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Write the program in c++ Write a program that prints the day number of the year,

ID: 3793103 • Letter: W

Question

Write the program in c++

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.

Explanation / Answer


// C++ code
#include <iostream>
#include <stdio.h>
#include <stdlib.h>
using namespace std;


int main()
{

int date,month,year;
int totalDays=0;

printf("Enter date mm-dd-yyyy: ");
scanf("%d-%d-%d",&month,&date,&year);

while(month--)
{
   switch(month)
   {
   case 11:totalDays = totalDays + 3   0;
          break;
   case 10:totalDays = totalDays + 31;
          break;
   case 9:totalDays = totalDays + 30;
          break;
   case 8:totalDays = totalDays + 31;
          break;
   case 7:totalDays = totalDays + 31;
          break;
   case 6:totalDays = totalDays + 30;
          break;
   case 5:totalDays = totalDays + 31;
          break;
   case 4:totalDays = totalDays + 30;
          break;
   case 3:totalDays = totalDays + 31;
          break;
   case 2:if((!(year%4))&&(year%100))
   totalDays = totalDays + 29;
   else if(!(year%400))
   totalDays = totalDays + 29;
   else
   totalDays = totalDays + 28;
   break;
   case 1:totalDays = totalDays + 31;
          break;
   case 0:break;
   }
}

cout << "totalDays: " << (totalDays+date) << endl;


return 0;
}

/*
output:

Enter date mm-dd-yyyy: 12-25-2006
totalDays: 359


Enter date mm-dd-yyyy: 12-31-1992
totalDays: 366

*/