Question

Today's challenge

Wap in JAVA to accept a given date, the program will return with the day falling on the given date without using array and fiction.
example...
if u enter today's date...i.e. 31 oct 2020
it will return Saturday​​

Answer 1

Answer:

String input_date="01/08/2012"; SimpleDateFormat format1=new SimpleDateFormat("dd/MM/yyyy"); Date dt1=format1. parse(input_date); DateFormat format2=new SimpleDateFormat("EEEE"); String finalDay=format2. format(dt1); Use this code for find the Day name from a input date.

Answer 2

Challenge Accepted.

Question:-

Write a program in Java to accept a given date, the program will return with the day falling on the given date without using array.

Program:-

import java.util.*;

public class Main

{

public static void main(String[] args)

{

Scanner sc=new Scanner(System.in);

System.out.print("Enter the date: ");

int K=sc.nextInt();

System.out.print("Enter the month: ");

int M=sc.nextInt();

System.out.print("Enter the year: ");

int y=sc.nextInt();

if(M>0&&M<3)

y--;

int D=y%100;

int C=y/100;

M-=2;

if(M==0)

M=12;

else if(M==-1)

M=11;

int F;

F=K+((13*M-1)/5)+D+(D/4)+(C/4) -(2*C);

F=F%7;

String x="";

switch(F)

{

case 1: x="Monday";

break;

case 2: x="Tuesday";

break;

case 3: x="Wednesday";

break;

case 4: x="Thursday";

break;

case 5: x="Friday";

break;

case 6: x="Saturday";

break;

case 0: x="Sunday";

break;

}

System.out.println("Its "+x+".");

}

}

How to solve?

This is solved by using zellers rule.

Zellers rule has only 1 formula which calculates the day on which the given date is falling.

Consider a date, say 06/08/1990

Here is the calculation part.

This is the required formula.

F = K + [(13xM – 1)/5] + D + [D/4] + [C/4] – 2C

where,

1) K = Date. (here, K=06)

In Zellers rule, months start from March.

2) M = Month no.

Remember that month Starts from March. So,

March = 1,

April = 2,

May = 3

.

.

.

.

Dec = 10,

Jan = 11

Feb. = 12

So, for 06/08/1990, M=6

3) D = Last two digits of the year

So, in our example of 6/08/1990 D=90

Remember that when you have to find day of the first or second month of any year, then Year=Given year-1

i.e. When you want to find Day of 15-2-1990.,

K=15,

Month=12,

D=Given Year-1=1990-1=1989=89

4) C = The first two digits of century

Here, C = 19.

Now, using the formula, we will calculate.

The formula is F = K + [(13xM – 1)/5] + D + [D/4] + [C/4] – 2C

Replacing the values in the formula, we get

F = 06 + [{(13 x 6)- 1}/5] + 90 + 90/4 + 19/4 – (2 x 19)

Therefore,

F = 06 + 77/5 + 90 + 90/4 + 19/4 – 38

Remember:- We have to calculate the integer part only.

F =06 + 15.4 + 90 + 22.5 + 4.75 – 38

>> F =06 + 15 + 90 + 22 + 4 – 38

Therefore, F = 99

Now, divide F by 7 and get the remainder.

These are the days. Put the values and find them.

1 = Monday

2 = Tuesday

3 = Wednesday

4 = Thursday

5 = Friday

6 = Saturday

0 = Sunday

Here, 1 represents Monday.

So by Zeller’s rule, 6th of August, 1990 was on a Monday.

In this way, the problem is solved.