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#include<stdio.h>
#include<conio.h>
void main()
{
 int  i,j,gen[4],rem[4],frl=8,genl=4,k,k1,fr[11];
 clrscr();
 printf("enter frame:");
 for(i=0;i<frl;i++)
{ 
scanf("%d",&fr[i]);
} 
for(i=frl;i<11;i++)
{ 
fr[i]=0;
}
printf("enter generator:");
 for(i=0;i<4;i++)
 scanf("%d",&gen[i]);
  for(k=0;k<frl;k++)
  {
   if(fr[k]==1)
   {
   k1=k;
    for(i=0,j=k;i<genl;i++,j++)
    {
     rem[i]=fr[j]^gen[i];
    }
 
    for(i=0;i<genl;i++)
    {
	  fr[k1]=rem[i];
     k1++;
    }
   }
    }
      printf("\nremainder is: ");
      for(i=0;i<4;i++)
     printf("%d",rem[i]);
     getch();
}

OUTPUT:
enter frame:
1 0 1 1 0 1 0 0
enter generator:
1 0 1 0
remainder is:
0 0 1 0

Logic: The program follows the mathematical cosine series, where cosine of the entered radian angle ? is,

The program asks the user to enter the Angle x° . Also it asks for the number of iterations ‘n’ that the cosine loop should iterate. As the iteration increases, the accuracy boosts up. After the desired loop, the resultant gets printed out. Similar algorithm can be set to find the Sine value of the given number.

#include<stdio.h>
#include<math.h>
void main()
{
float pow = 2.0, nr, dr = 1.0, x1, sum;
int i = 1,n,s = -1,x;
clrscr();
printf(“\n\n\t ENTER THE ANGLE…: “);
scanf(“%d”, &x);
x1 = 3.142 * (x / 180.0);
sum = 1.0;
nr = x1*x1;
printf(“\n\t ENTER THE NUMBER OF TERMS…: “);
scanf(“%d”,&n);
while(i<=n)
{
dr = dr * pow * (pow – 1.0);
sum = sum + (nr / (dr * s));
s = s * (-1);
pow = pow + 2.0;
nr = nr * x1 * x1;
i++;
}
printf(“\n\t THE SUM OF THE COS SERIES IS..: %0.3f”, sum);
getch();
}
Download exe and source code here.

Logic: The program follows the mathematical sine series, where cosine of the entered radian angle x  is,

The program asks the user to enter the Angle x° . Also it asks for the number of iterations ‘n’ that the sine loop should iterate. As the iteration increases, the accuracy boosts up. After the desired loop, the resultant gets printed out. Similar algorithm can be set to find the Cosine value of the given number.

#include<stdio.h>
#include<math.h>
void main()
{
int i = 2, n, s = 1, x, pwr = 1, dr;
float nr = 1, x1, sum;
clrscr();
printf(“\n\n\t ENTER THE ANGLE…: “);
scanf(“%d”, &x);
x1 = 3.142 * (x / 180.0);
sum = x1;
printf(“\n\t ENTER THE NUMBER OF TERMS…: “);
scanf(“%d”, &n);
while(i <= n)
{
pwr = pwr + 2;
dr = dr * pwr * (pwr – 1);
sum = sum + (nr / dr) * s;
s = s * (-1);
nr = nr * x1 * x1;
i+= 2;
}
printf(“\n\t THE SUM OF THE SINE SERIES IS..: %0.3f”,sum);
getch();
}
Download exe and source code here.

Logic : The program asks the user to enter the number, to find out the sum of its digits. Sets the flag ’sum’ to zero, implies that sum till now is Nil. Stores the entered number as Num. The while loop slices the number into digits. In each iteration, the current digit gets added up to the flag ‘Sum’ . Finally it prints out the number.

The same logic can be implemented for other programs too, where the situation comes to slice up the given number, as in the case of  Reversing the number.

#include<stdio.h>
#include<math.h>
void main()
{
long int num,sum = 0,dig;
clrscr();
printf(“\n\n\t ENTER A NUMBER…: “);
scanf(“%ld”,&num);
while(num>0)
{
dig = num % 10;
sum = sum + dig;
num = num / 10;
}
printf(“\n\t SUM OF DIGITS IS…: %ld”, sum);
getch();
}

Download exe and source code here.

Logic :  The program expects the user to enter the number to check if palindrome. The entered number is reversed and stored in another variable. Finally the reversed number is checked for equality with the entered number. If they are the same, then it is a palindrome.

A slight change in the code can check if the entered string is a palindrome.

