Program to calculate the number of odd days in given number of years

Given an integer N, the task is to find the number of odd days in the years from 1 to N. 
Odd Days: Number of odd days refer to those days that are left in a certain year(s) when it’s days gets converted into weeks. Say, an ordinary year has 365 days, that is 52 weeks and one odd day. This means, out of the 365 days in an ordinary year, 364 days will get converted into 52 weeks and one day will remain. This one day is referred to as 1 odd day. 

• Simply the modulus total number of days by 7(days in a week) gives us the number of odd days.
• It’s value lies between 0 to 6 only. [0, 6]
• Leap Year: Every year divisible either by 400 or by 4 but not 100
• Ordinary Year: Years Except Leap Years
• Every Ordinary Year has 1 odd day.
• Every Leap Year has 2 odd days.

Examples:

Input: N = 8 
Output: 3 
Out of the 8 years, 4 and 8 are the only leap years. 
(6 x 1) + (2 x 2) = 10 i.e. 1 week and 3 days
Input: N = 400 
Output: 0 

Approach:

1. Count number of ordinary years and number of leap years out of given N years.
2. Calculate the overall number of days.
3. Print the modulo(7) of the total number of days.

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Function to return the count of odd days
int oddDays(int N)
{

    // Count of years divisible
    // by 100 and 400
    int hund1 = N / 100;
    int hund4 = N / 400;

    // Every 4th year is a leap year
    int leap = N >> 2;
    int ord = N - leap;

    // Every 100th year is divisible by 4
    // but is not a leap year
    if (hund1) {
        ord += hund1;
        leap -= hund1;
    }

    // Every 400th year is divisible by 100
    // but is a leap year
    if (hund4) {
        ord -= hund4;
        leap += hund4;
    }

    // Total number of extra days
    int days = ord + leap * 2;

    // modulo(7) for final answer
    int odd = days % 7;

    return odd;
}

// Driver code
int main()
{

    // Number of days
    int N = 100;
    cout << oddDays(N);

    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
class GFG {

    // Function to return the count of odd days
    static int oddDays(int N)
    {
        // Count of years divisible
        // by 100 and 400
        int hund1 = N / 100;
        int hund4 = N / 400;

        // Every 4th year is a leap year
        int leap = N >> 2;
        int ord = N - leap;

        // Every 100th year is divisible by 4
        // but is not a leap year
        if (hund1 > 0) {
            ord += hund1;
            leap -= hund1;
        }

        // Every 400th year is divisible by 100
        // but is a leap year
        if (hund4 > 0) {
            ord -= hund4;
            leap += hund4;
        }

        // Total number of extra days
        int days = ord + leap * 2;

        // modulo(7) for final answer
        int odd = days % 7;

        return odd;
    }

    // Driver code
    public static void main(String args[])
    {

        // Number of days
        int N = 100;
        System.out.print(oddDays(N));
    }
}

Python3

# Python3 implementation of the approach

# Function to return the count of odd days
def oddDays(N):

    # Count of years divisible
    # by 100 and 400
    hund1 = N // 100
    hund4 = N // 400

    # Every 4th year is a leap year
    leap = N >> 2
    ordd = N - leap

    # Every 100th year is divisible by 4
    # but is not a leap year
    if (hund1):
        ordd += hund1
        leap -= hund1

    # Every 400th year is divisible by 100
    # but is a leap year
    if (hund4):
        ordd -= hund4
        leap += hund4

    # Total number of extra days
    days = ordd + leap * 2

    # modulo(7) for final answer
    odd = days % 7

    return odd

# Driver code

# Number of days
N = 100
print(oddDays(N))

# This code is contributed
# by mohit kumar

C#

// C# implementation of the approach
using System;

class GFG
{

    // Function to return the count of odd days
    static int oddDays(int N)
    {
        // Count of years divisible
        // by 100 and 400
        int hund1 = N / 100;
        int hund4 = N / 400;

        // Every 4th year is a leap year
        int leap = N >> 2;
        int ord = N - leap;

        // Every 100th year is divisible by 4
        // but is not a leap year
        if (hund1 > 0)
        {
            ord += hund1;
            leap -= hund1;
        }

        // Every 400th year is divisible by 100
        // but is a leap year
        if (hund4 > 0)
        {
            ord -= hund4;
            leap += hund4;
        }

        // Total number of extra days
        int days = ord + leap * 2;

        // modulo(7) for final answer
        int odd = days % 7;

        return odd;
    }

    // Driver code
    static void Main()
    {

        // Number of days
        int N = 100;
        Console.WriteLine(oddDays(N));
    }
}

// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach

// Function to return the count of odd days
function oddDays($N)
{

    // Count of years divisible
    // by 100 and 400
    $hund1 = floor($N / 100);
    $hund4 = floor($N / 400);

    // Every 4th year is a leap year
    $leap = $N >> 2;
    $ord = $N - $leap;

    // Every 100th year is divisible by 4
    // but is not a leap year
    if ($hund1)
    {
        $ord += $hund1;
        $leap -= $hund1;
    }

    // Every 400th year is divisible by 100
    // but is a leap year
    if ($hund4)
    {
        $ord -= $hund4;
        $leap += $hund4;
    }

    // Total number of extra days
    $days = $ord + $leap * 2;

    // modulo(7) for final answer
    $odd = $days % 7;

    return $odd;
}

// Driver code

// Number of days
$N = 100;

echo oddDays($N);

// This code is contributed by Ryuga
?>

Javascript

<script>

// JavaScript implementation of the approach

// Function to return the count of odd days
    function oddDays(N) {
        // Count of years divisible
        // by 100 and 400
        var hund1 = N / 100;
        var hund4 = N / 400;

        // Every 4th year is a leap year
        var leap = N >> 2;
        var ord = N - leap;

        // Every 100th year is divisible by 4
        // but is not a leap year
        if (hund1 > 0) {
            ord += hund1;
            leap -= hund1;
        }

        // Every 400th year is divisible by 100
        // but is a leap year
        if (hund4 > 0) {
            ord -= hund4;
            leap += hund4;
        }

        // Total number of extra days
        var days = ord + leap * 2;

        // modulo(7) for final answer
        var odd = days % 7;

        return odd;
    }

    // Driver code


        // Number of days
        var N = 100;
        document.write(oddDays(N).toFixed());

// This code is contributed by todaysgaurav

</script>

Output:

5

Time Complexity: O(1)
Auxiliary Space: O(1)