Hey!
Here is a 'Forever Calendar' I have made. It gives the calendar of any month
and year. Let's see which one of you is able to identify the logic of the
program first!! It is very tricky. Best of luck!
-Bharath
Program follows:
/*
This is a program for a calendar. You have to enter an year and a month
and
it will show you the calendar for the month.
*/
#include<iostream.h>
#include<conio.h>
int no_days(int,int);
int leap(int);
int step1 (unsigned int);
void month_name(unsigned int);
void main()
{
restart:
clrscr();
cout<<"Enter year : ";
unsigned int y,m;
cin>>y;
int x;
x=step1(y);
int month[14][12]= {1,4,4,7,2,5,7,3,6,1,4,6,
2,5,5,1,3,6,1,4,7,2,5,7,
3,6,6,2,4,7,2,5,1,3,6,1,
4,7,7,3,5,1,3,6,2,4,7,2,
5,1,1,4,6,2,4,7,3,5,1,3,
6,2,2,5,7,3,5,1,4,6,2,4,
7,3,3,6,1,4,6,2,5,7,3,5,
1,4,5,1,3,6,1,4,7,2,5,7,
2,5,6,2,4,7,2,5,1,3,6,1,
3,6,7,3,5,1,3,6,2,4,7,2,
4,7,1,4,6,2,4,7,3,5,1,3,
5,1,2,5,7,3,5,1,4,6,2,4,
6,2,3,6,1,4,6,2,5,7,3,5,
7,3,4,7,2,5,7,3,6,1,4,6};
cout<<"Enter month (1 - 12) : ";
month_input:
cin>>m;
if(m<1 || m>12)
{
cout<<"Enter a valid month (1 - 12) : ";
goto month_input;
}
cout<<"\n\n\t\t";
month_name(m);
cout<<' '<<y<<"\n\nMon\tTue\tWed\tThu\tFri\tSat\tSun\n";
int j=month[x-1][m-1], days=no_days(m,y);
for (int i=0; i<j-1; i++)
cout<<'\t';
for(i=1; i<=days; i++)
{
cout<<i<<'\t';
if(j==7)
{
cout<<'\n';
j=1;
}
else
j++;
}
cout<<"\n\nTry for another month? (y/n) : ";
char ch;
cin>>ch;
if(ch=='y' || ch=='Y')
goto restart;
}
int step1 (unsigned int y)
{
int x=y%7;
x+=(y-1)/4;
x-=(y-1)/100;
x+=(y-1)/400;
x=x%7;
if(x==0)
x=7;
if(leap(y))
x+=7;
return x;
}
void month_name(unsigned int m)
{
switch(m)
{
case 1 : cout<<"January";
break;
case 2 : cout<<"February";
break;
case 3 : cout<<"March";
break;
case 4 : cout<<"April";
break;
case 5 : cout<<"May";
break;
case 6 : cout<<"June";
break;
case 7 : cout<<"July";
break;
case 8 : cout<<"August";
break;
case 9 : cout<<"September";
break;
case 10: cout<<"October";
break;
case 11: cout<<"November";
break;
case 12: cout<<"December";
break;
}
}
int no_days(int mon, int y)
{
int d;
switch(mon)
{
case 1 :
case 3 :
case 5 :
case 7 :
case 8 :
case 10:
case 12: d=31;
break;
case 4 :
case 6 :
case 9 :
case 11: d=30;
break;
case 2 : if(leap(y))
d=29;
else
d=28;
break;
}
return d;
}
int leap (int y)
{
if (y%100==0)
if (y%400==0)
return 1;
else
return 0;
else
if (y%4==0)
return 1;
else
return 0;
}
[Non-text portions of this message have been removed]
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