Escribe una función que calcule el día de la semana para cualquier fecha en particular en el pasado o en el futuro. Una aplicación típica es calcular el día de la semana en que nació alguien o ocurrió algún otro evento especial.
La siguiente es una función simple sugerida por Sakamoto, Lachman, Keith y Craver para calcular el día. La siguiente función devuelve 0 para el domingo, 1 para el lunes, etc.
Understanding the Maths:
14/09/1998
dd=14
mm=09
yyyy=1998 //non-leap year
Step 1: Informations to be remembered.
Magic Number Month array.
For Year: {0,3,3,6,1,4,6,2,5,0,3,5}
DAY array starting from 0-6: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Century Year Value: 1600-1699 = 6
1700-1799 = 4
1800-1899 = 2
1900-1999 = 0
2000-2099 = 6..
Step 2: Calculations as per the steps
Last 2 digits of the year: 98
Divide the above by 4: 24
Take the date(dd): 14
Take month value from array: 5 (for September month number 9)
Take century year value: 0 ( 1998 is in the range 1900-1999 thus 0)
-----
Sum: 141
Divide the Sum by 7 and get the remainder: 141 % 7 = 1
Check the Day array starting from index 0: Day[1] = Monday
**If leap year it will be the remainder-1
C++
/* A program to find day of a given date */
#include <bits/stdc++.h>
using namespace std;
int dayofweek(int d, int m, int y)
{
static int t[] = { 0, 3, 2, 5, 0, 3,
5, 1, 4, 6, 2, 4 };
y -= m < 3;
return ( y + y / 4 - y / 100 +
y / 400 + t[m - 1] + d) % 7;
}
// Driver Code
int main()
{
int day = dayofweek(30, 8, 2010);
cout << day;
return 0;
}
// This is code is contributed
// by rathbhupendra
C
/* A program to find day of a given date */
#include<stdio.h>
int dayofweek(int d, int m, int y)
{
static int t[] = { 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 };
y -= m < 3;
return ( y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
}
/* Driver function to test above function */
int main()
{
int day = dayofweek(30, 8, 2010);
printf ("%d", day);
return 0;
}
Java
/*This code will return the string value and the exact date
* both for leap and non leap years*/
import java.util.*;
class FindDay {
static int dd;
static int mm;
static int yyyy;
FindDay(int dd, int mm, int yyyy)
{
this.dd = dd;
this.mm = mm;
this.yyyy = yyyy;
}
static int checkLeap(int y)
{
if ((y % 4 == 0 && y % 100 != 0) || (y % 400 == 0))
return 1;
else
return 0;
}
static void calculate()
{
// Checking Leap year. If true then 1 else 0.
int flag_for_leap = checkLeap(yyyy);
/*Declaring and initialising the given informations
* and arrays*/
String day[] = { "Sunday", "Monday", "Tuesday",
"Wednesday", "Thursday", "Friday",
"Saturday" };
int m_no[] = { 0, 3, 3, 6, 1, 4, 6, 2, 5, 0, 3, 5 };
/*Generalised check to find any Year Value*/
int j;
if ((yyyy / 100) % 2 == 0) {
if ((yyyy / 100) % 4 == 0)
j = 6;
else
j = 2;
}
else {
if (((yyyy / 100) - 1) % 4 == 0)
j = 4;
else
j = 0;
}
/*THE FINAL FORMULA*/
int total = (yyyy % 100) + ((yyyy % 100) / 4) + dd
+ m_no[mm - 1] + j;
if (flag_for_leap == 1) {
if ((total % 7) > 0)
System.out.println(day[(total % 7) - 1]);
else
System.out.println(day[6]);
}
else
System.out.println(day[(total % 7)]);
}
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
/*Take input in string format and then convert to
* integer values*/
String date = sc.next();
String[] values = date.split("/");
new FindDay(Integer.parseInt(values[0]),
Integer.parseInt(values[1]),
Integer.parseInt(values[2]));
calculate();
}
}
/*Contributed and written by Aniket Dey*/
Python3
# Python3 program to find day # of a given date def dayofweek(d, m, y): t = [ 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 ] y -= m < 3 return (( y + int(y / 4) - int(y / 100) + int(y / 400) + t[m - 1] + d) % 7) # Driver Code day = dayofweek(30, 8, 2010) print(day) # This code is contributed by Shreyanshi Arun.
C#
// C# program to find day of a given date
using System;
class GFG {
static int dayofweek(int d, int m, int y)
{
int []t = { 0, 3, 2, 5, 0, 3, 5,
1, 4, 6, 2, 4 };
y -= (m < 3) ? 1 : 0;
return ( y + y/4 - y/100 + y/400
+ t[m-1] + d) % 7;
}
// Driver Program to test above function
public static void Main()
{
int day = dayofweek(30, 8, 2010);
Console.Write(day);
}
}
// This code is contributed by Sam007.
PHP
<?php
// PHP program to find
// day of a given date
function dayofweek($d, $m, $y)
{
static $t = array(0, 3, 2, 5, 0, 3,
5, 1, 4, 6, 2, 4);
$y -= $m < 3;
return ($y + $y / 4 - $y / 100 +
$y / 400 + $t[$m - 1] + $d) % 7;
}
// Driver Code
$day = dayofweek(30, 8, 2010);
echo $day;
// This code is contributed by mits.
?>
Javascript
<script>
// Javascript program to find day of a given date
function dayofweek(d, m, y)
{
let t = [ 0, 3, 2, 5, 0, 3, 5, 1, 4, 6, 2, 4 ];
y -= (m < 3) ? 1 : 0;
return ( y + y/4 - y/100 + y/400 + t[m-1] + d) % 7;
}
// Driver Code
let day = dayofweek(30, 8, 2010);
document.write(Math.round(day));
</script>
Salida: 1 (lunes)
Complejidad de tiempo: O(1)
Complejidad espacial: O(1)
Consulte esto para obtener una explicación de la función anterior.
Referencias:
http://en.wikipedia.org/wiki/Determination_of_the_day_of_the_week
Este artículo fue compilado por Dheeraj Jain y revisado por el equipo de GeeksforGeeks. Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA