=='*' then LO= left( date('S'), 4) /*Not given? Use default.*/ if HI== | HI=="," then HI= LO /* " " " " */ if HI=='*' then HI= left( date('S'), 4) /*an asterisk ≡ current yr*/
do j=LO to HI /*process single yr or range of years.*/ say ' year ' j " is a " right( word('short long', weeks(j)-51),5) " year" end /*j*/
exit 0 /*stick a fork in it, we're all done. */ /*──────────────────────────────────────────────────────────────────────────────────────*/ pWeek: parse arg #; return (# + # % 4 - # % 100 + # % 400) // 7 weeks: parse arg y; if pWeek(y)==4 | pWeek(y-1)==3 then return 53; return 52</lang>
- output when using the inputs of: 1990 2030
(Shown at three-quarter size.)
year 1990 is a short year year 1991 is a short year year 1992 is a long year year 1993 is a short year year 1994 is a short year year 1995 is a short year year 1996 is a short year year 1997 is a short year year 1998 is a long year year 1999 is a short year year 2000 is a short year year 2001 is a short year year 2002 is a short year year 2003 is a short year year 2004 is a long year year 2005 is a short year year 2006 is a short year year 2007 is a short year year 2008 is a short year year 2009 is a long year year 2010 is a short year year 2011 is a short year year 2012 is a short year year 2013 is a short year year 2014 is a short year year 2015 is a long year year 2016 is a short year year 2017 is a short year year 2018 is a short year year 2019 is a short year year 2020 is a long year year 2021 is a short year year 2022 is a short year year 2023 is a short year year 2024 is a short year year 2025 is a short year year 2026 is a long year year 2027 is a short year year 2028 is a short year year 2029 is a short year year 2030 is a short year
Ring
<lang ring> see "long years 2000-2099: " for year = 2000 to 2100
num1 = (year-1900)%7 num2 = floor((year-1904)/4) num3 = (num1+num2+5)%7 if num3 = 0 or (num1 = 6 and num3 = 1) see "" + year + " " ok
next </lang>
- Output:
long years 2000-2099: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Ruby
<lang ruby> require 'date'
def long_year?(year = Date.today.year)
Date.new(year, 12, 28).cweek == 53
end
(2020..2030).each{|year| puts "#{year} is long? #{ long_year?(year) }." } </lang>
- Output:
2020 is long? true. 2021 is long? false. 2022 is long? false. 2023 is long? false. 2024 is long? false. 2025 is long? false. 2026 is long? true. 2027 is long? false. 2028 is long? false. 2029 is long? false. 2030 is long? false.
Rust
<lang Rust>extern crate time; // 0.2.16
use time::Date;
fn main() {
(2000..=2099) .filter(|&year| is_long_year(year)) .for_each(|year| println!("{}", year));
}
fn is_long_year(year: i32) -> bool {
Date::try_from_ymd(year, 12, 28).map_or(false, |date| date.week() == 53)
} </lang>
- Output:
2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Scala
- Output:
Best seen running in your browser by Scastie (remote JVM).
