C PROGRAMMING



Write a C Program to calculates the area (floating point number with two decimal places) of a Circle given it’s radius .

 Write a C program to check if a given Number is zero or Positive or Negative Using if..else statement.

Write a C program to check whether a given number (integer) is Even or Odd.

Write a C Program to find the Largest Number (integer)  among Three Numbers (integers) using IF and Logical && operator.



 Write a C Program to Find the Smallest Number  among Three Numbers (integer values) using Nested IF-Else statement.



 Write a C program to find whether a triangle can be formed or not. If not display “This Triangle is NOT possible.” If the triangle can be formed then check the triangle formed is equilateral, isosceles, scalene or a right-angled triangle.







Write a program to find the factorial of a given number using while loop.

Sample Test Cases                Input                                   Output

Test Case 1                              5                        The Factorial of 5 is : 120

Test Case 2                             10                       The Factorial of 10 is : 3628800

Test Case 3                             7                         The Factorial of 7 is : 5040

Test Case 4                            11                        The Factorial of 11 is : 39916800 


#include<stdio.h>
void main()
{
    int n;
    long int fact;  /* n is the number whose factorial we have to find and fact is the factorial */
    scanf("%d",&n);  /* The value of n is taken from test cases */

int i=1;
fact = 1;
while(i<=n)
    {
        fact*=i;
        i++;
    }
    printf("The Factorial of %d is : %ld",n,fact);
}





Write a Program to find the sum of all even numbers from 1 to N where the value of N is taken as input. 

Sample Test Cases                                  Input                                  Output

Test Case 1                                               10                                   Sum = 30

Test Case 2                                               30                                   Sum = 240

Test Case 3                                               15                                   Sum = 56

Test Case 4                                               25                                   Sum = 156 


#include<stdio.h>   
void main()
{
int N, sum=0; 
scanf("%d", &N); /* The value of N is taken from the test cases */

int i;
for (i = 1; i <= N; i++)
{
if (i % 2 == 0)
sum = sum + i;
}
printf("Sum = %d", sum);
}





 Write a program to find whether a given character is a Vowel or consonant. A character is taken as input.

Sample Test Cases                            Input                                 Output

Test Case 1                                          C                            C is a consonant.

Test Case 2                                          o                             o is a vowel.

Test Case 3                                          I                              I is a vowel.

Test Case 4                                          k                             k is a consonant. 


#include<stdio.h> 
int main()
{
    char ch;
    scanf("%c",&ch);  /* It reads a character from the input cases and store it in ch */

int Lower_case, Upper_case;
Lower_case = (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u');

   Upper_case = (ch == 'A' || ch == 'E' || ch == 'I' || ch == 'O' || ch == 'U');

    if (Lower_case|| Upper_case)
        printf("%c is a vowel.", ch);
    else
        printf("%c is a consonant.", ch);
    return 0;
}





Write a C program to check whether a given number (N) is a perfect number or not?

Sample Test Cases                                     Input                                  Output

Test Case 1                                                   6                          6 is a perfect number.

Test Case 2                                                  87                         87 is not a perfect number.

Test Case 3                                                  8000                     8000 is not a perfect number.

Test Case 4                                                  8128                     8128 is a perfect number. 




#include <stdio.h>
int main()
{
    int N; 
    scanf("%d",&N); /* An integer number is taken as input from test case */

int i, sum=0;
    for(i=1; i<N;i++)
    {
        if(N%i==0)
            sum+=i;
    }
 
    if(sum==N)
        printf("%d is a perfect number.",N);
    else
        printf("%d is not a perfect number.",N);
}









 Write a C program to count total number of digits of an Integer number (N). 

Sample Test Cases                            Input                                            Output

Test Case 1                                        3456                   The number 3456 contains 4 digits.

Test Case 2                                        30000                 The number 30000 contains 5 digits.

Test Case 3                                           57                    The number 57 contains 2 digits.

Test Case 4                                           909                  The number 909 contains 3 digits. 


#include<stdio.h> 
 int main()
{
    int N; 
    scanf("%d",&N); /*The number is accepted from the test case data*/

int temp, count; 
count=0;
    temp=N;
    while(temp>0)
 {
        count++;
        temp/=10;
    }
     printf("The number %d contains %d digits.",N,count);
}





Write a C program to check whether the given number(N) can be expressed as Power of Two (2) or not. For example 8 can be expressed as 2^3. 

Sample Test Cases                        Input                                         Output

Test Case 1                                       8              8 is a number that can be expressed as power of 2.

Test Case 2                                      46            46 cannot be expressed as power of 2.

Test Case 3                                    6572        6572 cannot be expressed as power of 2.

Test Case 4                                    1024        1024 is a number that can be expressed as power of 2. 


include<stdio.h> 
int main()
{
    int N;
    scanf("%d",&N); /* The value of N is taken from the test case data */
 
int temp, flag;
    temp=N;
    flag=0;
   
    while(temp!=1)
    {
        if(temp%2!=0){
            flag=1;
            break;
        }
        temp=temp/2;
    }
  
    if(flag==0)
        printf("%d is a number that can be expressed as power of 2.",N);
    else
        printf("%d cannot be expressed as power of 2.",N);
}





  Write a C program to find sum of following series where the value of N is taken as input  1+ 1/2 + 1/3 + 1/4 + 1/5 + .. 1/N 

Sample Test Cases                         Input                                        Output

Test Case 1                                       6                            Sum of the series is: 2.45

Test Case 2                                      50                           Sum of the series is: 4.50

Test Case 3                                    100                           Sum of the series is: 5.19

Test Case 4                                      20                           Sum of the series is: 3.60 


#include<stdio.h>
int main()
{
int N;
float sum = 0.0;
scanf("%d",&N); /*Read the value of N from test cases provided*/

int i;
for(i=1;i<=N;i++)
sum = sum + ((float)1/(float)i);
printf("Sum of the series is: %.2f\n",sum);
}





 Write a C program to print the following Pyramid pattern upto Nth row. Where N (number of rows to be printed) is taken as input. 


