When was the last time the date was divisible by n?

Question

A date can be represented by an unsigned integer as such: YYYYMMDD. What you need to do, is write the shortest program or function that figures out the most recent date whose number was divisible by a given number n (including today's date) and then returns that date in the format showed above. If there has never been a date (between 00000101 and today inclusive) divisible by the given integer, you should return -1.

Examples

Current Date     Input     Output

30 July, 2014      4       20140728
30 July, 2014      7       20140729
28 July, 2014      4       20140728
28 July, 2014      7       20140722
28 July,    5    90000     -1

Input

You can read from STDIN or take a function argument or even expect the input to be stored in a variable. The input will be an unsigned integer.

Output

Write to STDOUT or return (or save in a variable) the integer representing the date in the format YYYYMMDD.

Restrictions

You may use any standard library your language offers. Standard loopholes apply.

Winning conditions

This is a code-golf, so smallest program (in bytes) wins. In case of a tie, the answer with the most votes wins.

Answer 1

Mathematica, 93 60 bytes

For[i=0,(r=DatePlus@i--~FromDigits~100)>0&&!n∣r,];r~Max~-1

Expects the input to be stored in n.

Note that the vertical line ∣ is the unicode character for "divides", which I've counted as 3 bytes (UTF-8).

Edit: Found a neat trick to avoid the bloated DateString and format specification :).

Edit: Totally forgot about the -1 requirement. Fixed now.

Here is an explanation

For[i=0,            i--                        ,];         (* i is the number of days AFTER
                                                              today. Hence, we decrement 
                                                              it. *)
For[i=0,   DatePlus@i--                        ,];         (* If no reference date is
                                                              provided, DatePlus will add
                                                              the given number of days to
                                                              today's date. The result is a 
                                                              list of 3 integers, luckily 
                                                              in order {year,month,day} *)
For[i=0,   DatePlus@i--~FromDigits~100         ,];         (* Interpret these as the digits
                                                              of a base 100 number. The 
                                                              beauty is that FromDigits
                                                              doesn't care about digits 
                                                              greater than the base and 
                                                              just carries them over. *)
For[i=0,(r=DatePlus@i--~FromDigits~100)        ,];         (* Store the number in r. *)
For[i=0,(r=DatePlus@i--~FromDigits~100)>0      ,];         (* Make sure it's positive. *)
For[i=0,(r=DatePlus@i--~FromDigits~100)>0&&!n|r,];         (* And keep going while n does 
                                                              not divide r. *)
For[i=0,(r=DatePlus@i--~FromDigits~100)>0&&!n|r,];r~Max~-1 (* Clamp result to -1. *)

Note that I've used | instead of ∣ in the explanation, because the Unicode one messes with monospacing.

Answer 2

Python 2 - 150

import datetime as d,re
def f(n):
 t=d.date.today()
 while t:
    c=int(re.sub("-","",str(t)))
    if c%n<1:return c
    try:t-=d.timedelta(1)
    except:return-1

Thanks @chill0r for suggestion to remove days=, and Jason S for tip that the try block can be reduced to one line.

Answer 3

C# 136

With the revised specs, a function that takes an unsigned int and returns an int.

int F(uint n){var d=System.DateTime.Now;int i;try{while((i=int.Parse(d.ToString("yyyMMdd")))%n>0)d=d.AddDays(-1);}catch{i=-1;}return i;}

---

152 characters with variable input/output

Taking advantage of the loose input/output requirements, input is to be stored in the variable n (currently counting all characters except the integer literal), and output is provided with the variable s.

class P{static void Main(){var n=4;var d=System.DateTime.Now;string s;try{while(int.Parse(s=d.ToString("yyyMMdd"))%n>0)d=d.AddDays(-1);}catch{s="-1";}}}

---

204 characters with STDIN/STDOUT:

using System;class P{static void Main(){int n=int.Parse(Console.ReadLine());var d=DateTime.Now;string s;try{while(int.Parse(s=d.ToString("yyyMMdd"))%n>0)d=d.AddDays(-1);}catch{s="-1";}Console.Write(s);}}

Answer 4

T-SQL (2012) - 148

Assumes there is a free variable @n with the n value.

declare @ date=getdate()while convert(char,@,112)%@n>0 and'00010101'<@ set @=dateadd(d,-1,@)print iif(convert(char,@,112)%@n=0,convert(char,@),'-1')

