Find the GCD of two numbers

Using For Loop

#include<stdio.h>
int main(){
    int x,y,m,i;
    do
    {
        printf("Enter two number: ");
        scanf("%d%d",&x,&y);
        if(x==0 || y==0)
          printf("Please check the input and try again...\n");
    }while(x==0 || y==0);
    if(x>y)
         m=y;
    else
         m=x;

    for(i=m;i>=1;i--){
         if(x%i==0&&y%i==0){
             printf("\nGCD of %d and %d is %d",x,y,i) ;
             break;
         }
    }
    return 0;
}

Using Recursion

#include <stdio.h>

int gcd(int a, int b)
{

    if (a == b)
        return a;
    if (a > b)
        return gcd(a-b, b);
    return gcd(a, b-a);
}

int main()
{
    int x,y,m,i;
    do
    {
        printf("Enter two number: ");
        scanf("%d%d",&x,&y);
        if(x==0 || y==0)
          printf("Please check the input and try again...\n");
    }while(x==0 || y==0);
    printf("GCD of %d and %d is %d ", x, y, gcd(x, y));
    return 0;
}

   

Output
Enter two number: 0 99
Please check the input and try again...
Enter two number: 99 121

GCD of 99 and 121 is 11

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Find the Factorial of a number

Summary: Factorial is represented using '!', so five factorial will be written as (5!),n factorial as (n!).
n! = n*(n-1)*(n-2)*(n-3)...3.2.1 and zero factorial is defined as one i.e. 0! = 1.



Using While Loop

#include<stdio.h>
void main()
{
    int a,f,i;
    printf("Enter a number: ");
    scanf("%d",&a);
    f=1;
    i=1;
    while(i<=a)
    {
        f = f * i;
        i++;
    }
     printf("\nFactorial of %d is: %d",a,f);
}

Using For Loop

#include<stdio.h>
void main()
{
    int a,f,i;
    printf("Enter a number: ");
    scanf("%d",&a);
    f=1;
    for(i=1;i<=a;i++)
        f = f * i;
     printf("\nFactorial of %d is: %d",a,f);
}

Using Recursion

#include<stdio.h>
int fact(int);
int main(){
 int num,f;

 printf("\nEnter a number: ");
 scanf("%d",&num);

 f=fact(num);

 printf("\nFactorial of %d is: %d",num,f);
 return 0;
}

int fact(int n){
 if(n==1)
  return 1;
 else
  return(n*fact(n-1));
}

   

Output

Enter a number: 5

Factorial of 5 is: 120

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Solve All-Pairs Shortest Path Problem using Floyd - Warshall Algorithm

#include<stdio.h>
#include<conio.h>
#define inf 999

void main()
{
 int i,j,k,n,w[20][20];
 printf("\n Enter the no. of vertices : ");
 scanf("%d",&n);
 printf("\n Enter the weights : ");
   for(i=1;i<=n;i++)
    for(j=1;j<=n;j++)
      {
       printf("w[%d][%d]= ",i,j);
       scanf("%d",&w[i][j]);
    if(i!=j && (w[i][j]==0))
      w[i][j]=inf;
       }

 for(i=1;i<=n;i++)
 {
     printf("\n");
     for(j=1;j<=n;j++)
  printf("%d\t",w[i][j]);
 }


       for(k=1;k<=n;k++)
 {
    for(i=1;i<=n;i++)
     {
       for(j=1;j<=n;j++)
        {
   if(w[i][k]+w[k][j]<w[i][j])
       w[i][j]=w[i][k]+w[k][j];
        }
     }
 }


  printf("\n The resultant Matrix : \n");
    for(i=1;i<=n;i++)
 {
     printf("\n");
     for(j=1;j<=n;j++)
        printf("%d\t",w[i][j]);
 }

 getch();

}

   

Output

Enter the no. of vertices : 5

 Enter the weights : w[1][1]= 0
w[1][2]= 3
w[1][3]= 8
w[1][4]= 0
w[1][5]= 4
w[2][1]= 0
w[2][2]= 0
w[2][3]= 0
w[2][4]= 1
w[2][5]= 7
w[3][1]= 0
w[3][2]= 4
w[3][3]= 0
w[3][4]= 0
w[3][5]= 0
w[4][1]= 2
w[4][2]= 0
w[4][3]= 5
w[4][4]= 0
w[4][5]= 0
w[5][1]= 0
w[5][2]= 0
w[5][3]= 0
w[5][4]= 6
w[5][5]= 0