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



#include<stdio.h>

int main()

{

int N;

scanf("%d", &N); /*The value of N is taken as input from the test case */

int i,j;

for(i=N; i>0; i--)

  {

  for(j=0;j<i;j++)

    {

    printf("*");

    }

  printf("\n");

  }

} 

Write a C Program to find Largest Element of an Integer Array.

Sample Test Cases                       Input                                                    Output

Test Case 1                         5 10 50 40 30 20                              Largest element = 50

Test Case 2                         7 100 50 60 70 90 30 40                  Largest element = 100

Test Case 3                         4 -400 -800 -700 -50                        Largest element = -50

Test Case 4                         7 60 70 200 12 40 -90 60                 Largest element = 200 


#include<stdio.h> 

int main()
{
    int i, n, largest;
    int arr[100];

    scanf("%d", &n); /*Accepts total number of elements from the test data */

 for(i = 0; i < n; ++i)
    {
       scanf("%d", &arr[i]); /* Accepts the array element from test data */
    }
largest = arr[0];
for(i = 1; i < n; ++i)
    {
           if(largest < arr[i])
           largest = arr[i];
    }
    printf("Largest element = %d", largest);

    return 0;
}





Write a C Program to print the array elements in reverse order (Not reverse sorted order. Just the last element will become first element, second last element will become second element and so on)

Sample Test Cases                                   Input                                                 Output

Test Case 1                                            5 1 2 3 4 5                                         5  4  3  2  1

Test Case 2                                            4 45 65 35 25                                    25 35 65 45

Test Case 3                                            5 10 20 30 40 50                               50 40 30 20 10

Test Case 4                                            6 41 42 43 44 45 46                          46 45 44 43 42 41 


#include<stdio.h>
 
int main() {
   int arr[20], i, n;
 
   scanf("%d", &n); /* Accepts the number of elements in the array */

  for (i = 0; i < n; i++) 
     scanf("%d", &arr[i]); /*Accepts the elements of the array */
   
int j, temp;  
j = i - 1;   // last Element of the array
i = 0;       // first element of the array
 
   while (i < j) {
      temp = arr[i];
      arr[i] = arr[j];
      arr[j] = temp;
      i++;             
      j--;        
   }
 
for (i = 0; i < n; i++) {
      printf("%d\n", arr[i]); // For printing the array elements 
   }
 
   return (0);
}





Write a C program to read Two One Dimensional Arrays of same data type (integer type) and merge them into another One Dimensional Array of same type. 






InputOutput


Test Case 1
3
10
20
30
4
40
50
60
70

10
20
30
40
50
60
70



Test Case 2
4
9
7
6
5
2
30
50

9
7
6
5
30
50



Test Case 3
3
15
45
25
3
60
70
80

15
45
25
60
70
80



Test Case 4
4
90
80
10
30
2
25
75

90
80
10
30
25
75









#include<stdio.h>
int main() 
{
   int arr1[20], arr2[20], array_new[40], n1, n2, size, i;
 /*n1 size of first array (i.e. arr1[]), n2 size of second array(i.e. arr2[]), 
   size is the total size of the new array (array_new[]) */
 
   scanf("%d", &n1); //Get the size of first array from test data and store it in n1.
   
   for (i = 0; i < n1; i++)
      scanf("%d", &arr1[i]); //Accepts the values for first array 
 
   scanf("%d", &n2); //Get the size of second array from test data and store it in n2.
   
   for (i = 0; i < n2; i++)
      scanf("%d", &arr2[i]); //Accepts the values for second array 
//Marge two arrays
int j;
for (i=0;i<n1;++i)
array_new[i]=arr1[i];

size =  n1 + n2;

for(i=0, j=n1; j<size && i<n2; ++i, ++j)
array_new[j] = arr2[i];
//Printing after merging

for (i = 0; i < size; i++) {
      printf("%d\n", array_new[i]);
   }

}






Write a C Program to delete duplicate elements from an array of integers.   






InputOutput


Test Case 1
5
50
60
30
20
30

50
60
30
20



Test Case 2
6
40
20
50
30
20
10

40
20
50
30
10



Test Case 3
6
50
6
7
7
2
7

50
6
7
2



Test Case 4
7
2
4
2
6
4
2
4

2
4
6









#include<stdio.h>
 
int main() 
{
   int array[50], i, size;
 
   scanf("%d", &size); /*Accepts the size of array from test case data */

   for (i = 0; i < size; i++)
   scanf("%d", &array[i]); /* Read the array elements from the test case data */ 
   int j, k;
   for (i = 0; i < size; i++) {
      for (j = i + 1; j < size;) {
         if (array[j] == array[i]) {
            for (k = j; k < size; k++) {
               array[k] = array[k + 1];
            }
            size--;
         } else
            j++;
      }
   }
 
for (i = 0; i < size; i++) {
      printf("%d\n", array[i]);
   }

}





C Program to delete an element from a specified location of an Array starting from array[0] as the 1st position, array[1] as second position and so on.  






InputOutput


Test Case 1
5
10
20
30
40
50
4

10
20
30
50



Test Case 2
6
600
500
400
300
300
200
4

600
500
400
300
200



Test Case 3
4
50
60
70
80
1

60
70
80



Test Case 4
5
9
10
11
1
20
5

9
10
11
1





#include<stdio.h>
int main() {
int array[30], num, i, pos;  
/*num is number of elements in the array and loc is the position of the array to be deleted. 
starting from arr[0] as the first(1) position and so on. */

   scanf("%d", &num); /*Accepts the size of array from test case data */
 
   for (i = 0; i < num; i++) {
      scanf("%d", &array[i]);  /* Read the array elements from the test case data */
   }
 
   scanf("%d", &pos); /* Accepts the Position of the element to be deleted */
/* loop for the deletion  */
   while (pos < num) {
      array[pos - 1] = array[pos];
      pos++;
   }
   num--; /* No of elements reduced by 1 as 1 element is deleted */
/* Printing the array after deletion */

for (i = 0; i < num; i++)
      printf("%d\n", array[i]);
 
   return (0);
}





Write a C Program to Count Number of Uppercase and Lowercase Letters in a given string. The string may be a word or a sentence.  