<lang Scala>import java.time.temporal.TemporalAdjusters.firstInMonth import java.time.temporal.{ChronoField, IsoFields} import java.time.{DayOfWeek, LocalDate, Month}
import scala.util.{Failure, Try}
private object LongYear extends App {
private val (currentCentury, maxWeekNumber) = (LocalDate.now().getYear / 100, ChronoField.ALIGNED_WEEK_OF_YEAR.range().getMaximum) private val centuries = currentCentury * 100 until (currentCentury + 1) * 100 private val results = List( centuries.filter(isThursdayFirstOrLast), centuries.filter(year => maxIsoWeeks(year) == maxWeekNumber), centuries.filter(mostThursdaysInYear) )
// Solution 1, the first or respectively last day of the year is a Thursday. private def isThursdayFirstOrLast(_year: Int): Boolean = {
LocalDate.of(_year, Month.DECEMBER, 31).get(ChronoField.DAY_OF_WEEK) == DayOfWeek.THURSDAY.getValue || LocalDate.of(_year, Month.JANUARY, 1).get(ChronoField.DAY_OF_WEEK) == DayOfWeek.THURSDAY.getValue }
// Solution 2, if last week that contains at least four days of the month of December. private def maxIsoWeeks(_year: Int) = { // The last week that contains at least four days of the month of December. LocalDate.of(_year, Month.DECEMBER, 28).get(IsoFields.WEEK_OF_WEEK_BASED_YEAR) }
// Solution 3, if there are 52 Thursdays in a year private def mostThursdaysInYear(_year: Int) = { val datum = LocalDate.of(_year, Month.JANUARY, 1).`with`(firstInMonth(DayOfWeek.THURSDAY))
datum.plusDays(52 * 7).getYear == _year }
println(s"Years in this ${currentCentury + 1}st century having ISO week $maxWeekNumber :")
Try { // Testing the solutions assert(results.tail.forall(_ == results.head), "Discrepancies in results.") } match { case Failure(ex) => Console.err.println(ex.getMessage) case _ => }
results.zipWithIndex.foreach(solution => println(s"Solution ${solution._2}: ${solution._1.mkString(" ")}"))
}</lang>
Scheme
The demo code uses iota as defined in SRFI-1, so won't work in Schemes lacking that function. Racket requires the addition of (require srfi/1).
<lang scheme>(define (dec31wd year)
(remainder (apply + (map (lambda (d) (quotient year d)) '(1 4 -100 400))) 7))
(define (long? year) (or (= 4 (dec31wd year)) (= 3 (dec31wd (- year 1)))))
(display "Long years between 1800 and 2100:") (newline) (display (filter long? (iota 300 1800)))</lang>
- Output:
Long years between 1800 and 2100: (1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099)
Sidef
<lang ruby>func is_long_year(year) {
Date.parse("#{year}-12-28", "%Y-%m-%d").week == 53
}
say ( "Long years in the 20th century:\n", (1900..^2000).grep(is_long_year)) say ("\nLong years in the 21st century:\n", (2000..^2100).grep(is_long_year)) say ("\nLong years in the 22nd century:\n", (2100..^2200).grep(is_long_year))</lang>
- Output:
Long years in the 20th century: [1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953, 1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998] Long years in the 21st century: [2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099] Long years in the 22nd century: [2105, 2111, 2116, 2122, 2128, 2133, 2139, 2144, 2150, 2156, 2161, 2167, 2172, 2178, 2184, 2189, 2195]
Snobol
<lang snobol> DEFINE('DEC31WD(Year)') :(END_DEC31WD) DEC31WD DEC31WD = REMDR(Year + (Year / 4) - (Year / 100) + (Year / 400), 7) :(RETURN) END_DEC31WD
DEFINE('ISOLONG(Year)') :(END_ISOLONG)
ISOLONG EQ(DEC31WD(Year), 4) :S(RETURN)
EQ(DEC31WD(Year - 1), 3) :S(RETURN)F(FRETURN)
END_ISOLONG
DEFINE('ISODEMO(Start,End)') :(END_ISODEMO)
ISODEMO OUTPUT = 'ISO long years between ' Start ' and ' End ':'
Year = Start
LOOP OUTPUT = ISOLONG(Year) Year