0       3       8       999     4
999     0       999     1       7
999     4       0       999     999
2       999     5       0       999
999     999     999     6       0
 The resultant Matrix :

0       3       8       4       4
3       0       6       1       7
7       4       0       5       11
2       5       5       0       6
8       11      11      6       0

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Binary Search using Recursion

#include<stdio.h>

void bins(int lb,int ub,int item,int a[]);
void main()
 {
    int i,n,item,a[20];
    printf("Enter the size of an array: ");
    scanf("%d",&n);

    printf("Enter the elements of the array: " );
    for(i=0;i<n;i++)
        {
         scanf("%d",&a[i]);
        }

    printf("Enter the number to be search: ");
    scanf("%d",&item);


    bins(0,n-1,item,a);

 }

void bins(int lb,int ub,int item,int a[])
{
     int mid,flag=0;

     if(lb<=ub)
        {
          mid=(lb+ub)/2;
          if(item==a[mid])
            {
              flag=flag+1;
            }
          else if(item<a[mid])
          {
              return bins(lb,mid-1,item,a);
          }
          else
              return bins(mid+1,ub,item,a);
        }


       if(flag==0)
         printf("Number is not found.");
       else
         printf("Number is found at %d", mid+1);
}

   

Output
Enter the size of an array: 6
Enter the elements of the array: 90 95 1000 1001 1003 1006
Enter the number to be search: 95
Number is found at 2

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Find the maximum length of the subarray in increasing order.

For example we have an array {5,4,6,-2,-1,0,1,9}. The biggest sub-array in increasing order is: {-2,-1,0,1,9}. So the length is 5.

#include<stdio.h>
void main () {
int arr[8] = {5,4,6,-2,-1,0,1,9},i,arrn[8],count=1,j=0,max;
printf("The given array is: ");
for(i=0;i<8;i++)
    printf("  %d" ,arr[i]); // printing the actual array  - unnecessary step.
printf("\n");
for(i=0;i<7;i++){
        if(arr[i]<arr[i+1])
        {
            count++;
        }
        else
        {
            arrn[j]=count; // adding the length of sub array
            count=1;
            j++;
        }
}
arrn[j]=count; // adding the final length
max=arrn[0];
printf("Length of sub arrays in increasing order");
for(i=0;i<=j;i++){
        if(max<arrn[i])
            max=arrn[i];
        printf(" %d",arrn[i]);  // printing the new array  - unnecessary step.
}
        printf("\n Maximum Length of the substring: %d",max); // print max value
}

   

Output
The given array is:   5  4  6  -2  -1  0  1  9
Length of sub arrays in increasing order 1 2 5
Maximum Length of the substring: 5

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Spiral Matrix

#include <stdio.h>
#define R 3
#define C 3

void spiralPrint(int m, int n, int a[R][C])
{
    int i, k = 0, l = 0;

    /*  k - starting row index
        m - ending row index
        l - starting column index
        n - ending column index
        i - iterator
    */

    while (k < m && l < n)
    {
        /* Print the first row from the remaining rows */
        for (i = l; i < n; ++i)
        {
            printf("%d ", a[k][i]);
        }
        k++;

        /* Print the last column from the remaining columns */
        for (i = k; i < m; ++i)
        {
            printf("%d ", a[i][n-1]);
        }
        n--;

        /* Print the last row from the remaining rows */
        if ( k < m)
        {
            for (i = n-1; i >= l; --i)
            {
                printf("%d ", a[m-1][i]);
            }
            m--;
        }

        /* Print the first column from the remaining columns */
        if (l < n)
        {
            for (i = m-1; i >= k; --i)
            {
                printf("%d ", a[i][l]);
            }
            l++;
        }
    }
}

/* Program to test above functions */
int main()
{
    int a[R][C] = { {1,  2,  3},
                    {8,  9,  4},
                    {7,  6,  5}
    };

    spiralPrint(R, C, a);
    return 0;
}

   

Output
1 2 3 4 5 6 7 8 9

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Find the maximum occurrence of a character in a string