             InputOutput


Test         Case 1
Online NPTEL Course.

Uppercase Letters : 7
Lowercase Letters : 10


Test Case 2
National Programme on Technology Enhanced Learning is a joint initiative of the IITs and IISc.

Uppercase Letters : 11
Lowercase Letters : 68


Test Case 3
Problem Solving through Programming in C.

Uppercase Letters : 4
Lowercase Letters : 31


Test Case 4
AICTE Approved FDP Course

Uppercase Letters : 10
Lowercase Letters : 12








#include<stdio.h>
int main() {
   int upper = 0, lower = 0;
   char ch[100];
   scanf(" %[^\n]s", ch);  /*A word or a sentence is accepted from test case data */

/* Complete the remaining part of the code to store number of uppercase letters
in the variable upper and lowercase letters in variable lower.
The print part of already written. You can declare any variable if necessary */
int i = 0;
   while (ch[i] != '\0') {
      if (ch[i] >= 'A' && ch[i] <= 'Z')
         upper++;
      if (ch[i] >= 'a' && ch[i] <= 'z')
         lower++;
      i++;
   }
   printf("Uppercase Letters : %d\n", upper); /*prints number of uppercase letters */
   printf("Lowercase Letters : %d", lower); /*prints number of lowercase letters */
 
   return (0);
}



Write a C program to find the sum of all elements of each row of a matrix.  






InputOutput


Test Case 1
3
3
1
1
1
2
2
2
3
3
3

3
6
9



Test Case 2
2
3
1
2
3
4
5
6

6
15



Test Case 3
3
2
4
4
5
5
6
6

8
10
12



Test Case 4
3
4
1
-1
2
-2
5
-5
7
-7
8
-8
6
-6

0
0
0









#include<stdio.h> 
int main()
{
    int matrix[20][20];
    int i,j,r,c;

    scanf("%d",&r); //Accepts number of rows
    scanf("%d",&c); //Accepts number of columns 

    for(i=0;i< r;i++) //Accepts the matrix elements from the test case data
    {
        for(j=0;j< c;j++)
        {
            scanf("%d",&matrix[i][j]); 
        }
    } 
 int sum;
    for(i=0;i< r;i++)
    {
        sum=0;      
        for(j=0;j< c;j++)
        {
          //  printf("%d\t",matrix[i][j]);   
            sum     +=  matrix[i][j];
        }
        printf("%d\n",sum);
    }
 
}





Write a C program to find subtraction of two matrices i.e. matrix_A - matrix_B=matrix_C. 






InputOutput


Test Case 1
3
3
2
3
5
4
5
6
6
5
7
1
5
2
2
3
4
3
3
4

1 -2 3 
2 2 2 
3 2 3 



Test Case 2
3
4
5
6
7
8
3
2
5
6
1
3
9
5
2
9
3
1
2
5
1
2
2
3
4
1

3 -3 4 7 
1 -3 4 4 
-1 0 5 4 





#include <stdio.h>
int main()
{
 int matrix_A[20][20], matrix_B[20][20], matrix_C[20][20];
    int i,j,row,col;
    scanf("%d",&row); //Accepts number of rows
    scanf("%d",&col); //Accepts number of colums 
 
    /* Elements of first matrix are accepted from test data */
    for(i=0; i<row; i++)
    {
        for(j=0; j<col; j++)
        {
            scanf("%d", &matrix_A[i][j]);
        }
    }

     /* Elements of second matrix are accepted from test data */
    
    for(i=0; i<row; i++)
    {
        for(j=0; j<col; j++)
        {
            scanf("%d", &matrix_B[i][j]);
        }
    }
 /* 
    * Subtract both matrices and store the result in matrix C
  */
   for(i=0; i<row; i++)
    {
        for(j=0; j<col; j++)
        {
    
            matrix_C[i][j] = matrix_A[i][j] - matrix_B[i][j];
        }
    }

     for(i=0; i<row; i++)
    {
        for(j=0; j<col; j++)
        {
            printf("%d ", matrix_C[i][j]);
        }
        printf("\n");
    }

    return 0;
}




Write a C program to print lower triangle of a square matrix.   






InputOutput


Test Case 1
3
1
2
3
1
2
3
1
2
3

1 0 0 
1 2 0 
1 2 3 



Test Case 2
4
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4

1 0 0 0 
2 2 0 0 
3 3 3 0 
4 4 4 4 





#include <stdio.h>
int main()
{
int matrix[20][20];
int i,j,r;
   
scanf("%d", &r); //Accepts number of rows or columns

    for(i=0;i< r;i++) //Accepts the matrix elements from the test case data
    {
        for(j=0;j<r; j++)
        {
            scanf("%d",&matrix[i][j]);
        }
    }
 for(i=0; i<r; i++) {
  for(j=0; j<r; j++) {
   if(i>=j)
    printf("%d ", matrix[i][j]);
   else
    printf("%d ", 0);
  }
  printf("\n");
 }
return 0;
}





 Write a C program to print Largest and Smallest Word from a given sentence. If there are two or more words of same length then the first one is considered.   






InputOutput


Test Case 1
Problem Solving through Programming in C.

Largest Word is: Programming
Smallest word is: C



Test Case 2
NPTEL is a joint initiative of the IITs and IISc.

Largest Word is: initiative
Smallest word is: a



Test Case 3
AICTE Approved FDP Course.