Answer 5

Golflua 90 86

n=I.r()d="%Y%m%d"i=O.d(d)+0j=0@i>0?i%n==0w(i)O.q()$j=j+1i=O.d(d,O.t()-j*86400)+0$w(-1)

An ungolfed Lua version would be,

n = io.read()
d = "%Y%m%d"
i = os.date(d)+0   -- implicitly casts os.date(d) to int
j = 0
while i>0 do
   if i % n == 0 then
      print(i)
      os.exit()
   end
   j = j+1
   i = os.date(d,os.time()-j*86400)+0
end
print(-1)

Answer 6

MATLAB: 61

-1,s=str2num(datestr(1:now,'YYYYmmDD')),d=s(~mod(s,n)),d(end)

Assumes the divisor is stored in n. The result will be stored in a variable called ans.

---

Commented version:

-1                                     % Store -1 in ans in case we don't find anything
s=str2num(datestr(1:now,'YYYYmmDD'))   % Make a list of date numbers
d=s(~mod(s,n)),                        % Select only those who are dividable and prepend -1
d(end)                                 % Store last found value in ans, if anything is found

Will generate an error if no result is found, but the answer is still available in the variable despite that.

---

Error could be avoided at the cost of 2 extra chars:

s=str2num(datestr(1:now,'YYYYmmDD')),d=[-1;s(~mod(s,n))],d(end)

Answer 7

PHP (92=85+7)

Expects input to be stored in $n.

for($d=date("Ymd");!($d%$n==0&checkdate($d/100%100,$d%100,substr($d,0,4))|$d<0);$d--);echo$d

I just remembered why I do not like PHP anymore=)

EDIT: Now the -1 of the specifications is implemented too.

Answer 8

JavaScript (ES6) 115

Expects number in variable n, result stored in variable r. Each day is checked, starting with current date and decrementing - there must be a better way.
Moreover, using standard javascript date functions, all dates are gregorian down to year 1 (with leap years accordingly wrong before gregorian reform).

for(z=new Date,t=n+1;t>n&&t%n;)
  d=z.getDate(),
  t=z.getFullYear()*1e4+(z.getMonth()+1)*100+d,
  z.setDate(d-1);
r=t>n?t:-1

Answer 9

C# - 144 (Or 124 in LINQPad) + 1 for each digit in n

This expects the input to be in the variable n. By the end of the execution the desired value will be in the variable r. This considers 00010101 as the first date, though, because the date 00000101 does not exist. Suggestions for improvement are always welcome.

class P{static void Main(){int n=7,r;var d=System.DateTime.Now;try{for(;(r=int.Parse(d.ToString("yyyMMdd")))%n>0;d=d.AddDays(-1));}catch{r=-1;}}}

LINQPad version:

int n=7,r;var d=System.DateTime.Now;try{for(;(r=int.Parse(d.ToString("yyyMMdd")))%n>0;d=d.AddDays(-1));}catch{r=-1;}r.Dump();

Answer 10

Groovy - 301 300 chars

Very simple (and slow), with no tricks to hide the fact that it uses Joda Time.

Golfed:

@Grab(group='joda-time', module='joda-time', version='2.3')
import org.joda.time.*
import org.joda.time.format.*
f={DateTimeFormat.forPattern("yyyyMMdd").print(new LocalDate().minusDays(it)) as int}
n=args[0] as int;b=0;x=-1;c=0
while(!b){if(f(c++)%n==0){x=f(--c);b=1};if(f(0)-c<=101){b=1}}
println x

Example run (on 7/30/2014):

$ groovy D.groovy  7
20140729
$ groovy D.groovy  16
20140720
$ groovy D.groovy  90000
-1

Ungolfed:

@Grab(group='joda-time', module='joda-time', version='2.3')

import org.joda.time.*
import org.joda.time.format.*

f = { DateTimeFormat.forPattern("yyyyMMdd").print(new LocalDate().minusDays(it)) as int }

n = args[0] as int
b = 0 
x = -1
c = 0

while (!b) {
    if(f(c++)%n==0) { x=f(--c); b=1}
    if(f(0)-c<=101){b=1}
}

println x

Answer 11

R, 146 139

D=function(n){
z=as.double(gsub("-","",y<-Sys.Date()))
d=F
while(z>100&!d){
y=y-1 
z=as.double(gsub("-","",y))
d=!z%%n}
ifelse(z>100,z,-1)}

