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29 changes: 29 additions & 0 deletions DayNumber.cpp
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/*Write a program that prints the day number of the year, given the date in the form month-day-year.*/
#include<iostream>
using namespace std;
int main()
{
int mm, dd, yyyy, days_count=0;
int days_in_lp[]={0,31,29,31,30,31,30,31,31,30,31,30,31}; //number of days in a month of a leap year
int days_in_nlp[]={0,31,28,31,30,31,30,31,31,30,31,30,31}; //number of days in a month of a non-leap year
cout<<"Enter the date in the mm dd yyyy format: ";
cin>>mm>>dd>>yyyy;
if ((yyyy%4==0 && yyyy%100!=0)||(yyyy%400==0))
{
for (int i=1;i<mm;i++)
{
days_count+=days_in_lp[i]; //Adding days of previous months for a leap year
}
days_count+=dd; //Adding days of current month if the year is a leap year
}
else
{
for (int i=1;i<mm;i++)
{
days_count+=days_in_nlp[i]; //Adding days of previous months for a non- leap year
}
days_count+=dd; //Adding days of current month for a non- leap year
}
cout<<"The day number for "<<mm<<"-"<<dd<<"-"<<yyyy<<" is "<<days_count; //printing the number of days
return 0;
}
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19 changes: 19 additions & 0 deletions DigitReverser.cpp
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/*Write a function, reverseDigit, that takes an integer as a parameter and returns the number with its digits reversed. For example, the value of reverseDigit(12345) is 54321; the value of reverseDigit(5600) is 65; the value of reverseDigit(7008) is 8007; and the value of reverseDigit(-532) is -235.*/
#include<iostream>
using namespace std;
int main()
{
long int n, to_be_rev; //decleration of variables
int rem=0,rev=0;
cout<<"Enter the number to be reversed: ";
cin>>n;
to_be_rev=n; //Number to be reversed
while (n!=0)
{
rem=n%10;
rev=rev*10+rem; //Creation of reversed number
n=n/10;
}
cout<<"The value of reverseDigit("<<to_be_rev<<") is "<<rev; //printing of old number and its new reversed form
return 0;
}
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18 changes: 18 additions & 0 deletions Reverse of number.cpp
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#include<iostream>
using namespace std;
int main()
{
long int n, to_be_rev; //decleration of variables
int rem=0,rev=0;
cout<<"Enter the number to be reversed: ";
cin>>n;
to_be_rev=n; //Number to be reversed
while (n!=0)
{
rem=n%10;
rev=rev*10+rem; //Creation of reversed number
n=n/10;
}
cout<<"Reverse of ("<<to_be_rev<<") is "<<rev; //printing of old number and its new reversed form
return 0;
}