Year = Year + 1 LE(YEAR, 2045) :S(LOOP) F(RETURN)
END_ISODEMO
ISODEMO(1995, 2045)
END</lang>
- Output:
ISO long years between 1995 and 2045: 1998 2004 2009 2015 2020 2026 2032 2037 2043
Swift
<lang swift>func isLongYear(_ year: Int) -> Bool {
let year1 = year - 1 let p = (year + (year / 4) - (year / 100) + (year / 400)) % 7 let p1 = (year1 + (year1 / 4) - (year1 / 100) + (year1 / 400)) % 7
return p == 4 || p1 == 3
}
for range in [1900...1999, 2000...2099, 2100...2199] {
print("\(range): \(range.filter(isLongYear))")
}</lang>
- Output:
1900...1999: [1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953, 1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998] 2000...2099: [2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099] 2100...2199: [2105, 2111, 2116, 2122, 2128, 2133, 2139, 2144, 2150, 2156, 2161, 2167, 2172, 2178, 2184, 2189, 2195]
Tcl
<lang Tcl>
proc p {year} {
return [expr {($year + ($year/4) - ($year/100) + ($year/400)) % 7}]
}
proc is_long_year {year} {
return [expr {[p $year] == 4 || [p [expr {$year - 1}]] == 3}]
}
proc print_long_years {from to} {
for {set year $from; set count 0} {$year <= $to} {incr year} { if {[is_long_year $year]} { if {$count > 0} { puts -nonewline [expr {($count % 10 == 0) ? "\n" : " "}] } puts -nonewline $year incr count } }
}
puts "Long years between 1800 and 2100:" print_long_years 1800 2100 puts ""</lang>
- Output:
Long years between 1800 and 2100: 1801 1807 1812 1818 1824 1829 1835 1840 1846 1852 1857 1863 1868 1874 1880 1885 1891 1896 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099
Terraform
Contents of main module:
<lang terraform>module "iso-long-years" {
source = "./iso-long-years" start_year = 1995 end_year = 2045
}
output "long-years" {
value = module.iso-long-years.long-years
}</lang>
Contents of iso-long-years module:
<lang terraform>variable start_year {
type = number
}
variable end_year {
type = number
}
locals {
year_list = range(var.start_year, var.end_year+1)
}
module "iso-long-year" {
for_each = toset([for y in local.year_list: tostring(y)]) source = "../iso-long-year" year = each.key
}
output "long-years" {
value = compact([for y in [for n in local.year_list: tostring(n)]: module.iso-long-year[y].isLong ? y : ""])
}</lang>
Contents of iso-long-year module:
<lang terraform>variable year {
type = string default = ""
}
locals {
ystr = var.year != "" ? var.year : split("-",timestamp())[0] y = tonumber(local.ystr) e = local.y - 1 dec31 = local.y * 365 + floor(local.y/4) - floor(local.y/100) + floor(local.y/400) jan1 = local.e * 365 + floor(local.e/4) - floor(local.e/100) + floor(local.e/400) + 1
}
output isLong {
value = (local.dec31 % 7 == 4 || local.jan1 % 7 == 4)
}</lang>
- Output:
Apply complete! Resources: 0 added, 0 changed, 0 destroyed. Outputs: long-years = [ "1998", "2004", "2009", "2015", "2020", "2026", "2032", "2037", "2043", ]
TypeScript
<lang typescript>const isLongYear = (year: number): boolean => {
const jan1: Date = new Date(year, 0, 1); const dec31: Date = new Date(year, 11, 31); return (4 == jan1.getDay() || 4 == dec31.getDay())
}
for (let y: number = 1995; y <= 2045; y++) {
if (isLongYear(y)) { console.log(y) }
}</lang>
- Output:
1998 2004 2009 2015 2020 2026 2032 2037 2043
UNIX Shell
Thursdays check using cal(1) and grep(1)
<lang sh>long_year() {
cal 1 $1 | grep -q ' 3 *$' && return 0 cal 12 $1 | grep -q ' 26 *$'
}</lang>
Straightforward check using GNU date(1)
<lang sh>long_year() {
expr $(date -d "$1-12-28" +%V) = 53 >/dev/null
}</lang>
Direct computation with built-in arithmetic in newer shells
<lang sh>dec31wd() {
typeset -i y=$1 echo $(( (y + y / 4 - y / 100 + y / 400) % 7 ))
