Rollover Calendar

Question

This is inspired by one of Downgoat's questions in Sandbox, where I suggested that he include April 31 as Pi day for people who use day/month format, only for him to inform me that there is no April 31!

Given a date string in month/day format that might be invalid, output the correct date using rollover. (First rollover the month, then rollover the day).

Examples:

"15/43" - This reads as the 43rd day of the 15th month. First, we roll over the month to the next year, so we end up with 3 (March). Now, since March only has 31 days, we rollover the extra days to April, so we output the actual date as "4/12" (April 12th).

"3/16" - This is a valid date (March 16th). Return it as is.

"12/64" - Ah, so many fond memories from December 64th... December has 31 days, January has 31 days, so what I really mean is "2/2" (February 2nd).

"19/99" - First, the 19 becomes a 7 (July). July has 31 days, August has 31 days, September has 30 days, so the output is "10/7" (October 7th).

"1/99999" - A year has 365 days. 99999 (mod 365) = 354. The 354 day of the year is "12/20".

"9999999/10" - Apparently, 9999999 (mod 12) = 3, so this is "3/10" (March 10th).

Criteria:

Input month is an integer > 0. Input day is an integer > 0. Year never needs to be specified, as such there are no leap years to deal with.

Update:

As I think it would overly simplify the challenge, calendar functions, such as those in the Java Calendar class, are banned. Date parsing/formatting functions are still allowed though.

Answer 1

LabVIEW, 32 LabVIEW Primitives

Answer 2

C#, 269 223

string v(string n){var x=new[]{31,28,31,30,31,30,31,31,30,31,30,31};var s=n.Split('/');Func<string,int> p=int.Parse;var m=p(s[0]);var d=p(s[1]);m=m>=12?m%12:m;while(d>x[m]){d-=x[m];m=++m>=12?m%12:m;}return(m==0?1:m)+"/"+d;}

Edit: Fixed to work for cases like 24/1, 36/1, etc. and golfed a little. Thanks for the comments, there are several places I saved a bit!

Ungolfed:

string v(string n)
{
    var x = new [] { 31 ,28, 31, 30, 31, 30,31, 31, 30, 31, 30, 31 };

    var s = n.Split('/');
    Func<string,int> p = int.Parse;
    var m = p(s[0]);
    var d = p(s[1]);
    m = m >= 12 ? m % 12 : m;
    while (d > x[m])
    {
        d -= x[m];
        m = ++m >= 12 ? m % 12 : m;
    }
    return (m==0?1:m) + "/" + d;
}

Answer 3

R, 208 182 bytes

m=c(31,28,31,30,31,30,31,31,30,31,30,31)
e=scan(sep="/");n=(e[1]/12-1)*12;if(!n%%12)n=12;if(n<0)n=e[1];j=e[2];while((j<-j-m[n])>0){n=n+1;if(n>12)n=1};j=m[n]+j;cat(n,j,sep="/")

Get the month by dividing by 12, then loop, removing the number of days of the current month until you get a negative number., inverse last step and print.

On multiple lines (need to use a file and source it):

m=c(31,28,31,30,31,30,31,31,30,31,30,31)
e=scan(sep="/")
n=(e[1]/12-1)*12
if(!n%%12)n=12
if(n<0)n=e[1]
j=e[2]
while((j<-j-m[n])>0){n=n+1;if(n>12)n=1}
j=m[n]+j;cat(n,j,sep="/")

Answer 4

JavaScript (ES6), 106 bytes

s=>eval('q="030101001010";p=s.split`/`;for(d=i=p[1],m=p[0]-1;i--;d>n&&(m++,d-=n))n=31-q[m%=12];m+1+"/"+d')

Explanation

s=>
  eval(`              // use eval to enable for loop without needing to write {} or return
    q="030101001010"; // q = array of 31 - days in each month
    p=s.split\`/\`;   // p = array of [ month, day ]
    for(
      d=i=p[1],       // d = day
        m=p[0]-1;     // m = month - 1
      i--;            // loop for each day, this is more iterations than needed but extra
                      //     iterations do not affect the result and it's the shortest way
                      //     to guarantee all months have been subtracted from d, it also
                      //     ensures the loop runs at least once to get m % 12
      d>n&&(m++,d-=n) // if too many days, subtract the month's days and increment month
    )
      n=31-q[m%=12];  // n = number of days in month, get m % 12
    m+1+"/"+d         // return the result
  `)

Test

var solution = s=>eval('q="030101001010";p=s.split`/`;for(d=i=p[1],m=p[0]-1;i--;d>n&&(m++,d-=n))n=31-q[m%=12];m+1+"/"+d')

<input type="text" id="input" value="19/99" />
<button onclick="result.textContent=solution(input.value)">Go</button>
<pre id="result"></pre>

Answer 5

PHP >= 5.5, 181 bytes

list($m,$d)=explode("/",$argv[1]);$m%=12;$d%=365;$i=0;while($d>100)$d-=[31,28,31,30,31,30,31,31,30,31,30,31][$i++];$m+=$i;echo date_create_from_format("m/d","$m/$d")->format("n/j");

PHP almost supports rollover with date parsing and formatting instructions alone. For example:

echo date_create_from_format("m/d","12/64")->format("n/j"); // Output: 2/2

However, once either number gets bigger than 100, PHP rejects parsing it and returns an error (probably for some arbitrary reason). So, the theory with this answer is to get it back down to where PHP will parse it, then submit it to date_create_from_format().

Ungolfed:

list($month, $day) = explode("/", $argv[1]);
$month = $month % 12;
$day = $day % 365;
$i = 0;
$days = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31];
while($day > 31) $day -= $days[$i++];
$month += $i;
echo DateTime::createFromFormat("m/d", $month . "/" . $day)->format("n/j");

Try it online