Largest Word is: Approved
Smallest word is: FDP





#include<stdio.h>
#include<string.h>
int main()
{
    char str[100]={0},substr[100][100]={0}; 
//str[100] is for storing the sentence and substr[50][50] is for storing each word.
    
scanf("%[^\n]s", str); //Accepts the sentence from the test case data.

int i=0,j=0,k=0,a,minIndex=0,maxIndex=0,max=0,min=0;
char c; 
while(str[k]!='\0')  //for splitting sentence into words 
    {
        j=0;
        while(str[k]!=' '&&str[k]!='\0' && str[k]!='.')
        {
            substr[i][j]=str[k];
            k++;
            j++;
        }
        substr[i][j]='\0';
        i++;
        if(str[k]!='\0')
        {
            k++;
        }        
    }
    int len=i;
    max=strlen(substr[0]);
    min=strlen(substr[0]);

    //After splitting getting length of string and finding its index having max length and index having min length
    for(i=0;i<len;i++)
    {
       a=strlen(substr[i]);
       if(a>max)
        {
            max=a;
            maxIndex=i;
        }
        if(a<min)
        {
            min=a;
            minIndex=i;
        }
    }    
  printf("Largest Word is: %s\nSmallest word is: %s\n",substr[maxIndex],substr[minIndex]);
  return 0;  
}




Write a C Program to find GCD of 4 given numbers using recursive function  






InputOutput


Test Case 1
100
30
50
70

The GCD is 10


Test Case 2
49
56
147
21

The GCD is 7


Test Case 3
50
455
60
200

The GCD is 5


Test Case 4
67
89
45
41

The GCD is 1









#include<stdio.h>
int gcd(int, int); //You have to write this function which calculates the GCD. 
  
int main()
{
   int a, b, c, d, result;
   scanf("%d %d %d %d", &a, &b, &c, &d); /* Takes 4 number as input from the test data */
   result = gcd(gcd(a, b), gcd(c,d));
   printf("The GCD is %d", result);
}

//Complete the rest of the program to calculate GCD
int gcd(int x, int y)
{
   while (x != y)
   {
      if (x > y)
       {
           return gcd(x - y, y);
       }
       else
       {
       return gcd(x, y - x);
       }
    }
    return x;
}





Write a C Program to find power of a given number using recursion.  






InputOutput


Test Case 1
5
2

5^2 is 25


Test Case 2
10
4

10^4 is 10000


Test Case 3
5
3

5^3 is 125


Test Case 4
16
3

16^3 is 4096





#include<stdio.h>   
long power(int, int);
int main()
{
int pow, num;
long result;

scanf("%d", &num); //The number taken as input from test case data 

scanf("%d", &pow); //The power is taken from the test case 
result = power(num, pow);
printf("%d^%d is %ld", num, pow, result);
return 0;
}
long power(int num, int pow)
{
if (pow)
{
return (num * power(num, pow - 1));
}
return 1;
}





 Write a C Program to print Binary Equivalent of an Integer using Recursion.






InputOutput


Test Case 1
10

The binary equivalent of 10 is 1010



Test Case 2
257

The binary equivalent of 257 is 100000001



Test Case 3
30

The binary equivalent of 30 is 11110



Test Case 4
111
The binary equivalent of 111 is 1101111





#include<stdio.h> 
int binary_conversion(int); //function to convert binary to decimal number
int main()
  {
  int num, bin;  //num is the decimal number and bin is the binary equivalent for the number

  scanf("%d", &num); //The decimal number is taken from the test case data
  bin = binary_conversion(num); //binary number is stored in variable bin
  printf("The binary equivalent of %d is %d\n", num, bin);
  }
int binary_conversion(int num)
  { 
     if (num == 0)
        {
            return 0;
        }
        else
        {
            return (num % 2) + 10 * binary_conversion(num / 2);
        }
  }





Write a C Program to reverse a given word using function. 






InputOutput


Test Case 1
computer

The string after reversing is: retupmoc


Test Case 2
NPTEL

The string after reversing is: LETPN


Test Case 3
India

The string after reversing is: aidnI


Test Case 4
Kharagpur

The string after reversing is: rupgarahK








#include<stdio.h>
#include<string.h>

void reverse(char[], int, int);
    int main()
    {
        char str1[20];
        scanf("%s", str1); //The string is taken as input form the test data. 
     
        int size;
 size = strlen(str1);
        reverse(str1, 0, size - 1);
        printf("The string after reversing is: %s", str1);  
        return 0;
    }

    void reverse(char str1[], int index, int size)
    {
        char temp;
        temp = str1[index];
        str1[index] = str1[size - index];
        str1[size - index] = temp;
        if (index == size / 2)
        {
            return;
        }
        reverse(str1, index + 1, size);
    }




Write a C program to print a triangle of prime numbers upto given number of lines of the triangle.  






InputOutput


Test Case 1
3

2 
3 5 
7 11 13 



Test Case 2
5

2 
3 5 
7 11 13 
17 19 23 29 
31 37 41 43 47 



Test Case 3
4

2 
3 5 
7 11 13 
17 19 23 29 



Test Case 4
6

2 
3 5 
7 11 13 
17 19 23 29 
31 37 41 43 47 
53 59 61 67 71 73 







#include<stdio.h>
 
int prime(int num); //Function to find whether the number is prime or not.
int main() {
   int lines;
   scanf("%d", &lines); //Number of lines of the triangle is taken from test data.
int i, j;
int num = 2;
for (i = 0; i < lines; i++) {
      
   for (j = 0; j <= i; j++) {
      while (!prime(num)) 
      {
         num++;
      }
         printf("%d\t", num++);  
      }
      printf("\n");
   }
   return (0);
}
 
int prime(int num) {
   int i, flag;
   for (i = 2; i < num; i++) {
      if (num % i != 0)
         flag = 1;
      else {
         flag = 0;
         break;
      }
   }
 
   if (flag == 1 || num == 2)
      return (1);
   else
      return (0);
}





Write a program to print all the locations at which a particular element(taken as input) is found in a list and also print the total number of times it occurs in the list. The location starts from 1.






InputOutput


Test Case 1
4
5
6
5
7
5

5 is present at location 1.
5 is present at location 3.
5 is present 2 times in the array.



Test Case 2
5
67
80
45
97
100
50

50 is not present in the array.



Test Case 3
7
30
50
90
30
70
30
30
30

30 is present at location 1.
30 is present at location 4.
30 is present at location 6.
30 is present at location 7.
30 is present 4 times in the array.