}
long_year() {
typeset -i y=$1 (( 4 == $(dec31wd $y) || 3 == $(dec31wd $(( y - 1 ))) ))
}</lang>
Demo code for any of the above:
<lang sh>for y in $(seq 1995 2045); do
if long_year $y; then echo $y fi
done | column </lang>
- Output:
1998 2004 2009 2015 2020 2026 2032 2037 2043
Visual Basic
<lang vb>Option Explicit
Function IsLongYear(ByVal Year As Integer) As Boolean
Select Case vbThursday Case VBA.DatePart("w", VBA.DateSerial(Year, 1, 1)), _ VBA.DatePart("w", VBA.DateSerial(Year, 12, 31)) IsLongYear = True End Select
End Function
Sub Main() 'test Dim l As Long
For l = 1990 To 2021 Select Case l Case 1992, 1998, 2004, 2009, 2015, 2020 Debug.Assert IsLongYear(l) Case Else Debug.Assert Not IsLongYear(l) End Select Next l
End Sub </lang>
Visual Basic for DOS
Translated from Delohi <lang vb>OPTION EXPLICIT
DECLARE FUNCTION p (Yr AS INTEGER) AS INTEGER DECLARE FUNCTION LongYear (Yr AS INTEGER) AS INTEGER
DIM iYi AS INTEGER, iYf AS INTEGER, i AS INTEGER
CLS PRINT "This program calculates which are 53-week years in a range." PRINT INPUT "Initial year"; iYi INPUT "Final year (could be the same)"; iYf IF iYf >= iYi THEN
FOR i = iYi TO iYf IF LongYear(i) THEN PRINT i; " "; END IF NEXT i
END IF PRINT PRINT PRINT "End of program." END
FUNCTION p (Yr AS INTEGER) AS INTEGER
p = (Yr + INT(Yr / 4) - INT(Yr / 100) + INT(Yr / 400)) MOD 7
END FUNCTION
FUNCTION LongYear (Yr AS INTEGER) AS INTEGER
LongYear = (p(Yr) = 4) OR (p(Yr - 1) = 3)
END FUNCTION </lang>
Wren
<lang ecmascript>import "/date" for Date
var centuries = ["20th", "21st", "22nd"] var starts = [1900, 2000, 2100] for (i in 0...centuries.count) {
var longYears = [] System.print("\nLong years in the %(centuries[i]) century:") for (j in starts[i]...starts[i]+100) { var t = Date.new(j, 12, 28) if (t.weekOfYear[1] == 53) { longYears.add(j) } } System.print(longYears)
}</lang>
- Output:
Long years in the 20th century: [1903, 1908, 1914, 1920, 1925, 1931, 1936, 1942, 1948, 1953, 1959, 1964, 1970, 1976, 1981, 1987, 1992, 1998] Long years in the 21st century: [2004, 2009, 2015, 2020, 2026, 2032, 2037, 2043, 2048, 2054, 2060, 2065, 2071, 2076, 2082, 2088, 2093, 2099] Long years in the 22nd century: [2105, 2111, 2116, 2122, 2128, 2133, 2139, 2144, 2150, 2156, 2161, 2167, 2172, 2178, 2184, 2189, 2195]
XPL0
<lang xpl0> \Long year code Rem=2, CrLf=9, IntIn=10, IntOut=11, Text=12, Clear=40; integer S, E, Y;
function integer Weekday(Y); \Weekday of Y-12-31, 0 Sunday integer Y; return Rem((Y + Y / 4 - Y / 100 + Y / 400) / 7);
function integer IsLongYear(Y); integer Y; return 4 = Weekday(Y) ! 3 = Weekday(Y - 1);
begin Clear; Text(0, "**** List of ISO long years ****"); CrLf(0); Text(0, "Start year: "); S:= IntIn(0); Text(0, "End year: "); E:= IntIn(0); CrLf(0); for Y:= S, E do
if IsLongYear(Y) then [IntOut(0, Y); Text(0, " ")];
CrLf(0); end </lang>
- Output:
**** List of ISO long years **** Start year: 1995 End year: 2045 1998 2004 2009 2015 2020 2026 2032 2037 2043
zkl
<lang zkl>fcn isLongYear(y){ Time.Date.weeksInYear(y)==53 } foreach nm,y in (T(T("20th",1900), T("21st",2000), T("22nd",2100))){
println("\nLong years in the %s century:\n%s".fmt(nm, [y..y+99].filter(isLongYear).concat(" ")));
}</lang>
- Output:
Long years in the 20th century: 1903 1908 1914 1920 1925 1931 1936 1942 1948 1953 1959 1964 1970 1976 1981 1987 1992 1998 Long years in the 21st century: 2004 2009 2015 2020 2026 2032 2037 2043 2048 2054 2060 2065 2071 2076 2082 2088 2093 2099 Long years in the 22nd century: 2105 2111 2116 2122 2128 2133 2139 2144 2150 2156 2161 2167 2172 2178 2184 2189 2195
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