#include<stdio.h>
#include<conio.h>

int main() {
   char str[20], ch;
   int count = 0, i,j=0 , max;
   int arr[10];
   printf("\nEnter a string : ");
   scanf("%s", &str);

   printf("\nEnter the character to be searched : ");
   scanf("%c", &ch);
   for (i = 0; str[i] != '\0'; i++) {
      if (str[i] == ch)
         count++;
      else if(count != 0)
      {
          arr[j] = count;
          j++;
          count = 0;
      }
   }
   arr[j]=count;
   max = 0;
       for(i=0;i<=j;i++)
        if(max<arr[i])
         max=arr[i];
   if (max == 0)
      printf("\nCharacter '%c'is not present", ch);
   else
   {
       printf("\nMaximum occurrence of character '%c' : %d", ch, max);
   }
   return (0);
}

   

Output

Enter a string : aabbbaaab

Enter the character to be searched :
Maximum occurrence of character 'a' : 3

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Basic Structure of a C Program

A C program generally has the following parts:

§  Preprocessor Commands

§  Functions

§  Variables

§  Statements & Expressions

§  Comments

Let us consider a C Program:

#include<stdio.h>
void main()
{ 
 printf("Hello, World!");
}

Preprocessor Commands: These commands tells the compiler to do some preprocessing before executing the actual compilation. For example, “#include <stdio.h>” is a preprocessor command which tells a C compiler to include stdio.h file before going to actual compilation.

Functions:  These are main building blocks of any C Program. Every C Program will at least have one function which is a mandatory function called main() function. This function is prefixed with keyword int or void or any other datatype. This defines the return type of the function. For int main() the program will return an integer value to the terminal (system).

The C Programming language provides a set of built-in functions. In the above example printf() is a C built-in function which is used to print anything on the screen.

Variables: These are used to hold numbers, strings and complex data for manipulation.

Statements & Expressions: Expressions combine variables and constants to create new values. Statements are expressions, assignments, function calls, or control flow statements which make up C programs.

Comments: are used to give additional useful information inside a C Program. All the comments will be put inside /*...*/

Note
•C is a case sensitive programming language. It means in C printf and Printf will have different meanings.
•C has a free-form line structure. End of each C statement must be marked with a semicolon.
•Multiple statements can be on the same line.
•White Spaces (ie tab space and space bar ) are ignored.
•Statements can continue over multiple lines.

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Get your hands dirty

The only way to learn a new programming language is by writing programs in it. The first program to write is the same for all languages: Print the words hello, world!

In C, the program to print ``Hello, world!'' is

#include <stdio.h>
main()
{
  printf("Hello, world!\n");
}

Just how to run this program depends on the system you are using.

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Basics of C

What Is C?

C is a highly flexible and adaptable language. Since its creation in 1970, it's been used for a wide variety of programs including firmware for micro-controllers, operating systems, applications, and graphics programming.

When was C developed?

C was developed at Bell Laboratories in 1972 by Dennis Ritchie along with Ken Thompson. Many of its principles and ideas were taken from the earlier language B and B's earlier ancestors BCPL and CPL.

Who is the inventor of C?

Dennis Ritchie invented C, the computer-programming language that underlies Microsoft Windows, the Unix operating system and much of the other software running on computers around the world. Mr. Ritchie was a longtime research scientist at Bell Labs, originally AT&T's research division.

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Accept Array elements

#include<stdio.h>
void main() {
   int arr[10], i, j, k, size;
   printf("\nEnter array size : ");
 scanf("%d", &size);

   printf("\nAccept Numbers : \n");
   for (i = 0; i < size; i++)
 {
 printf("Data [%d]: ",i+1);
 scanf("%d", &arr[i]);
 }

   printf("\nThe Array is : ");
   for (i = 0; i < size; i++)
 printf("%d ", arr[i]);

    getch();
}

   

Output

Enter array size : 5

Accept Numbers :
Data [1]: 2
Data [2]: 6
Data [3]: 7
Data [4]: 1
Data [5]: 2

The Array is : 2 6 7 1 2

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Clear Screen without clrscr() function call

#include<stdio.h>
void clearmyscreen();
void main()
 {
  printf("Hello");
  getch();
  clearmyscreen();
  printf("World");
  getch();
 }

void clearmyscreen()
 {
  system("cls");
 }

   

Output
World

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Find Perimeter of a Circle. Take the value of the radius from user.