#include<stdio.h>
#include<math.h>
void main()
{
long int n, num, rev = 0, dig;
clrscr();
printf(“\n\n\t ENTER A NUMBER…: “);
scanf(“%ld”, &num);
n = num;
while(num>0)
{
dig = num % 10;
rev = rev * 10 + dig;
num = num / 10;
}
if (n == rev)
printf(“\n\t GIVEN NUMBER IS A PALINDROME”);
else
printf(“\n\t GIVEN NUMBER NOT A PALINDROME”);
getch();
}
Download exe and source code here.

Logic :  The main idea here is to trace the entered number till its length, and slicing up in each of the iterations. The program asks the user to enter the number to reverse. Sets two variable flags to hold each digits and the final reversed number.
The while loop slices down the given number to digits, and appends to the reversed flag ‘rev’. Finally it prints out the reversed number after exiting from the while loop.

The similar logic for slicing up the given number is used in other programs like, finding the sum of digits and, palindrome checking programs

#include<stdio.h>
#include<math.h>
void main()
{
long int num, rev = 0, dig;
clrscr();
printf(“\n\n\t ENTER A NUMBER…: “);
scanf(“%ld”, &num);
while(num>0)
{
dig = num % 10;
rev = rev * 10 + dig;
num = num / 10;
}
printf(“\n\t REVERSED NUMBER IS…: %ld”, rev);
getch();
}
Download exe and source code here.

Definition : Armstrong numbers are the integer numbers , where -the sum of third  power of each digits – equals to the given number itself.

Logic :  Here we traced the upper limit through a for loop, where in each of the iterations, we are checking if the sum of cubes of its digits equals that integer. If so, displaying the number.

The while loop is  used to slice up a number, which, with a slight modification, can also help us in  many other math related program segments. It needs to include the header “math.h”.

Program to find the armstrong numbers below a given number

#include<stdio.h>
#include<math.h>
void main()
{
int lim_up,n,dig,sum,num;
clrscr();
printf(“\n\n\t ENTER THE UPPER LIMIT…: “);
scanf(“%d”,&lim_up);
printf(“\n\n\t ARMSTRONG NUMBERS ARE…: “);
for(n=1;n<lim_up;n++)
{
sum = 0;
num = n;
while(num>0)
{
dig = num%10;
sum = sum+pow(dig,3);
num = num/10;
}
if(sum == n)
printf(“\n\n\t\t\t%d”,n);
}
getch();
}

Download exe and source code here.

Logic: This is advanced version of the previous program. Here, user need to enter two numbers as the lower and upper limits for the iteration loop to find the prime number in between. The outer for loop traces the iteration till the limit, wherein each of iteration inner for loop checks the present number is prime or not, with the prime number’s logic. If it is, it prints out the present number.

In this program, both the lower limit and the upper limit are variables, and so is flexible. These three programs show, how we can upgrade the logic to make the code flexible one.

The similar growth can be seen in the case of finding the perfect numbers.

Program to find the prime numbers between a given range

#include<stdio.h>
void main()
{
int i, prime, lim_up, lim_low, n;
clrscr();
printf(“\n\n\t ENTER THE LOWER LIMIT…: “);
scanf(“%d”, &lim_low);
printf(“\n\n\t ENTER THE UPPER LIMIT…: “);
scanf(“%d”, &lim_up);
printf(“\n\n\t PRIME NUMBERS ARE…: “);
for(n=lim_low+1; n<lim_up; n++)
{
prime = 1;
for(i=2; i<n; i++)
if(n%i == 0)
{
prime = 0;
break;
}
if(prime)
printf(“\n\n\t\t\t%d”, n);
}
getch();
}
Download exe and source code here.

Logic: This is a slightly modified version of the previous program. Here, user need to enter the number as the upper limit for the for the iteration loop to find the prime number The outer for loop traces the iteration till the limit,, wherein each of iteration inner for loop checks the present number is prime or not, with the previous program’s logic. If it is, it prints out the present number.

In this program, the lower limit is constant zero. We can change that also, to make the program fully flexible, and to print out all the perfect number between the given range.

Program to find the prime numbers below a given number.

#include<stdio.h>
void main()
{
int i, prime, lim_up, n;
clrscr();
printf(“\n\n\t ENTER THE UPPER LIMIT…: “);
scanf(“%d”, &lim_up);
printf(“\n\n\t PRIME NUMBERS ARE…: “);
for(n=1; n<lim_up; n++)
{
prime = 1;
for(i=2; i<n; i++)
if(n%i == 0)
{
prime = 0;
break;
}
if(prime)
printf(“\n\n\t\t\t%d”,n);
}
getch();
}
Download exe and source code here.