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Complex C Programs

1. C program to check whether a number is a neon number or not

In this program, we will be given a number as input, and we have to tell whether the number is a neon number or not.  A neon number is a unique type of number where the sum of the digits of the number multiplied by the number itself is equal to the original number. Our program should return “The given number is a neon number” or “The given number is not a neon number” in response to a number we give input to the program.

Example 1:

Input: 1

Output: 1 is a neon number.

Explanation: 1 multiplied by 1 will give 1, and the sum of the digits of 1 is also 1; hence, it is a neon number (as the sum of the digits of the number multiplied by the number itself is the number itself).

Example 2:

Input: 225

Output: 225 is not a neon number.

Explanation: - The Square of the number 225 is 50625, and the sum of digits of 50625 is 5 + 0 + 6 + 2 + 5 = 18, which is not equal to 225. Hence, 225 is not a neon number.

Algorithm:

At first, take the number as input and find the square of the number.

Now, extract the digits of the obtained number one by one and sum them.

At last, compare the sum of the digits with the original number. If they are the same, print“%d is a neon number”; otherwise, print “%d is not a neon number”.

Example in C:

#include <stdio.h>
  int main() {
    int num, square_num, digit, sum = 0;


// Take the number to be checked as input to the program
    printf("Enter a number to check if it is a neon number: ");
    scanf("%d", &num);


    // Calculate the square of the number taken as input
    square_num = num * num;


    // Extract the digits of the square and sum them up
    while (square_num> 0) {
        digit = square_num % 10;
        sum += digit;
        square_num /= 10;
    }


    // Calculate the sum of the digits and check for the neon number
    if (sum == num) {
        printf("%d is a neon number.\n", num);
    } else {
        printf("%d is not a neon number.\n", num);
    }


    return 0;
}

Input:

Enter a number to check if it is a neon number: 1

Output:

1 is a neon number.

Time Complexity:  

O(logn)

Because calculating the square of the number will take O(logn) and extract the digits, and summing them up will take O(logn). Hence the overall time complexity will be O(logn)+O(logn) = O(logn).

2. Write a C program to print a square of asterisks inside another hollow square

In this program, we have to print a hollow square of asterisks, and inside that, we have to print another non-touching hollow square of asterisks.

Example 1:

Input: 5

Output:

* * * * *

* * * * *

* *   * *

* * * * *

* * * * *

Example 2:

Input: 9

Output:

* * * * * * * * *

* * * * * * * * *

* *           * *

* *           * *

* *           * *

* *           * *

* *           * *

* * * * * * * * *

* * * * * * * * *

Algorithm:

At first, take the size of the outer square as input.

Try printing the square of asterisks using nested for loops.

For each row and column, check if it is on the border or near the border of the square, and print * if it is on the border or near the border; otherwise, print a space.

Example in C:

#include <stdio.h>
int main() {
    int n;


    // Take the size of the outer square as input
    printf("Enter the size of the outer square: ");
    scanf("%d", &n);


    // Print both outer and inner square
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i == 1 || i == n || j == 1 || j == n || i == 2 || i == n-1 || j == 2 || j == n-1) {
                printf("* ");
            } else {
                printf("  ");
            }
        }
        printf("\n");
    }


    return 0;
}

Input:

Enter the size of the outer square: 9

Output:  

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

Time Complexity:  

O(n^2)

Because the nested for loops, used to print the inner and outer square, will give a time complexity of O(n^2).

3. Write a C program to print the Compound Interest (CI), given the principal amount, time duration, and rate of interest

In this program, we will have been given the principal amount, time duration, and the rate of Interest as input. And we have to return the total compound interest earned as output. Compound interest is defined as the interest calculated on the initial money given, as well as interest on the summed interest of past time.

Formula to calculate compound interest (CI):  Total amount = Principal (1+Rate/100) ^ t. Where, Interest = Amount – Principal.

Example: 1

Input:

Principal amount: $10,000

Time period: 2 years

Rate: 5% Annually

Output: Compound Interest: $ 102.5

At the end of the first year, the account will have $1,050, where $ 50 is the interest in the first year; now, for the second year, the interest will be 1050x5x1/100 = 52.5. Hence after two years, the total amount will be $ 1102.5, and the total interest will be $102.5.

Example: 2

Input:

Principal amount: $50,000

Time period: 1 year

Rate: 5% Annually

Output: Compound Interest: $ 2500

At the end of the first year, the account will have $52,500, where $ 2500 is the interest in the first year,

Algorithm:

Declare the variables: principal, rate, time, and compound interest.

Take input principal, rate, and time and also read these variables.