Good luck with a date that doesn't work. microbenchmark reports it takes about half a second to go back 15 days. As of July 31, 2014, this will take something like 20 million seconds (~23 days) to spit out -1, at least according to the back of the envelope.

edit: some shortcuts in the comments

Answer 12

Matlab 104

function d=f(v);for d=fix(now):-1:1 d=str2num(datestr(d,'YYYYmmDD'));if~mod(d,v)return;end;end;d=-1;end

Ungolfed :

function d = f(v)
   for d=fix(now):-1:1
       d = str2num(datestr(d,'YYYYmmDD'));
       if ~mod(d,v)
          return; 
       end
   end
   d = -1;
end

EDIT: I managed to optimise it a bit, but @DennisJaheruddin has the real solution here

Answer 13

Python 3 - 151 148 bytes, generators

from datetime import*
t=date.today()
f=lambda n:next((y for y in(int((t-timedelta(o)).strftime("%Y%m%d"))for o in range(t.toordinal()))if y%n<1),-1)

Thanks @nyuszika7h for import* suggestion

Answer 14

Ruby 103

require'date'
f=->{d=Date.today 
(s=d.strftime('%Y%m%d').to_i
return s if s%n<1
d-=1)while d.year>0
-1}

Input

Expects the divisor value to be present in variable n.

Output

The return value of the f function

Online example: http://ideone.com/LoYxG4

Answer 15

Java : 373 chars

This is a port of the Groovy answer, and uses Joda Time.

Golfed:

import org.joda.time.*;
import org.joda.time.format.*;
public class D {
static int f(int i){return Integer.parseInt(DateTimeFormat.forPattern("yyyyMMdd").print(new LocalDate().minusDays(i)));}
public static void main(String[] args){
int n=Integer.parseInt(args[0]);int b=0,c=0,x=-1;
while(b!=1){if(f(c++)%n==0){x=f(--c);b=1;};if(f(0)-c<=101){b=1;}}
System.out.println(x);}}

Sample runs (with joda-time-2.4.jar on classpath:

$ java D 7
20140729
$ java D 4
20140728
$ java D 16
20140720
$ java D 90000
-1

Ungolfed:

import org.joda.time.*;
import org.joda.time.format.*;

public class D {
    static int f(int i) {
        return Integer.parseInt(DateTimeFormat.forPattern("yyyyMMdd").print(new LocalDate().minusDays(i)));
    }

    public static void main(String[] args) {
        int n = Integer.parseInt(args[0]);
        int b = 0,c = 0,x = -1;

        while(b!=1) {
            if(f(c++)%n==0) { x=f(--c);b=1; }
            if(f(0)-c<=101) { b=1; }
        }

        System.out.println(x);
    }
}

Answer 16

Bash+coreutils (8.21), 67 bytes

seq -f-%gday $[9**9]|date -f- +[pq]sp[_1pq]sq%Y%m%ddA1=qd$1%%0=p|dc

• seq generates integers from 1 to 9⁹, one per line, and formats it as -<x>day
• pipe this to date -f which interprets each line and outputs the date formatted into a dc expression such as [pq] sp [_1pq] sq 20140728 d A1 =q d 7% 0=p (spaces added for readability)
  • [pq] define a macro to print the top of stack, then quit
  • sp save macro in register p
  • [pq] define a macro to push -1, print the top of stack, then quit
  • sq save macro in register q
  • 20140728 embedded date integer
  • d duplicate top of stack
  • A1 push 101 (00000101)
  • =q pop top 2 stack values: compare date and 101, and call macro q if equal
  • 7 push divider
  • % pop divider and dividee, the divide and push the remainder
  • 0 push 0
  • =p pop top 2 stack values: compare remainder and 0, and call macro p if equal
  • d duplicate top of stack
  • macro p is called: prints date integer and quits dc entirely
• dc expressions are piped to dc for evaluation. Once dc prints the right value and quits, the rest of the pipeline is torn down

Output:

$ ./lastdivdate.sh 4
20140728
$ ./lastdivdate.sh 7
20140729
$ ./lastdivdate.sh 123456
17901120
$ ./lastdivdate.sh 77777
19910912
$ ./lastdivdate.sh 7777777
-1
$ 

---

Since this program generates integers from 1 to 9⁹, it will be valid up to just over 1 million years into the future. I hope this limitation is acceptable ;-)

---

Thanks @WumpusQ.Wumbley for shortening the return of -1.