Test Case 4
4
50
60
20
10
80

80 is not present in the array.









#include<stdio.h> 
int main()
{
   int array[100], search, n, count = 0;
   //"search" is the key element to search and 'n' is the total number of element of the array
   // "count" is to store total number of elements

 scanf("%d", &n); //Number of elements is taken from test case

 int c;
   for (c = 0; c < n; c++)
      scanf("%d", &array[c]);

   scanf("%d", &search); // The element to search is taken from test case

  for (c = 0; c < n; c++)
    {
      if (array[c] == search)
      {
         printf("%d is present at location %d.\n", search, c+1);
         count++;
      }
    }
   if (count == 0)
      printf("%d is not present in the array.\n", search);
   else
      printf("%d is present %d times in the array.\n", search, count);

   return 0;
}



Write a C program to search a given element from a 1D array and display the position at which it is found by using linear search function.  






InputOutput


Test Case 1
5
78
90
34
54
98
90

90 is present at location 2.



Test Case 2
6
30
40
50
20
90
60
90

90 is present at location 5.



Test Case 3
4
45
65
85
25
95

95 is not present in the array.



Test Case 4
5
6
9
5
4
7
6

6 is present at location 1.













#include<stdio.h> 
int linear_search(int[], int, int);
int main()
{
   int array[100], search, c, n, position;
   /* search - element to search, c - counter, n - number of elements in array,
   position - The position in which the element is first found in the list. */

    scanf("%d", &n); // Number of elements in the array is read from the test case data

    for (c = 0; c < n; c++)
    scanf("%d", &array[c]); //Elements of array is read from the test data

    scanf("%d", &search);  //Element to search is read from the test case data

   position = linear_search(array, n, search);

   if (position == -1)
      printf("%d is not present in the array.\n", search);
   else
      printf("%d is present at location %d.\n", search, position+1);
   return 0;
}

int linear_search(int a[], int n, int find) {
   int c;
   for (c = 0 ;c < n ; c++ )
    {
      if (a[c] == find)
         return c;
    }
   return -1;
}





Write a C program to search a given number from a sorted 1D array and display the position at which it is found using binary search algorithm.  






InputOutput


Test Case 1
5
10
20
30
40
50
40

40 found at location 4.



Test Case 2
4
44
55
66
77
10

Not found! 10 isn't present in the list.



Test Case 3
6
1
2
3
4
5
6
2

2 found at location 2.



Test Case 4
7
40
50
60
70
80
90
100
100

100 found at location 7.





#include<stdio.h> 
int main()
{
 int c, n, search,
 array[100];
 scanf("%d",&n); //number of elements in the array

 for (c = 0; c < n; c++)
 scanf("%d",&array[c]);

 scanf("%d", &search); //The element to search is read from test case.

   int first, last, middle;
   first = 0;
   last = n - 1;
   middle = (first+last)/2;

   while (first <= last) {
      if (array[middle] < search)
         first = middle + 1;
      else if (array[middle] == search) {
         printf("%d found at location %d.\n", search, middle+1);
         break;
      }
      else
         last = middle - 1;

      middle = (first + last)/2;
      }
   if (first > last)
      printf("Not found! %d isn't present in the list.\n", search);

     return 0;
      }





 Write a C program to reverse an array by swapping the elements and without using any new array.  






InputOutput


Test Case 1
4
10
20
30
40

Reversed array elements are:
40
30
20
10



Test Case 2
5
50
60
40
30
20

Reversed array elements are:
20
30
40
60
50



Test Case 3
7
8
9
10
6
4
7
11

Reversed array elements are:
11
7
4
6
10
9
8









#include<stdio.h> 
int main() {
  int array[100], n, c;
  scanf("%d", &n); // n is number of elements in the array.
  for (c = 0; c < n; c++) {
    scanf("%d", &array[c]);
  }
 int temp, end;
   end = n - 1;
   for (c = 0; c < n/2; c++) {
    temp       = array[c];
    array[c]   = array[end];
    array[end] = temp;

    end--;
  }
 printf("Reversed array elements are:\n");

  for (c = 0; c < n; c++) {
    printf("%d\n", array[c]);
  }
  return 0;
}



Write a C program to marge two given sorted arrays (sorted in ascending order).  






InputOutput


Test Case 1
3
10
20
30
4
40
50
60
70

Sorted array:
10
20
30
40
50
60
70



Test Case 2
2
40
60
4
30
50
70
80

Sorted array:
30
40
50
60
70
80



Test Case 3
5
20
30
60
70
80
3
10
50
100

Sorted array:
10
20
30
50
60
70
80
100





#include<stdio.h> 
void merge(int a[], int m, int b[], int n, int sorted[]);
int main() 
{
int a[100], b[100], m, n, c, sorted[200];

/* a[100] and b[100] are the two given arrays and m and n are the their respective sizes. c is a counter and sorted[200] is the final sorted array */

scanf("%d", &m); //Number of elements in the first array

  for (c = 0; c < m; c++) 
    {
    scanf("%d", &a[c]); //Elements of first array is read
    }

  scanf("%d", &n); //Number of elements in second array

  for (c = 0; c < n; c++) 
  {
    scanf("%d", &b[c]); //Elements of second array is read
  }

 merge(a, m, b, n, sorted); 

//The merged function is called where the two arrays are merged and sorted.

  printf("Sorted array:\n");

  for (c = 0; c < m + n; c++) {
    printf("%d\n", sorted[c]);
  }

  return 0;
}
void merge(int a[], int m, int b[], int n, int sorted[]) {
  int i, j, k;

  j = k = 0;

  for (i = 0; i < m + n;) {
    if (j < m && k < n) {
      if (a[j] < b[k]) {
        sorted[i] = a[j];
        j++;
      }
      else {
        sorted[i] = b[k];
        k++;
      }
      i++;
    }
    else if (j == m) {
      for (; i < m + n;) {
        sorted[i] = b[k];
        k++;
        i++;
      }
    }
    else {
      for (; i < m + n;) {
        sorted[i] = a[j];
        j++;
        i++;
      }
    }
  }
}