#include<stdio.h>
#define pi 3.14

void main()
{
    int r;
    float p;
    printf("Enter the value of radius: ");
    scanf("%d", &r); 
    p=2*pi*r;
    printf("Perimeter=%f",p);
}

   

Output
Enter the value of radius: 9
Perimeter=56.520000

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Use Macro to define the value of PI and Find the Perimeter of a Circle

#include<stdio.h>
#define pi 3.14
void main()
{
    int r;
    r=7;
    printf("Perimeter=%f",2*pi*r);
}

   

Output
Perimeter=43.960000

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Print Pattern 5

#include <stdio.h>
void main()
{
    int i, j, k;
    for(i=5;i>=1;i--)
    {
 for(k=5;k>=i;k--)
        {
            printf(" ");
        }
        for(j=1;j<i;j++)
        {
            printf("*");
        }
        
 printf("\n");
    }
    getch();
}

   

Output

 ****
  ***
   **
    *

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Print Pattern 4

#include <stdio.h>
void main()
{
    int i, j, k;
    for(i=5;i>=1;i--)
    {
 for(j=1;j<=i;j++)
     printf("*");
 printf("\n");
    }
    getch();
}

   

Output
*****
****
***
**
*

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Print Pattern 3

#include <stdio.h>
void main()
{
    int i, j, k;
    for(i=5;i>=1;i--)
    {
        for(j=1;j<i;j++)
        {
            printf(" ");
        }
        for(k=5;k>=i;k--)
        {
            printf("*");
        }
 printf("\n");
    }
    getch();
}

   

Output

    *
   **
  ***
 ****
*****

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Print Pattern 2

#include <stdio.h>
void main()
{
    int i,j;
    for(i=1;i<=5;i++)
    {
 for(j=1;j<=i;j++)
 {
     printf("*");
 }
 printf("\n");
    }
    getch();
}

   

Output
*
**
***
****
*****

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Print Pattern 1

#include<stdio.h>
int main()
{
 int num,r,c;
 printf("Enter number of rows/columns: ");
 scanf("%d",&num);
 for(r=1; r<=num; r++)
 {
  for(c=1; c<=num; c++)
     printf("* ");
  printf("\n");
 }
 getch();
 return 0;
}

   

Output
Enter number of rows/columns: 5
* * * * *
* * * * *
* * * * *
* * * * *
* * * * *

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List all the Prime Numbers

#include<stdio.h>
void main()
{
 int n,i,j,flag; 
 printf("\n\nEnter the limit\t:"); 
 scanf("%d",&n);
 printf("\nThe prime numbers within the given limit:\n");
 for(i=1;i<=n;i++)
  { 
  flag=0;
  for(j=2;j<=i/2;j++)
   { 
   if(i%j==0) 
    { 
    flag=1;
    break;
    }
   }
  if(flag==0)
   printf("%d\t",i);
  }
}

   

Output
Enter the limit :60

The prime numbers within the given limit:
1       2       3       5       7       11      13      17      19      23
29      31      37      41      43      47      53      59

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Solve Quadratic Equation

#include <stdio.h>
#include <math.h>
void main() 
{
 float a,b,c,d,temp,r1,r2; 
 printf("\nEnter the value of a\t:");
 scanf("%f",&a); 
 printf("Enter the value of b\t:");
 scanf("%f",&b);
 printf("Enter the value of c\t:");
 scanf("%f",&c);

 d=((b*b)-(4*c*a));
 
 if(d==0)
 {
  printf("\nRoots are real and equal\n");
  r1=-b/(2*a);
  printf("\t X1=%f \n \t X2=%f ", r1, r1 );
 }

  else if(d>0)
 {
  printf("\nRoots are real and distinct\n");
  temp=sqrt(d);

  r1=(-b+temp)/(2*a);
  r2=(-b-temp)/(2*a);

  printf("\n\tX1=%f\n\tX2=%f",r1,r2);
 }

  else
 {
  printf("\nRoots are Imaginary\n");
  d*=-1;
  temp=sqrt(d);
  r1=-b/(2*a);
  r2=temp/(2*a);
  printf("\n\tX1=(%f + %fi)\n\tX2=(%f - %fi)",r1,r2,r1,r2);
 }
}

  

Output
Enter the value of a    :1
Enter the value of b    :-11
Enter the value of c    :30

Roots are real and distinct

        X1=6.000000
        X2=5.000000

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Find the greatest of two numbers using Conditional Operator

