How many friday the 13th in a year?

Question

Your challenge is to write a program which, given a year, outputs the number of "Friday 13ths" in it.

Rules & Details:

• You can take input via STDIN, or as an argument passed to your program.
• You should output the result to STDOUT.
• You may assume that input will be a valid year, and does not pre-date the Gregorian calendar (undefined behaviour is permitted in these cases).
• Calendar/Date libraries are allowed.

This is a code-golf, so the shortest code (in bytes) wins.

(Related challenge link)

Answer 1

Lua, 95 bytes

n=0 for i=1,12 do n=os.date('%w',os.time{month=i,day=13,year=...})=='5'and n+1or n end print(n)

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Answer 2

Scala 3, 107 bytes

Golfed version. Attempt This Online!

y=>(1 to 12).map{m=>if(java.time.LocalDate.of(y,m,6).getDayOfWeek==java.time.DayOfWeek.FRIDAY)1 else 0}.sum

Ungolfed version. Attempt This Online!

import java.time._

object Main {
  def main(args: Array[String]): Unit = {
    println(countFridays(1776))
    println(countFridays(2012))
    println(countFridays(2013))
    println(countFridays(2014))
  }

  def countFridays(year: Int): Int = {
    (1 to 12).map { month =>
      val firstDay = LocalDate.of(year, month, 6).getDayOfWeek
      if (firstDay == DayOfWeek.FRIDAY) 1 else 0
    }.sum
  }
}

Answer 3

JavaScript, 56 47 bytes

f=(y,m=12)=>m&&f(y,--m)+!new Date(y,m).getDay()

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Answer 4

Japt -x, 10 8 bytes

CÆ!ÐUX e

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CÆ5¥ÐUXD e     :Implicit input of integer U
C              :12
 Æ             :  Map each X in the range [0,C) (months are 0-indexed in JavaScript)
  5¥           :  Check if 5 is equal to
    ÐUXD       :  new Date(U,X,13)
         e     :  0-based index of day of the week
               :Implicitly reduce by addition and output

Answer 5

JavaScript (Node.js), 47 bytes

f=(y,m=12)=>m&&f(y,--m)+!new Date(y,m).getDay()

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-2 bytes thanks to l4m2

Answer 6

Go, 114 bytes

import."time"
func f(y int)(c int){for i:=0;i<12;i++{if Date(y,Month(i+1),1,0,0,0,0,UTC).Weekday()<1{c++}}
return}

Counts months starting with Sundays.

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Answer 7

Rust, 103 bytes

Golfed version. Run it on Rust Playground!

fn c(y:i32)->i32{(1..=12).filter(|m|NaiveDate::from_ymd(y,*m,6).weekday()==Weekday::Fri).count()as i32}

Ungolfed version. Run it on Rust Playground!

use chrono::{Datelike, Weekday, NaiveDate};

fn main() {
    println!("{}", count_fridays(1776));
    println!("{}", count_fridays(2012));
    println!("{}", count_fridays(2013));
    println!("{}", count_fridays(2014));
}

fn count_fridays(year: i32) -> i32 {
    let mut count = 0;
    for month in 1..=12 {
        let date = NaiveDate::from_ymd(year, month, 6);
        if date.weekday() == Weekday::Fri {
            count += 1;
        }
    }
    count
}