คำตอบที่สั้นที่สุดชนะ

มันจะต้องถูกจัดเรียงและในเวลา 24 ชั่วโมง บรรทัดสุดท้ายไม่มีเครื่องหมายจุลภาค

ผลลัพธ์ควรเป็นดังนี้:

'00:00',
'00:30',
'01:00',
'01:30',
'02:00',
'02:30',
'03:00',
'03:30',
'04:00',
'04:30',
'05:00',
'05:30',
'06:00',
'06:30',
'07:00',
'07:30',
'08:00',
'08:30',
'09:00',
'09:30',
'10:00',
'10:30',
'11:00',
'11:30',
'12:00',
'12:30',
'13:00',
'13:30',
'14:00',
'14:30',
'15:00',
'15:30',
'16:00',
'16:30',
'17:00',
'17:30',
'18:00',
'18:30',
'19:00',
'19:30',
'20:00',
'20:30',
'21:00',
'21:30',
'22:00',
'22:30',
'23:00',
'23:30'

Pyth, 26 ไบต์

Pjbm%"'%02d:%s0',"d*U24"03

สาธิต.

เราเริ่มต้นด้วยผลิตภัณฑ์คาร์ทีเซียนของrange(24)(U24 ) "03"กับสตริง

จากนั้นเราจับคู่ค่าเหล่านี้กับการแทนที่สตริงที่เหมาะสมในการจัดรูปแบบ (m%"'%02d:%s0',"d )

จากนั้นสตริงผลลัพธ์จะถูกรวมเข้ากับอักขระบรรทัดใหม่ (jb )

สุดท้ายเราลบเครื่องหมายจุลภาคต่อท้าย ( P) และพิมพ์

Befunge-93, 63 bytes

0>"'",:2/:55+/.55+%.":",:2%3*v
 ^,+55,","<_@#-*86:+1,"'".0. <

^{(ภาพเคลื่อนไหวที่สร้างด้วยBefunExec )}

Bash: 58 47 46 ตัวอักษร

h=`echo \'{00..23}:{0,3}0\'`
echo "${h// /,
}"

CJam, 31 30 29 ไบต์

24,[U3]m*"'%02d:%d0',
"fe%~7<

สิ่งนี้ค่อนข้างตรงไปตรงมาโดยใช้การจัดรูปแบบ printf:

24,                              e# get the array 0..23
   [U3]                          e# put array [0 3] on stack
       m*                        e# do a cartesian product between 0..23 and [0 3] array
                                 e# now we have tuples like [[0 0], [0 3] ... ] etc
         "'%02d:%d0',
"fe%                             e# this is standard printf formatting. What we do here is
                                 e# is that we format each tuple on this string
    ~7<                          e# unwrap and remove comma and new line from last line
                                 e# by taking only first 7 characters

ลองออนไลน์ได้ที่นี่

Python 2, 58 56 ไบต์

for i in range(48):print"'%02d:%s0',"[:57-i]%(i/2,i%2*3)

เช่นเดียวกับคำตอบของ sentiaoแต่ใช้forloop โดยมีการแบ่งเพื่อลบเครื่องหมายจุลภาค ขอบคุณ @grc สำหรับการปิดสองไบต์

Java - 119 ไบต์

ฉันเริ่มต้นด้วย StringJoiner ของ Java 8 แต่นั่นหมายถึงการรวมคำสั่งการนำเข้าดังนั้นฉันจึงตัดสินใจทำแบบเก่า:

void f(){String s="";for(int i=0;i<24;)s+=s.format(",\n'%02d:00',\n'%02d:30'",i,i++);System.out.print(s.substring(2));}

บางทีสิ่งนี้สามารถปรับปรุงได้โดยการกำจัดสิ่งที่เกิดขึ้นStringและSystemคำค้นหา

ทับทิม, 94 61 56 51 51

$><<(0..47).map{|i|"'%02d:%02d'"%[i/2,i%2*30]}*",
"

ขอบคุณ @blutorange (อีกครั้ง) สำหรับความช่วยเหลือของเขาในการเล่นกอล์ฟ!