#include<stdio.h>
void main()
{
 int n1,n2,g;
 printf("Enter Numbers (n1, n2) : ");
 scanf("%d %d",&n1,&n2);
 g=n1>n2?n1:n2;
 printf("Greatest=%d",g);
}

  

Output
Enter Numbers (n1, n2) : 5 6 

Greatest=6

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Generate the Fibonacci Series

#include<stdio.h>

void main() 
{
 int t,i,m=0,n=1,f=1; 
 printf("\n\nEnter the number of terms:");
 scanf("%d",&t); 
 printf("\n The Fibonacci Series of %d terms:\n",t);
 for(i=1;i<=t;i++)
  {
   printf("%d\t",f);
   f=m+n;
   m=n;
   n=f;
  }
}

  

Output
Enter the number of terms:10

 The Fibonacci Series of 10 terms:
1       1       2       3       5       8       13      21      34      55

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Print 1-10 using do-while loop

#include<stdio.h>
void main()
{
    int i=1;
    do{
        printf("%d\n",i);
        i++;
      }while(i=100);
}

  

Output

1
2
3
4
5
6
7
8
9
10

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Find the perimeter of a Circle

#include <stdio.h>

void main()
{
    int r;
    float pi=3.14;
    r=7;
    printf("Perimeter=%f",2*pi*r);
}

  

Output
Perimeter=43.960003

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Check whether an alphabet is Vowel or not

#include <stdio.h>

void main()
 {
  char in;
  printf("Enter Character: ");
  scanf("%c", &in);
  if(in>=65&&in<=90 || in>=97&& in<=122)
  {
   if(in<97)
   in=in+32;
   if(in=='a'||in=='e'||in=='i'||in=='o'||in=='u')
   printf("vowel");
   else
         printf("consonent");
  }
  else
   printf("Not an valid Alphabet");
 }

  

Output
Enter Character: A 
vowel

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Print the ASCII values of all the English Alphabets

#include<stdio.h>
void main()
{
 char chl='a';
 char chu='A';
 int i;
 printf("\n===========================");
 for(i=1;i<=26;i++,chu++,chl++)
    printf("\n|%c |  %d \t|| %c| %d  |",chl,chl, chu, chu);   
 printf("\n===========================");
}

  

Output
===========================
|a |  97        || A| 65  |
|b |  98        || B| 66  |
|c |  99        || C| 67  |
|d |  100       || D| 68  |
|e |  101       || E| 69  |
|f |  102       || F| 70  |
|g |  103       || G| 71  |
|h |  104       || H| 72  |
|i |  105       || I| 73  |
|j |  106       || J| 74  |
|k |  107       || K| 75  |
|l |  108       || L| 76  |
|m |  109       || M| 77  |
|n |  110       || N| 78  |
|o |  111       || O| 79  |
|p |  112       || P| 80  |
|q |  113       || Q| 81  |
|r |  114       || R| 82  |
|s |  115       || S| 83  |
|t |  116       || T| 84  |
|u |  117       || U| 85  |
|v |  118       || V| 86  |
|w |  119       || W| 87  |
|x |  120       || X| 88  |
|y |  121       || Y| 89  |
|z |  122       || Z| 90  |
===========================

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Find Area and Circumference of a Circle

#include<stdio.h>
#define PI 3.142 

void main() 
{
 float r,a,c;
 printf("\nEnter the radius of circle:\t");
 scanf("%f",&r);
 a=PI*r*r; 
 printf("\nThe area of the circle:\t%f",a);
 c=2*PI*r; 
 printf("\nThe circumference of the circle: %f",c);
}

  

Output
Enter the radius of circle: 2
The area of the circle: 12.568000
The circumference of the circle: 12.568000
Enter the radius of circle: 1
The area of the circle: 3.142000
The circumference of the circle: 6.284000