 




InputOutput


Test Case 1
0.001


Root = 1.7197



Test Case 2
0.01

Root = 1.7266



Test Case 3
0.1

Root = 1.6875








#include<stdio.h>
float fun (float x); //Function fun returns the function value of f(x)
void bisection (float *x, float a, float b, int *itr); // This function computes the root of f(x) using bisection method

int main ()
{
    int itr = 0, maxmitr=10;
    float x, a=1.0, b=2.0, allerr, x1; // x is the value of root in each iteration, x1 is the final value of the root 
   // a and b are the initial range for calculating the root using bisection method
      
scanf("%f", &allerr);  // allerr is the allowable error taken from test case data 
    bisection (&x, a, b, &itr);

/* Use the printf statement as given below to print the root
printf("Root = %1.4f\n", x1); */

    do
    {
        if (fun(a)*fun(x) < 0)
            b=x;
        else
            a=x;
        bisection (&x1, a, b, &itr);
        if (((x1-x)<0 -="" allerr="" x1-x="">0 && (x1-x)< allerr))
        {
            printf("Root = %1.4f\n", x1);
            return 0;
        }
        x=x1;
    }
    while (itr < maxmitr);
    return 1;
}
float fun (float x)
{
    return (2*x*x*x - 3*x - 5);
}
void bisection (float *x, float a, float b, int *itr)
/* this function performs and prints the result of one iteration */
{
    *x=(a+b)/2;
    ++(*itr);
}























InputOutput


Test Case 1
2

Root = 3.465754


Test Case 2
5

Root = 1.908690



Test Case 3
10

Root = 1.893289



Test Case 4
4

Root = 2.059000









#include<stdio.h>
float f(float x);
float df (float x);

int main()
{
    int itr, maxmitr; // itr is the iteration number and maxitr is the maximum allowable iteration 
    float x0=1.0, x1; // x0 is the initial value and x1 is result 
    scanf("%d", &maxmitr); // Taken from the test cases 

float h;
    for (itr=1; itr<=maxmitr; itr++)
    {
        h=f(x0)/df(x0);
        x1=x0-h;
        x0=x1;
    }
    printf("Root = %8.6f\n", x1);
    return 0;
}
float f(float x)
{
    return x*x*x - 2*x  - 3;
}
float df (float x)
{
    return 3*x*x-2;
}







 Write a C program to sort a given 1D array using pointer in ascending order. 






InputOutput


Test Case 1
5
20
40
50
30
10

10
20
30
40
50



Test Case 2
8
90
70
30
-10
-40
20
100
50

-40
-10
20
30
50
70
90
100







#include <stdio.h>
int main()
{
    int a[100],i, n;
    scanf("%d",&n);  //Number of elements of the array is taken from the test case data 
   
   for (i=0; i<n; i++)
    {
        scanf("%d",a+i); // Input the array elements
    }
int j,t;
for (i=0; i<(n-1); i++) 
    {
        for (j=i+1; j<n; j++)
        {
            if (*(a+i)>*(a+j))
            {
                t=*(a+i);
                *(a+i)=*(a+j);
                *(a+j)=t;
            }
        }
    }
//   Printing sorted array in ascending order 
    for (i=0; i<n; i++)
    {
        printf("%d\n",*(a+i));
    }
    return 0;
   }






Write a C program to sort a 1D array using pointer by applying Bubble sort technique.  






InputOutput


Test Case 1
6
-10
90
30
20
-100
50

-100
-10
20
30
50
90



Test Case 2
7
70
40
80
10
200
30
60

10
30
40
60
70
80
200







#include<stdio.h>
void sort(int *a,int n);
int main()
{
    int a[20];
    int n,i; 
    scanf("%d",&n); // Enter number of elements to sort is taken from test case data
   
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]); // The elements of the array is taken from the test data
    }

sort(a, n); // Calling the sorting function

    //Printing the sorted array 
    for(i=0;i<n;i++)
    {
        printf("%d\n",a[i]);
    }
   return 0;
}

void sort(int *a, int n)
{
    int i,temp,j;
    for(i=1;i<n;i++)
    {
        for(j=0;j<n-i;j++)
        {
           if(*(a+j)>=*(a+j+1))
        {
            temp = *(a+j);
            *(a+j)= *(a+j+1);
            *(a+j+1)= temp;
        }
        }
    }
}









Write a C code to check if a 3 x 3 matrix is invertible. A matrix is not invertible if its determinant is 0.  






InputOutput


Test Case 1
1
2
3
4
5
6
7
8
9

The given matrix is not invertible


Test Case 2
1
0
5
2
1
6
3
4
0

The given matrix is invertible


Test Case 3
4
5
6
7
8
9
1
2
3

The given matrix is not invertible


Test Case 4
1
2
3
0
1
4
5
6
0

The given matrix is invertible








#include<stdio.h>
 int main()
{
  int a[3][3], i, j;
  long determinant;
 //  9 elements of matrix is taken as input from test data
  for(i = 0 ; i < 3; i++)
      for(j = 0; j < 3; j++)
           scanf("%d", &a[i][j]);
determinant = a[0][0] * ((a[1][1]*a[2][2]) - (a[2][1]*a[1][2])) -a[0][1] * (a[1][0]
   * a[2][2] - a[2][0] * a[1][2]) + a[0][2] * (a[1][0] * a[2][1] - a[2][0] * a[1][1]);
   if ( determinant == 0)
       printf("The given matrix is not invertible");
   else
       printf("The given matrix is invertible");
   return 0;
} 
