Perl, 52 50 48 45B

$,=",
";say map<\\'$_:{0,3}0\\'>,"00".."23"

With help from ThisSuitIsBlackNot :)

JAVA 95 94 bytes

I love the fact that printf exists in Java:

void p(){for(int i=0;i<24;)System.out.printf("'%02d:00',\n'%02d:30'%c\n", i,i++,(i<24)?44:0);}

Ungolfed

void p(){
    for(int i=0;i<24;)
        System.out.printf("'%02d:00',\n'%02d:30'%c\n", i,i++,(i<24)?44:0);
}

EDIT Replaced the ',' with 44

Pyth, 32 31 bytes

I golfed something in python but it turned out to be exactly the same as Sp3000's answer. So I decided to give Pyth a try:

V48<%"'%02d:%d0',",/N2*3%N2-54N

It's a exact translation of Sp3000 answer:

for i in range(48):print"'%02d:%d0',"[:57-i]%(i/2,i%2*3)

It's my first go at Pyth, so please do enlighten me about that 1 byte saving.

PHP, 109 bytes

foreach(new DatePeriod("R47/2015-05-07T00:00:00Z/PT30M")as$d)$a[]=$d->format("'H:i'");echo implode(",\n",$a);

Ruby, 54 51 bytes

puts (0..23).map{|h|"'#{h}:00',
'#{h}:30'"}.join",
"

C, 116,115,101,100,95,74,73, 71

May be able to scrape a few more bytes off this...

main(a){for(;++a<50;)printf("'%02d:%d0'%s",a/2-1,a%2*3,a<49?",\n":"");}

T-SQL, 319 307 305 bytes

WITH t AS(SELECT t.f FROM(VALUES(0),(1),(2),(3),(4))t(f)),i AS(SELECT i=row_number()OVER(ORDER BY u.f,v.f)-1FROM t u CROSS APPLY t v),h AS(SELECT i,h=right('0'+cast(i AS VARCHAR(2)),2)FROM i WHERE i<24)SELECT''''+h+':'+m+CASE WHEN i=23AND m='30'THEN''ELSE','END FROM(VALUES('00'),('30'))m(m) CROSS APPLY h

Un-golfed version:

WITH
t AS(
    SELECT
        t.f
    FROM(VALUES
         (0),(1),(2),(3),(4)
    )t(f)
),
i AS(
    SELECT
        i = row_number() OVER(ORDER BY u.f,v.f) - 1
    FROM t u 
    CROSS APPLY t v
),
h AS(
    SELECT
        i,
        h = right('0'+cast(i AS VARCHAR(2)),2)
    FROM i
    WHERE i<24
)
SELECT
    '''' + h + ':' + m + CASE WHEN i=23 AND m='30' 
                              THEN '' 
                              ELSE ',' 
                         END
FROM(
    VALUES('00'),('30')
)m(m)
CROSS APPLY h

Pyth, 34 bytes

j+\,bmjk[\'?kgd20Z/d2\:*3%d2Z\')48

This can definitely be improved.

Try it online: Pyth Compiler/Executor

Explanation:

     m                          48   map each d in [0, 1, ..., 47] to:
        [                      )       create a list with the elements:
         \'                              "'"
           ?kgd20Z                       "" if d >= 20 else 0
                  /d2                    d / 2
                     \:                  ":"
                       *3%d2             3 * (d%2)
                            Z            0
                             \'          "'"
      jk                               join by "", the list gets converted into a string
j+\,b                                join all times by "," + "\n"

Python 2, 69 bytes

print',\n'.join(["'%02d:%s0'"%(h,m)for h in range(24)for m in'03'])

Quite obvious, but here's an explanation:

• using double-for-loop
• alternating between '0' and '3' in string format is shorter than a list
• %02d does the padding for h
• mdoesn't need padding as the alternating character is on a fixed position
• '\n'.join() solves the final-line requirements

I have no idea if it can be done shorter (in Python 2).

by Sp3000, 61 bytes : print',\n'.join("'%02d:%s0'"%(h/2,h%2*3)for h in range(48))

Haskell, 85 bytes

putStr$init$init$unlines$take 48['\'':w:x:':':y:"0',"|w<-"012",x<-['0'..'9'],y<-"03"]

Unfortunately printf requires a 19 byte import, so I cannot use it.