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Print the Truth Table

#include<stdio.h>

int OR(int,int);
int AND(int,int);
int XOR(int,int);
int NOT(int);

void main()
{
 int a,b,ch,o;

 printf("\n-:MENU:-\n1.AND\n2.OR\n3.NOT\n4.XOR\n0.Exit\n");
 do{
    printf("\nEnter Choice:");
    scanf("%d",&ch);

    switch(ch)
     {
      case 1:
             printf ("\nTruth Table for AND\nA B  \tO");
             for(a=0;a<=1;a++)
             {
              for(b=0;b<=1;b++)
                { 
                  o=AND(a,b);
                  printf("\n%d %d \t%d",a,b,o);
                }
             }
              break;
      case 2:
             printf ("\nTruth Table for OR\nA B  \tO");
             for(a=0;a<=1;a++)
             {
              for(b=0;b<=1;b++)
                {
                o=OR(a,b);
                printf("\n%d %d \t%d",a,b,o);
                }
             }
              break;
      case 3: 
             printf ("\nTruth Table for NOT\nA  \tO");
             for(a=0;a<=1;a++)
             {
             o=NOT(a);
             printf("\n%d  \t%d",a,o);
             }
              break;
      case 4:
             printf ("\nTruth Table for XOR\nA B  \tO");
             for(a=0;a<=1;a++)
             {
              for(b=0;b<=1;b++)
                {
                o=XOR(a,b);
                printf("\n%d %d \t%d",a,b,o);
                }
             }
              break;

      case 0: exit();
              break;
             default: printf("\nError");
             }
      }while(ch!=0);

}


int OR(int a, int b)
{
if(a==0&&b==0)
   return 0;
else
   return 1;
}

int AND(int a, int b)
{
if (a==1&&b==1)
   return 1;
else
   return 0;
}

int NOT(int a)
{
if (a==0)
   return 1;
else
   return 0;
}

int XOR(int a,int b)
{
if (a==0&&b==1||a==1&&b==0)
   return 1;
else
   return 0;
}

  

Output
-:MENU:-
1.AND
2.OR
3.NOT
4.XOR
0.Exit
Enter Choice: 3 
Truth Table for NOT
A   O
1   0
0   1

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Simple Interest

#include<stdio.h>
void main()
{
 float p;
 float ir,t;
 float in;

  printf("Principle amount : ");
  scanf("%f",&p);
  printf("\nInterest : ");
  scanf("%f",&ir);
  printf("\nTime (in year) : ");
  scanf("%f",&t);
  in=p*ir*t/100;
  printf("\nYou will get interest : Rs. %.2f",in);

}

  

Output
Principle amount : 5000
Interest : 10
Time (in year) : 3
You will get interest :Rs. 1500.00

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Print the Multiplication Table

#include<stdio.h>

void main()
{
 int x=1,num,res;
 printf("Enter a Number : ");
 scanf("%d",&num);
 while(x<=10)
 {
   res=num*x;
   printf("\n%d x %d = %d",num,x,res);
   x++;
 } 
}

  

Output
Enter a Number : 6
6 x 1 = 6
6 x 2 = 12
6 x 3 = 18
6 x 4 = 24
6 x 5 = 30
6 x 6 = 36
6 x 7 = 42
6 x 8 = 48
6 x 9 = 54
6 x 10 = 60

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Find the Greatest of Three Numbers

#include<stdio.h>
void main()
{
 int x,y,z;
 printf("Enter values of x, y and z : ");
 scanf("%d,%d,%d",&x,&y,&z);
 if(x>=y && x>=z)
   printf("\n%d is greatest",x);
 else if(y>=z)
   printf("\n%d is greatest",y);
 else
   printf("\n%d is greatest",z);
}

  

Output
Enter values of x, y and z : 4, 9, 2
9 is greatest

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Find Greatest number using Conditional Operator

#include<stdio.h>

void main()
{
 int n1,n2,n3;
 printf("Enter Numbers (n1 n2 n3) : ");
 scanf("%d %d %d",&n1,&n2,&n3);
 printf("Greatest= %d",n1>n2?n1>n3?n1:n3:n2>n3?n2:n3);
}

  

Output
Enter Numbers (n1 n2 n3) : 4  9  2
Greatest= 9

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Print 1 to 10 using For Loop

#include<stdio.h>
int main()
{
 int x=1;
 
 for(x=1; x<=10; x++)
    printf("\n%d",x);

 return 0;
}

  

Output
  1
  2
  3
  4
  5
  6
  7
  8
  9
  10

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Print 1 to 10 using While Loop

#include<stdio.h>
#include<conio.h>
int main()
{
 int x=1;
 while(x<=10)
 {
   printf("\n%d",x);
   x++;
 }
 getch();
 return 0;
}

  