InputOutput


Test Case 1
25

The respective value of the variable v is: 56.42


Test Case 2
16

The respective value of the variable v is: 28.74


Test Case 3
20

The respective value of the variable v is: 41.62








<5 0="" b="b*(a-t[j]);" c="c*(t[i]-t[j]);" f="" for="" he="" i="" if="" is:="" j="" k="" of="" pre="" printf="" respective="" return="" the="" v="" value="" variable="">#include<stdio.h>
int main()
{
  float t[100]={10,15,18,22,30}, v[100]={22,26,35,48,68};
  float a; //Value of the t to find the respective value of v(t)
  scanf("%f", &a);  // This will be taken from test cases
int i,j;
float b, c, k =0;
for(i=0; i<5; i++)
   {
      b=1;
      c=1;
      for(j=0; j<5; j++)
        {
           if(j!=i)
             {
               b=b*(a-t[j]);
               c=c*(t[i]-t[j]);
              }
            }
        k=k+((b/c)*v[i]);
        }
printf("The respective value of the variable v is: %.2f", k);
  return 0;
}


<5 0="" b="b*(a-t[j]);" c="c*(t[i]-t[j]);" f="" for="" he="" i="" if="" is:="" j="" k="" of="" pre="" printf="" respective="" return="" the="" v="" value="" variable="">












InputOutput


Test Case 1
0
1

The integral is: 0.335000



Test Case 2
1
3

The integral is: 8.680000



Test Case 3
-2
1

The integral is: 3.045000





#include<stdio.h>
float func(float x);
int main()
{
  int n=10; //Taking n=10 sub intervals
  float a,b,integral; //integral is the integration result
  scanf("%f",&a); // initial limit taken from test case 
  scanf("%f",&b); // Final limit taken from test case

int i;
float h,x, sum=0;  
if(b>a)
  h=(b-a)/n;
  else
  h=-(b-a)/n;
  for(i=1;i<n;i++){
    x=a+i*h;
    sum=sum+func(x);
  }
  integral=(h/2)*(func(a)+func(b)+2*sum);
  printf("The integral is: %0.6f\n",integral);
  return 0;
}
float func(float x)
{
  return x*x;
}



























InputOutput


Test Case 1
0.9

y=1.4625


Test Case 2
1.2

y=-0.3062


Test Case 3
0.6

y=3.2313


Test Case 4
1

y=-0.3062





#include<stdio.h>
float func(float x,float y);
int main()
{
   float xn; 
    scanf("%f",&xn); // xn (i.e. the value of x) will be taken from test cases

//Use the printf statement as: printf("y=%.4f",y);  
float x0=0.3,y0=5,m1,m2,m3,m4,m,y,x,h=0.3;    
x=x0;
    y=y0;
    while(x<xn)
    {
        m1=func(x0,y0);
        m2=func((x0+h/2.0),(y0+m1*h/2.0));
        m3=func((x0+h/2.0),(y0+m2*h/2.0));
        m4=func((x0+h),(y0+m3*h));
        m=((m1+2*m2+2*m3+m4)/6);
        y=y+m*h;
        x=x+h;
    }
    
 printf("y=%.4f",y);  // Final output
 return 0;
}
float func(float x,float y)
{
    float m;
    m=(x*(x+1)-3*y*y*y)/10;
    return m;
}




  Write a C program to check whether the given input number is Prime number or not using recursion. So, the input is an integer and output should print whether the integer is prime or not. Note that you have to use recursion.






InputOutput


Test Case 1
13

13 is a prime number



Test Case 2
40

40 is not a prime number



Test Case 3
51

51 is not a prime number



Test Case 4
29

29 is a prime number









#include<stdio.h> 
int checkPrime(int, int); //Function to check prime or not 

int main()
{
    int num, check;
    scanf("%d", &num); //The number is taken from test case data
    check = checkPrime(num, num/2);
    if (check == 1)
    {
        printf("%d is a prime number\n", num);
    }
    else
    {
        printf("%d is not a prime number\n", num);
    }
    return 0;
}
int checkPrime(int num, int i)
{
    if (i == 1)
    {
        return 1;
    }
    else
    {
       if (num % i == 0)
       {
         return 0;
       }
       else
       {
         return checkPrime(num, i - 1);
       }
    }
}







Write a C program to reverse a word using Recursion. Input to the program is a string that is to be taken from the user and output is reverse of the input word. Note that you have to use recursion.  






InputOutput


Test Case 1
KHARAGPUR

The reversed string is : RUPGARAHK



Test Case 2
INDIAN

The reversed string is : NAIDNI



Test Case 3
NPTELONLINE

The reversed string is : ENILNOLETPN





#include<stdio.h> 
#define MAX 100
char *reverse(char[]);

int main()
{
    char str[MAX], *rev;
    //printf("Enter a String: ");
    scanf("%s", str);
    rev = reverse(str); //You have to write this function
    printf("The reversed string is : %s\n", rev);
    return 0;
}
char* reverse(char str[])
{
    static int i= 0;
    static char rev[MAX];
    if (*str)
    {
        reverse(str+1);
        rev[i++]= *str;
    }
    return rev;
}







Write a program in C to find the factorial of a given number using pointers.  






InputOutput


Test Case 1
5

The Factorial of 5 is : 120



Test Case 2
10

The Factorial of 10 is : 3628800



Test Case 3
15

The Factorial of 15 is : 1307674368000









#include<stdio.h> 
void findFact(int, long int*);
int main()
{
        long int fact; //factorial of the number
        int num1; 
  scanf("%d",&num1); //The number is taken from test data

         findFact(num1, &fact);
         printf("The Factorial of %d is : %ld\n",num1, fact);
         return 0;
        }
void findFact(int n, long int *f)
  {
        int i;

       *f =1;
       for(i=1;i<=n;i++)
       *f=*f*i;
       }







Write a C program to print the Record of the Student Merit wise Here a structure variable is defined which contains student rollno, name and score.






