C Exercises: Find Disarium numbers between 1 to 1000

Last update on December 20 2024 13:06:54 (UTC/GMT +8 hours)

Write a program in C to find Disarium numbers between 1 and 1000.

Sample Solution:

C Code:

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

// Function to count the number of digits in a given number
int DigiCount(int n)
{
    int ctr_digi = 0;
    int tmpx = n;
    // Count the number of digits
    while (tmpx)
    {
        tmpx = tmpx / 10;
        ctr_digi++;
    }
    return ctr_digi; // Return the count of digits
}

// Function to check if a number is a Disarium Number
bool chkDisarum(int n)
{
    int ctr_digi = DigiCount(n); // Get the count of digits in the number
    int s = 0; 
    int x = n;
    int pr;
    // Calculate the sum of powers of digits
    while (x)
    {
        pr = x % 10;
        s = s + pow(pr, ctr_digi--); // Add the power of the digit to the sum
        x = x / 10; // Move to the next digit
    }
    return (s == n); // Return true if the sum equals the original number
}

// Main function
int main()
{
    int i;
    printf("\n\n Find Disarium Numbers between 1 to 1000: \n");
    printf(" ---------------------------------------------\n");
    printf(" The Disarium numbers are: \n");

    // Check for Disarium Numbers in the range of 1 to 1000
    for (i = 1; i <= 1000; i++)
    {
        if (chkDisarum(i)) // If i is a Disarium Number
            printf("%d ", i); // Print the Disarium Number
    }	

    printf("\n");
    return 0;
}

Sample Output:

 The Disarium numbers are:                                                                                    
1 2 3 4 5 6 7 8 9 89 135 175 518 598 

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