//LOGIC OF THE PROGRAM

/*if a number >= 2^n its given 1 else 0. its similar to writing the following

for example: binary of 120 would be 1111000

decimal: 128  64  32  16  8  4  2  1

binary:   0    1  1   1   1  0  0  0

just add up the numbers which have 1 underneath them,it gives 120

all you need to do to find binary equivalent is dat make such a combo that
powers having 1 under them add to form the decimal number*/


#include<conio.h>
#include<math.h>
void main()
{
        unsigned long i=0;
        int m=0,comp,n=0,j=0;
        clrscr();
        printf("Enter the decimal number (upto 1024): ");
        scanf("%d",&n);

        for(m=10;m>=0;m--)
        {
                comp=pow(2,m);  //calculates 2^n like 2,4,8,16...
                if(n>=comp)    //comparison with 2^n
                {
                        j=1;   //true so 1 is given
                        n=n-comp;   //subtracting as we are going down
                }                   //the order 128->64->32....
                else j=0;
                i=10*i+j;          //generating the binary number
        }
        printf("\nThe decimal equivalent is: %ld",i);
        getch();
}