Answer 17

PYTHON: 134 bytes

Not going to be able to beat the current leader, and it's not that much better than the best Python answer, but I decided to post my best Python solution.

from datetime import*
def y(a,n):
 s=a.strftime("%Y%m%d")
 if int(s)%n==0:yield s
 try:x=y(a-timedelta(1),n)
 except:yield -1
 yield x

Ungolfed:

from datetime import *
def y(a, n):
    s=int(a.strftime("%Y%m%d"))
    if s%n==0:
        yield s
    try:
        x=y(a-timedelta(1), n)
    except:
        yield -1
    yield x

Answer 18

JavaScript (ES5) - 94

It expects the input in variable x, and places the output in o.

for(i=Date.now();i>-7e13&&(o=(new Date(i)).toISOString().replace(/-|T.*/g,''))%x;i-=864e5)o=-1

Answer 19

k4 (84) (73)

f:{f d@*|&~.q.mod[(f:{$[^x;-1;.($x)@&~"."=$x]})'d:{"d"$x+!1+"i"$y-x}[-730457;.z.D];x]}

This is just an initial cut with the first algorithm that came to mind; I'm sure better is possible in both performance and length.

This version hardcodes the "today" part (that's the .z.D); change it to a date literal (yyyy.mm.dd) or an integer in the q date system (days since January 1, 2000) to run the test cases. (q won't parse date literals earlier than the early eighteenth century, so for dates before that, you'll need to work out the value and use the appropriate integer directly. January 1, "A.D. 0", from the spec, turns out to be -730457, which is used in the function code. July 28, A.D. 5, from the last test case, turns out to be -728450.)

The given test cases:

  {f d@*|&~.q.mod[(f:{$[^x;-1;.($x)@&~"."=$x]})'d:{"d"$x+!1+"i"$y-x}[-730457;2014.07.30];x]}4
20140728
  {f d@*|&~.q.mod[(f:{$[^x;-1;.($x)@&~"."=$x]})'d:{"d"$x+!1+"i"$y-x}[-730457;2014.07.30];x]}7
20140729
  {f d@*|&~.q.mod[(f:{$[^x;-1;.($x)@&~"."=$x]})'d:{"d"$x+!1+"i"$y-x}[-730457;2014.07.28];x]}4
20140728
  {f d@*|&~.q.mod[(f:{$[^x;-1;.($x)@&~"."=$x]})'d:{"d"$x+!1+"i"$y-x}[-730457;2014.07.28];x]}7
20140722
  "d"$-728450
0005.07.28
  {f d@*|&~.q.mod[(f:{$[^x;-1;.($x)@&~"."=$x]})'d:{"d"$x+!1+"i"$y-x}[-730457;-728450];x]}90000
-1

edit:

g:.,/$`\:`$$:;f:{$[Z=r:{(z>x)&.q.mod[g z]y}[Z:-730458;y]{x-1}/x;-1;g"d"$r]}

This is a different approach which uses one of the convergence operators to decrement the date until either it finds a divisible one or it crosses the 1/1/0000 boundary. It also does the conversion from date to integer slightly differently.

The test cases, this time all at once:

  g:.,/$`\:`$$:;{$[Z=r:{(z>x)&.q.mod[g z]y}[Z:-730458;y]{x-1}/x;-1;g"d"$r]}'[2014.07.30 2014.07.30 2014.07.28 2014.07.28,"d"$-728450;4 7 4 7 90000]
20140728 20140729 20140728 20140722 -1

Answer 20

VBA 343 bytes (module)

Sub divD(i As Long)
a = Now()
b = Format(a, "yyyymmdd")
    Do While b / i <> Int(b / i)
    a = DateAdd("d", -1, a)
    b = Format(a, "yyyymmdd")
        If b = "01000101" Then
            MsgBox -1
            Exit Sub
        End If
    Loop
MsgBox b
End Sub

Answer 21

PowerShell - 76

This depends on the number being stored in the variable $n.

try{@(0..$n|%{'{0:yyyyMMdd}'-f(date).AddDays(-$_)}|?{!($_%$n)})[0]}catch{-1}