C CODING Write a program that prints the day number of the y

C++ CODING!!!

Write a program that prints the day number of the year, given the date in the form month-day-year. For example, if the input is 1-1-2006, the day number is 1; if the input is 12-25-2006, the day number is 359. The program should check for a leap year. A year is a leap year if it is divisible by 4, but not divisible by 100. For example, 1992 and 2008 are divisible by 4, but not by 100. A year that is divisible by 100 is a leap year if it is also divisible by 400. For example, 1600 and 2000 are divisible by 400. However, 1800 is not a leap year because 1800 is not divisible by 400.

//code has been tested on gcc compiler

#include <iostream>

using namespace std;

int main()
{
int month,date,year; //variable for storing month,date and yesr
int day_number; //variable for storing day number


cout<<\"Enter the date in the format mm-dd-yyyy\" <<endl; //asking the user to enter in format mm-dd-yyyy,ex:1-1-2006
cin>>month;
cin.get();
cin>>date;
cin.get();
cin>>year;






if(year%4==0 && year%100 != 0) //checking the leap year condition
{
switch(month) //applying switch on month
{ case 1: //if month=1 ,i.e january
day_number=date; //day number =date entered by yser


break;
case 2: //if month=2,i.e feburary
day_number=31+date; //day number=total day in january + current date,i.e 31+date
//follow the same logic in other months

break;
case 3:
day_number=60+date;

break;
case 4:
day_number=91+date;

break;
case 5:
day_number=121+date;

break;
case 6:
day_number=152+date;

break;
case 7:
day_number=182+date;

break;
case 8:
day_number=213+date;

break;
case 9:
day_number=244+date;

break;
case 10:
day_number=274+date;

break;
case 11:
day_number=305+date;

break;
case 12:
day_number=335+date;

break;

}
}
else if(year%100==0 && year%400 == 0) //checking leap year
{
switch(month)
{ case 1:
day_number=date;

break;
case 2:
day_number=31+date;

break;
case 3:
day_number=60+date;

break;
case 4:
day_number=91+date;

break;
case 5:
day_number=121+date;

break;
case 6:
day_number=152+date;

break;
case 7:
day_number=182+date;

break;
case 8:
day_number=213+date;

break;
case 9:
day_number=244+date;

break;
case 10:
day_number=274+date;

break;
case 11:
day_number=305+date;

break;
case 12:
day_number=335+date;

break;

}
}
else //if not a leap year
{
switch(month)
{case 1:
day_number=date;

break;
case 2:
day_number=31+date;

break;
case 3:
day_number=59+date;

break;
case 4:
day_number=90+date;

break;
case 5:
day_number=120+date;

break;
case 6:
day_number=151+date;

break;
case 7:
day_number=181+date;

break;
case 8:
day_number=212+date;

break;
case 9:
day_number=243+date;

break;
case 10:
day_number=273+date;

break;
case 11:
day_number=304+date;

break;
case 12:
day_number=334+date;

break;
}
}

cout<<\"the day number is \"<<day_number<<endl; //print output
return 0;
}
//end of code
//core logic of the code:day=current date+no.of days in previous months