Calculate the compound interest using the formula: compound interest = principal * pow (1+rate/100, time) – principal;

Example in C:

//Start of the program
#include <stdio.h>
#include <math.h>


int main() {
    float principal_amount, rate, time, compound_interest;
    printf("Enter the principal amount: ");
    scanf("%f", &principal_amount);
    printf("Enter the rate of interest: ");
    scanf("%f", &rate);
    printf("Enter the time duration (in years): ");
    scanf("%f", &time);


    // Calculate total compound interest
    compound_interest = principal_amount * pow(1 + rate/100, time) - principal_amount;


    printf("Compound interest = %.2f", compound_interest);
    return 0;
}

Input:

Enter the initial amount: 100
Enter the rate of interest: 4
Enter the time span (in years): 1

Output:

Compound interest (CI) = 4

Time Complexity:

O(1)

This is the program that has a fixed number of operations irrespective of the size of the input given.

4. Write a C program to print a Hut/House using stars/ asterisks

In this program, you have to take a number N as input and then print a Hut having width N.

Algorithm:

This C program prints the Hut/House pattern using loops and printf statements. It defines a function called “printHutStar” to print the pattern, and calls it from the main function. The pattern is composed of an upper triangle of stars and lower rectangles of asterisks.

Example in C:

// C program to print the Hut 


#include <stdio.h>


// Function to print the village House
void printHutStar(int n)
{
    int i, j;


    // displaying the upper triangle
    for (i = 0; i < n; i++) {


        // Left situated triangle
        for (j = i + 1; j < n; j++)
            printf(" ");


        // star’s triangle in center
        for (j = 0; j < (2 * i + 1); j++)
            printf("*");


        printf("\n");
    }


    // displaying Lower rectangles
    for (i = 0; i < 3; i++) {


        // rectangle in left
        for (j = 0; j < 3; j++)
            printf("*");


        // Center situated rectangle
        for (j = 0; j < (2 * n - 7); j++)
            printf(" ");


        // Right rectangle
        for (j = 0; j < 3; j++)
            printf("*");


        printf("\n");
    }
}


int main()
{
    int n = 5;


    // function calling
    printHutStar(n);


    return 0;
}

Output:

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

Time Complexity:

O(n)

Time complexity is O(n) because the number of operations directly depends on the size of the input.

5. Write a C program to add two given complex numbers:

Example: 1

Input:

Enter the first complex number: - 2 + 3j

Enter the second complex number: -2 + 3j

Output: ( 2+ 3j) + (2 + 3j) = 4 + 6j

Example: 2

Input:

Enter the first complex number: - 4 - 5j

Enter the second complex number: - 2 + 3j

Output: -(4 - 5j) + (2 + 3j) = 6 - 2j

Algorithm:

  1. Format the structure for real and imag, parts of the complex numbers.
  2. Input two complex numbers and add them using a function that returns the sum.
  3. Print the sum of the complex numbers in the standard a + bi form.

Example in C:

#include <stdio.h>
// Formatting the structure for real and imag, parts of the complex numbers: 
typedef struct {
float imaginary;
    float real;


} complex;
//Function to add two complex numbers
complex addnumbers(complex num1, complex num2) {
    complex result;
    result.real = num1.real + num2.real;
    result.imaginary = num1.imaginary + num2.imaginary;
    return result;
}


int main() {
    complex num1, num2, result;
    //Taking imaginary and real parts of the first complex number as input
    printf("Enter imaginary and real parts of the first complex number:\n");
    scanf("%f %f", &num1.imagianry, &num1.real);
    //Taking imaginary and real parts of the complex number which is 2nd as input
    printf("Enter imaginary and real parts of 2nd complex number:\n");
    scanf("%f %f", &num2.imaginary, &num2.real);
    //call the add function to add the two complex numbers.
    result = addnumbers(num1, num2);
    printf("Sum = %.2f + %.2fi", result.real, result.imaginary);


    return 0;
}

Input:

Enter imaginary and real parts of the 1st complex number:
4
5
Enter imaginary and real parts of the 2nd complex number:
6
9

Output:

The Addition of the given complex numbers= 10.00 + 14.00i

Time Complexity:

O(1)

This is because the program simply takes two complex numbers as input, performs a constant number of arithmetic operations (i.e., addition) on them, and then outputs the result. The time taken by the program is not dependent on the size of the input; hence the time complexity is O(1).

6. Write a C program to print the date and time of the system in AM-PM formats

The program gets the time and date from the system using the time() and localtime() functions. It formats the date and time in a 12-hour format using the strftime() function with the ‘%m/%d/%Y %I:%M %p’ format specifier and prints the date and time to the desktop, also called console.

Algorithm:

  1. Get the time and date from the system using the time() function.
  2. Convert the current time to a local time struct using the localtime() function.
  3. Format the local time struct using the strftime() function with %m/%d/%Y %I:%M %p format specifier.
  4. Print the time and date to the console using the printf() function.

Example in C:

#include <stdio.h>
#include <time.h>


int main() {
    time_t t = time(NULL);
    struct tm *tm = localtime(&t);


    char datetime_str[50];
//fetch the system time and date using the strft function
strftime(datetime_str, sizeof(datetime_str), "%m/%d/%Y %I:%M %p", tm);


//Print the system date and time in 12-hour format.
    printf("Current date and time in 12-hour format: %s\n", datetime_str);


    return 0;
}

Input:

No specific input is being given to the program.

