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Illusion [34]
3 years ago
9

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.
Computers and Technology
1 answer:
vlabodo [156]3 years ago
4 0

Answer:

C++:

C++ Code:

#include <iostream>

#include <string>

using namespace std;

struct date

{

  int d,m,y;

};

int isLeap(int y)

{

  if(y%100==0)

  {

      if(y%400==0)

      return 1;

      return 0;

  }

  if(y%4==0)

  return 1;

  return 0;

}

int day_no(date D)

{

  int m = D.m;

  int y = D.y;

  int d = D.d;

  int i;

  int mn[13] = {0,31,28,31,30,31,30,31,31,30,31,30,31};

  for(i=0;i<m;i++)

  {

      d += mn[i];

  }

  if(isLeap(y))

  {

      if(m>2)

      d++;

  }

  return d;

}

date get_info(string s)

{

  date D;

  int i,p1,p2,l = s.length();

  for(i=0;i<l;i++)

  {

      if(s[i] == '-')

      {

      p1 = i;

      break ;

      }

  }

  for(i=p1+1;i<l;i++)

  {

      if(s[i] == '-')

      {

      p2 = i;

      break ;

      }

  }

 

  D.m = 0;

  for(i=0;i<p1;i++)

  D.m = (D.m)*10 + (s[i]-'0');

 

  D.d = 0;

  for(i=p1+1;i<p2;i++)

  D.d = (D.d)*10 + (s[i]-'0');

 

  D.y = 0;

  for(i=p2+1;i<l;i++)

  D.y = (D.y)*10 + (s[i]-'0');

 

  return D;

 

}

int main()

{

  string s1 = "4-5-2008";

  string s2 = "12-30-1995";

  string s3 = "6-21-2000";

  string s4 = "1-31-1500";

  string s5 = "7-19-1983";

  string s6 = "2-29-1976";

 

  cout<<"Date\t\tDay no\n\n";

  cout<<s1<<"\t"<<day_no(get_info(s1))<<endl;

  cout<<s2<<"\t"<<day_no(get_info(s2))<<endl;

  cout<<s3<<"\t"<<day_no(get_info(s3))<<endl;

  cout<<s4<<"\t"<<day_no(get_info(s4))<<endl;

  cout<<s5<<"\t"<<day_no(get_info(s5))<<endl;

  cout<<s6<<"\t"<<day_no(get_info(s6))<<endl;

 

 

  return 0;

}

Explanation:

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Explanation:

Please go through the comments in the code to give an explanation of the program. The source code file has also been added to this response. Please download it and go through it.

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