Julia: 65 64 61 characters

[@printf("'%02d:%d0'%s
",i/2.01,i%2*3,i<47?",":"")for i=0:47]

Julia: 64 characters

(Kept here to show Julia's nice for syntax.)

print(join([@sprintf("'%02d:%d0'",h,m*3)for m=0:1,h=0:23],",
"))

Fortran 96

do i=0,46;print'(a1,i2.2,a,i2.2,a2)',"'",i/2,":",mod(i,2)*30,"',";enddo;print'(a)',"'23:30'";end

Standard abuse of types & requirement only for the final end for compiling. Sadly, due to implicit formatting, the '(a)' in the final print statement is required. Still, better than the C and C++ answers ;)

JavaScript (ES6), 77 86+1 bytes

Didn't realize there had to be quotes on each line (+1 is for -p flag with node):

"'"+Array.from(Array(48),(d,i)=>(i>19?"":"0")+~~(i/2)+":"+3*(i&1)+0).join("',\n'")+"'"

old solution:

Array.from(Array(48),(d,i)=>~~(i/2).toFixed(2)+":"+3*(i&1)+"0").join(",\n")

ungolfed version (using a for loop instead of Array.from):

var a = [];
// there are 48 different times (00:00 to 23:30)
for (var i = 0; i < 48; i++) {
    a[i] =
        (i > 19 ? "" : "0") +
            // just a ternary to decide whether to pad
            // with a zero (19/2 is 9.5, so it's the last padded number)
        ~~(i/2) +
            // we want 0 to 24, not 0 to 48
        ":" +  // they all then have a colon
        3*(i&1) +
            // if i is odd, it should print 30; otherwise, print 0
        "0" // followed by the last 0
}
console.log(a.join(",\n"));

golflua 52 51 chars

~@n=0,47w(S.q("'%02d:%d0'%c",n/2,n%2*3,n<47&44|0))$

Using ascii 44 = , and 0 a space saves a character.

An ungolfed Lua version would be

for h=0,47 do
   print(string.format("'%02d:%d0'%c",h/2,h%2*3, if h<47 and 44 or 0))
end

The if statement is much like the ternary operator a > b ? 44 : 0.

C# - 120 bytes

class P{static void Main(){for(var i=0;i<24;i++)System.Console.Write("'{0:00}:00',\n'{0:00}:30'{1}\n",i,i==23?"":",");}}

Python, 60 58 64 bytes

for i in range(24):print("'%02d:00,\n%02d:30'"%(i,i)+', '[i>22])

Ungolfed:

for i in range(24):
    if i <23:
        print( ('0'+str(i))[-2:] + ':00,\n' + str(i) + ':30,')
    else:
        print( ('0'+str(i))[-2:] + ':00,\n' + str(i) + ':30')

Try it online here.

𝔼𝕊𝕄𝕚𝕟, 39 chars / 67 bytes (non-competing)

↺;Ḁ⧺<Ḱ;)ᵖ`'⦃Ḁ<Ḕ?0:⬯}⦃0|Ḁ/2}:⦃Ḁ%2*3}0',”

Try it here (Firefox only).

Not a single alphabetical character in sight...

PHP, 69 70 62 bytes

for($x=-1;++$x<47;)printf("'%02d:%d0',
",$x/2,$x%2*3)?>'23:30'

Try it online

Outputting '23:30' at the end is a bit lame, and so is closing the php context using ?> without opening or re-opening it. An cleaner alternative (but 65 bytes) would be:

for($x=-1;++$x<48;)printf("%s'%02d:%d0'",$x?",
":'',$x/2,$x%2*3);

Try it online

Thank you @Dennis for the tips. Alternative inspired by the contribution of @ismael-miguel.

Swift, 74 bytes

Updated for Swift 2/3...and with new string interpolation...

for x in 0...47{print("'\(x<20 ?"0":"")\(x/2):\(x%2*3)0'\(x<47 ?",":"")")}

Javascript, 89 bytes

for(i=a=[];i<24;)a.push((x="'"+("0"+i++).slice(-2))+":00'",x+":30'");alert(a.join(",\n"))

Python 2: 64 bytes

print ',\n'.join(['%02d:00,\n%02d:30'%(h,h) for h in range(24)])

Ruby - 52 bytes

puts (0..47).map{|i|"'%02d:%02d'"%[i/2,i%2*30]}*",
"

Python 2, 74 65 bytes

We generate a 2 line string for each hour, using text formatting:

print',\n'.join("'%02u:00',\n'%02u:30'"%(h,h)for h in range(24))

This code is fairly clear, but the clever indexing and integer maths in the answer by Sp3000 gives a shorter solution.