Output:

Current date and time in 12-hour format: 04/18/2023 06:42 PM

Time Complexity:

O(1)

The time complexity of this program is O (1), which means that the time it takes to execute the program does not depend on the input size or any iterative or recursive operations. The program simply retrieves the current date and time from the system using the time() and localtime() functions, formats it using the strftime() function, and then prints it to the console.

7. Write a C program to find the value of Pi using Monte Carlo

This program estimates the value of Pi using Monte Carlo Parallel Computing Method.

Algorithm:

  1. Set a counter to zero to keep track of the number of points which are there within the circle.
  2. Create large number of random points in a square of side length 2 units, centered at (0,0).
  3. Check if each point falls within the circle of radius 1 unit, also centered at (0,0). If so, increment the counter.
  4. Estimate the value of pi as it is equal to four times the ratio of the number of points which are there inside the circle to the total number of points generated.

Example in C:

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


#define NUM_POINTS 1000000


int main()
{
    int i, count_inside = 0;
    double x, y, pi;


    srand(time(NULL));


    for (i = 0; i < NUM_POINTS; i++) {
        x = (double) rand() / RAND_MAX * 2 - 1;
        y = (double) rand() / RAND_MAX * 2 - 1;
        if (x * x + y * y <= 1) {
            count_inside++;
        }
    }


    pi = 4.0 * count_inside / NUM_POINTS;


    printf("Estimated value of pi = %f\n", pi);


    return 0;
}

Output:

Estimated value of pi = 3.142044

Time Complexity:

O(N)

This is because the program loops N times to generate N random points and perform the computation to estimate the value of pi.

8. Write a C program to calculate the angle between the minute hand and hour hand of an analog clock, when time is specified in 24 hour format

This program should calculate the angle (in degrees) between the minute hand and hour hand of the analog clock. The time will be given in the 24 hour format, and you have to give the angle as output.

Example 1:  Enter the hours (0-23): 9

Enter the minutes (0-59): 15

The angle between the hour and minute hand is: 172.50 degrees

Example 2:  Enter the hours (0-23): 13

Enter the minutes (0-59): 15

The angle between the hour and minute hand is: 52.50 degrees

Algorithm:

  1. Ask the user to enter the minutes and hour in 24 hour format of the time on the analog clock.
  2. Calculate the angle of the minute hand and the angle of the hour hand using the following formulas: hour angle = 0.5 * ((hours % 12) * 60 + minutes) and minute angle = 6 * minutes.
  3. Calculate the difference between the minute angle and the hour angle and keep it absolute to neglect the negative values produced.
  4. If the angle is greater than 180 degrees, subtract it from 360 to get the smaller angle, because an angle greater than 180 degrees is not possible.
  5. Print the angle between the minute and hour hand to the console for the user.

Example in C:

#include <stdio.h>
#include <math.h>


int main()
{   
    //Setting up all the variables
    int hours, minutes;
    float hour_angle, minute_angle, angle;


    // Take the hour as input
    printf("Enter the hours (0-23): ");
    scanf("%d", &hours);


    // Take the minutes as input.
    printf("Enter the minutes (0-59): ");
    scanf("%d", &minutes);


    // calculate the hour hand angle
    hour_angle = 0.5 * ((hours % 12) * 60 + minutes);


    // calculate the minute hand angle
    minute_angle = 6 * minutes;


    // calculate the absolute angle between the hour and minute hand
    angle = fabs(hour_angle - minute_angle);


    // adjust angle to be between 0 and 180 degrees so that it will not exceed.
    if (angle > 180) {
        angle = 360 - angle;
    }
    //Print the angle in degrees
    printf("The angle between the hour and minute hand is: %.2f degrees\n", angle);


    return 0;
}

Input:

Enter the hours (0-23): 9
Enter the minutes (0-59): 30

Output:

The angle between the hour and minute hand is: 105.00 degrees.

Time Complexity:

O(1)

This is because the program doesn’t have any recursive instructions to complete. And the instructions in the program complete in constant time, which results in O(1) time complexity.

9. Write a C program to print the given text within double quotes:

The problem with printing double quotes is that the printf function in C programming already uses these quotes to print.

Example 1:  "HR, JavaTpoint!"

Print the text like this in double quotes.

Algorithm:

1. Create one character array to store the text given.

2. Print the text using printf() and %s format specifier with escaped double quotes.

Example in C:

#include <stdio.h>


int main() {
    char text[] = "HR, JavaTpoint!";
    printf("\"%s\"", text);
    return 0;
}

Input:

No Input Given

Output:

"HR, JavaTpoint!"

Time Complexity:

O(1)

This is because the program doesn’t have any recursive instructions to complete.