InputOutput


Test Case 1
3
1
Santanu
700
2
Aparna
550
3
Vivek
900

The Merit List is :
3  Vivek  900
1  Santanu  700
2  Aparna  550



Test Case 2
4
1
Pradip
900
2
Asutosh
600
3
Santosh
750
4
Sandip
500

The Merit List is :
1  Pradip  900
3  Santosh  750
2  Asutosh  600
4  Sandip  500





#include<stdio.h>
struct student
{
int rollno;
char name[20];
int score;
};
void main()
{
struct student s[20];
int i, n;

scanf("%d" ,&n); //No. of Students taken from test data
// Roll no., Name and Score of n students are taken from test data
for(i=0;i<n;i++)
{
scanf("%d", &s[i].rollno);
scanf("%s", s[i].name);
scanf("%d", &s[i].score);
}
//Complete the program so that merit list is printed in descending order
struct student temp;
int j;
for(i=0;i<n-1;i++)
{
for(j=0;j<n-1;j++)
{
if(s[j].score<s[j+1].score)
{
temp=s[j];
s[j]=s[j+1];
s[j+1]=temp;
}
}
}

printf("The Merit List is :\n");
for(i=0;i<n;i++)
{
printf("%d", s[i].rollno);
printf("  %s", s[i].name);
printf("  %d\n", s[i].score);
}

}




Write a C program to store n elements using Dynamic Memory Allocation - calloc() and find the Largest element   






InputOutput


Test Case 1
5
200
300
500
100
400

Largest element = 500.00


Test Case 2
6
68.90
34.79
35.86
94.98
40.06
88.70

Largest element = 94.98





#include <stdio.h>
#include <stdlib.h>

int main()
{
    int n; 
    float *element;

    scanf("%d", &n); //Total number of elements

    // Allocate the memory for 'n' number of elements. 
    //Then take the elements as input from test data
    
    element = (float*) calloc(n, sizeof(float));

    if(element == NULL)
    {
        printf("Error!!! memory not allocated.");
        exit(0);
    }

    // Stores the number entered by the user.
    int i;
    for(i = 0; i < n; ++i)
    {
        scanf("%f", element + i);
    }

 // find the largest
    for(i = 1; i < n; ++i)
    {
       if(*element < *(element + i))
         *element = *(element + i);
    }

    printf("Largest element = %.2f", *element);

    return 0;
}




Write a C program to add two distance given as input in feet and inches.   






InputOutput


Test Case 1
40
11
10
11

Sum of distances = 51 feet 10 inches


Test Case 2
10
5
20
5

Sum of distances = 30 feet 10 inches


Test Case 3
5
11
9
1

Sum of distances = 15 feet 0 inches







#include<stdio.h>

struct Distance{
    int feet;
    int inch;
}d1,d2,sum;

int main()
    {
    //Enter 1st distance
    scanf("%d",&d1.feet); 
    scanf("%d",&d1.inch);
    //Enter 2nd distance
    scanf("%d",&d2.feet);
    scanf("%d",&d2.inch);
sum.feet=d1.feet+d2.feet;
    sum.inch=d1.inch+d2.inch;
 
/* If inch is greater than 12, converting it to feet. */
    if (sum.inch>=12)
    {
        sum.inch=sum.inch-12;
        ++sum.feet;
    }
 printf("Sum of distances = %d feet %d inches",sum.feet,sum.inch);
    return 0;
}






Write a C program to find the sum of two 1D integer arrays ‘A’  and ‘B’ of same size and store the result in another array ‘C’, where the size of the array and the elements of the array are taken as input.  






InputOutput


Test Case 1
5
10
20
30
40
50
1
2
3
4
5

Result is
11
22
33
44
55



Test Case 2
4
100
200
300
400
400
300
200
100

Result is
500
500
500
500



Test Case 3
6
15
25
35
45
55
65
5
5
5
5
5
5

Result is
20
30
40
50
60
70





#include<stdio.h>
#include<stdlib.h>

void main()
{
 int i,n;
 
//The number of elements in each array is taken from test case data 

 scanf("%d", &n);
int *a,*b,*c;
 a = (int *) malloc(n*sizeof(int));
 b = (int *) malloc(n*sizeof(int));
 c = (int *) malloc(n*sizeof(int));

// Input Elements of First List;
 for(i=0;i<n;i++)
 {
  scanf("%d",a+i);
 }

 //Input Elements of Second List;
 for(i=0;i<n;i++)
 {
  scanf("%d",b+i);
 }

 for(i=0;i<n;i++)
 {
  *(c+i) = *(a+i) + *(b+i);
 }
printf("Result is\n");

 for(i=0; i<n; i++)
 {
  printf("%d\n",*(c+i));
 }

}    




















































8 comments:








  1. UnknownMarch 18, 2019 at 12:29 AM
    this blog very helpful for me
    ReplyDelete
  2. HKvbApril 24, 2019 at 9:51 PM
    very helpfulll..thankss
    ReplyDelete
  3. Kartik AgrawalOctober 4, 2019 at 10:00 PM
    one program is wrong .
    of root program (2x cube - 3x - 5 ) .
    please correct this one
    ReplyDelete
    Replies
    1. DEVANSH RAIOctober 5, 2019 at 12:07 PM
      r8
      Delete
    2. DEVANSH RAIOctober 7, 2019 at 2:08 AM
      ek kam kijiye do loop se pehle math ki header file inlude kr lijiye(#include). ar jis if me error aa rha hai us if me condition daliye if(fabs(x1-x)<allerr).fir program run karaiye.
      Delete
  4. UnknownOctober 11, 2019 at 7:44 PM
    one program some mistake bisection method program
    ReplyDelete
  5. UnknownOctober 11, 2019 at 9:32 PM
    thanks
    ReplyDelete
  6. Anurag KumarFebruary 23, 2020 at 8:23 PM
    Write a  C program to print all prime number between 1 to 300 using for loop.

    Code of given problem:

    int main()
    {
        int a,b;
        printf("List of prime numbers between 1 to 300:\n");
        printf("%d\t",a=2);
        for(a=1;a<=300;a++)
            {
                for(b=2;b<=a-1;b++)
                    if(a%b==0)
                        break;
                    else if(b==a-1)
                {
                    printf("%d\t",a);
                }
            }
    return 0;
    }

    Very informative blog! 
    Please take some time to visit my blog @
    Top 10 programming examples in C

    Thanks!
    ReplyDelete




























        

      









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