Output
  1
  2
  3
  4
  5
  6
  7
  8
  9
  10

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Swap two Numbers

#include<stdio.h>

void main()
{
 int x,y,z;
 printf("Enter Values x and y : ");
 scanf("%d%d",&x,&y);
 z=y;
 y=x;
 x=z;
 printf("\nValue of x=%d",x);
 printf("\nValue of y=%d",y);
}

  

Output
Enter Values x and y : 5 4
Value of x= 4
Value of y= 5

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Sum of two numbers

#include<stdio.h>

void main()
{
 int a,b,sum;
 printf("Enter Numbers a and b : ");
 scanf("%d %d",&a,&b);
 sum=a+b;
 printf("\n%d + %d = %d",a,b,sum);
}

   

Output
Enter Numbers a and b :5 4
5 + 4 = 9 

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Print "Hello, World!" without semicolon

Solution: 1

#include<stdio.h>
void main(){
    if(printf("Hello world")){
    }
}

Solution: 2

#include<stdio.h>
void main(){
    while(!printf("Hello world")){
    }
}

Solution: 3

#include<stdio.h>
void main(){
    switch(printf("Hello world")){
    }
}

  

Output
Hello, World!

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Print "Hello, World!"

#include<stdio.h>

int main()
{
 printf("\nHello, World! ");
 return 0;
}

  

Output
Hello, World!

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Accept Name, Age from User and Print them on the Screen

#include<stdio.h>
int main()
{
 char nam[20];
 int age;
 printf("Enter Your Name : ");
 scanf("%s",&nam); 
 printf("Enter Your Age : ");
 scanf("%d",&age);
 printf("\nYour Name is %s ",nam); 
 printf("and your age is %d",age);
 return 0;
}

  

Output
Enter Your Name : Sumit
Enter Your Age : 21

Your Name is Sumit and your age is 21

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Normalization

Normalization is the process of splitting relations into well-structured relations that allow users to insert, delete, and update tuples without introducing database inconsistencies. The focus of normalization is to reduce redundant data to the minimum.  Normalization is also called “Bottom-up approach”, because this technique requires very minute details like every participating attribute and how it is dependant on the key attributes, is crucial. If you add new attributes after normalization, it may change the normal form itself.

Through Normalization, the collection of data in a single table is distributed into multiple tables with specific relation. Without normalization many problems can occur when trying to load an integrated conceptual model into the DBMS. These problems arise from relations that are generated directly from user views are called anomalies. There are three types of anomalies:

RollNo	StudentName	CourseNo	CourseName	Instructor
120	SKL	CS-75	DBMS	SB
89	KBC	CS-75	DBMS	SB
25	ABC	CS-75	DBMS	SB
48	XYZ	CS-30	CA	MC
57	TKC	CS-80	OS	JB
120	SKL	CS-80	OS	JB
68		CS-90	FLAT	LD

Update anomaly: An update anomaly is a data inconsistency that results from data redundancy and a partial update. For example, if there is a change in the name of the instructor for CS-75, we need to make change for all the rows. If we forget to update a single row the database will show two instructor names for the Course CS-75.

Deletion anomaly:  This means loss of useful information. In some cases it may occur that some useful data is lost. For example, consider the row for RollNo 48. XYZ is the only student who has opted for CS-30. If XYZ leaves the institute and the data is deleted for XYZ, the associated data for Course will also be deleted.

Insertion anomaly: An insertion anomaly is the inability to add data to the database due to absence of other data. For example, assume that the above table is defined so that null values are not allowed. If a course is added but not immediately students opt for the course, then this course could not be entered into the database (68). This results in database inconsistencies due to omission.

In database tables are normalized for the following reasons:

To allow data retrieval at an optimal speed

Data maintenance through update, insertion and deletion.

To reduce the need to restructure tables for new applications

A functional dependency is a constraint between two sets of attributes in a relation from a database. It occurs when an attribute in a relation uniquely determines another attribute.

In a given relation R, X and Y are attributes. Attribute Y is functionally dependent on attribute X if each value of X determines EXACTLY ONE value of Y, which is represented as X → Y (X can be composite in nature).

We say here “X determines Y” or “y is functionally dependent on x”.

 X→Y does not imply Y→X.

If the value of an attribute “Marks” is known then the value of an attribute “Grade” is determined since Marks→ Grade

RollNo	StudentName	Marks	Grade
59	ABC	90	A
96	XYZ	70	B

Types of functional dependencies:

1.       Full Functional dependency: X and Y are attributes. X Functionally determines Y (Note: Subset of X should not functionally determine Y)

2.       Partial Functional dependency:  X and Y are attributes. Attribute Y is partially dependent on the attribute X only if it is dependent on a sub-set of attribute X.

3.       Transitive dependency: X Y and Z are three attributes. X -> Y, Y-> Z => X -> Z

Now consider the following Relation

REPORT (STUDENT#, COURSE#, CourseName, IName, Room#, Marks, Grade)

• STUDENT# - Student Number

• COURSE# - Course Number

• CourseName - Course Name

• IName - Name of the Instructor who delivered the course

• Room# - Room number which is assigned to respective Instructor

• Marks - Scored in Course COURSE# by Student STUDENT#

• Grade - obtained by Student STUDENT# in Course COURSE#

The Functional Dependencies are:

• STUDENT# COURSE# → Marks

• COURSE# → CourseName,

• COURSE# → IName (Assuming one course is taught by one and only one

Instructor)

• IName → Room# (Assuming each Instructor has his/her own and non-shared room)

• Marks → Grade

           In above example Marks is fully functionally dependent on STUDENT# COURSE# and not on subset of STUDENT# COURSE#. This means Marks cannot be determined either by STUDENT# OR COURSE# alone. It can be determined only using STUDENT# AND COURSE# together. Hence Marks is fully functionally dependent on STUDENT# COURSE#. CourseName is not fully functionally dependent on STUDENT# COURSE# because subset of STUDENT# COURSE# i.e only COURSE# determines the CourseName and STUDENT# does not have any role in deciding CourseName. Hence CourseName is not fully functionally dependent on STUDENT# COURSE#.

Now CourseName, IName, Room# are partially dependent on composite attributes STUDENT# COURSE# because COURSE# alone defines the CourseName, IName, Room#.

Again, Room# depends on IName and in turn IName depends on COURSE#. Hence Room# transitively depends on COURSE#. Similarly Grade depends on Marks, in turn Marks depends on STUDENT# COURSE# hence Grade depends Fully transitively on STUDENT# COURSE#.

Now consider this table:

Student Data		Course Details				Result
STUDENT#	StudentName	COURSE#	CourseName	IName	Room#	Marks	Grade

First normal form (1NF)

A relation schema is in 1NF if:

·         If and only if all the attributes of the relation R are atomic in nature.

·         Atomic: the smallest level to which data may be broken down and remain meaningful.

In relational database design it is not practically possible to have a table which is not in 1NF.

STUDENT#	StudentName	COURSE#	CourseName	IName	Room#	Marks	Grade

Second normal form (2NF)

A Relation is said to be in Second Normal Form if and only if:

·         It is in the First normal form, and

·         No partial dependency exists between non-key attributes and key attributes.

An attribute of a relation R that belongs to any key of R is said to be a prime attribute and that which doesn’t is a non-prime attribute To make a table 2NF compliant, we have to remove all the partial dependencies

Note: - All partial dependencies are eliminated

STUDENT#	StudentName

COURSE#	CourseName	IName	Room#

STUDENT#	COURSE#	Marks	Grade

Third normal form (3 NF)

A relation R is said to be in the Third Normal Form (3NF) if and only if:

·         It is in 2NF and

·         No transitive dependency exists between non-key attributes andkey attributes.

• STUDENT# and COURSE# are the key attributes.

• All other attributes, except grade are nonpartially, non-transitively dependent on key attributes.

STUDENT#	StudentName

COURSE#	CourseName	IName	Room#

STUDENT#	COURSE#	Marks

Marks (Lower Bound)	Marks (Upper Bound)	Grade

Boyce-Codd Normal form (BCNF)

A relation is said to be in Boyce Codd Normal Form (BCNF)

·         If and only if all the determinants are candidate keys.

BCNF relation is a strong 3NF, but not every 3NF relation is BCNF.

Consider the table:

Student#	EmailID	Course#	Marks
ABC	abc@sumitkar.in	CS-75	98

Candidate Keys for the relation are [C#, S#] and [C#, Email]. Since Course # is overlapping, it is referred as Overlapping Candidate Key. Valid Functional Dependencies are:

S#  → EmailID

EmailID → S#

S#, C# → Marks

C#, EmailID →Marks

Student#	EmailID
ABC	abc@sumitkar.in

Student#	Course#	Marks
ABC	CS-75	98

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