C++ Exercises: Check if a number is Keith or not
C++ Numbers: Exercise-41 with Solution
Write a C++ program to check if a number is Keith or not (with explanation).
Sample Output:
Input a number : 197
1 + 9 + 7 = 17
9 + 7 + 17 = 33
7 + 17 + 33 = 57
17 + 33 + 57 = 107
33 + 57 + 107 = 197
The given number is a Keith Number.
Sample Solution:
C++ Code:
#include<bits/stdc++.h> // Including the complete set of standard C++ libraries
using namespace std; // Using the standard namespace
// Function to count the number of digits in a number
int lenCount(int nm) {
int ctr = 0; // Counter initialized to count the digits
while (nm > 0) { // Loop to count digits
nm = nm / 10; // Remove the last digit
ctr++; // Increment the counter for each digit removed
}
return ctr; // Return the count of digits
}
// Main function
int main() {
int num1 = 0, arr1[10], num2 = 0, flg = 0, i = 0, sum = 0, kk; // Declaration of integer variables
cout << "\n\n Check whether a number is Keith or not: \n"; // Displaying purpose
cout << " Sample Keith numbers: 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385..\n"; // Displaying purpose
cout << " -----------------------------------------------------------------------\n"; // Displaying purpose
cout << " Input a number : "; // Prompting user for input
cin >> num1; // Taking user input
num2 = num1; // Storing the original number
// Loop to store individual digits of the number into an array in reverse order
for (i = lenCount(num2) - 1; i >= 0; i--) {
arr1[i] = num1 % 10; // Extract the last digit
num1 /= 10; // Remove the last digit
}
// Loop to check if the number is a Keith number
while (flg == 0) {
for (i = 0; i < lenCount(num2); i++)
sum += arr1[i]; // Calculate the sum of the digits in the array
// If the sum matches the original number, it's a Keith number
if (sum == num2) {
flg = 1; // Set the flag to terminate the loop
kk = 1; // Set kk to 1 indicating the number is a Keith number
}
// If the sum exceeds the original number, it's not a Keith number
if (sum > num2) {
flg = 1; // Set the flag to terminate the loop
kk = 0; // Set kk to 0 indicating the number is not a Keith number
}
// Loop to reconstruct the array by shifting elements and adding the new sum
for (i = 0; i < lenCount(num2); i++) {
cout << " " << arr1[i]; // Display the array elements
if (i != lenCount(num2) - 1) {
arr1[i] = arr1[i + 1]; // Shift elements in the array
cout << " + "; // Displaying '+' between elements
} else {
arr1[i] = sum; // Store the new sum at the last index of the array
cout << " = " << arr1[i]; // Display the sum
}
}
cout << endl; // Move to the next line
sum = 0; // Reset the sum for the next iteration
}
// Displaying whether the number is a Keith number or not based on kk value
if (kk == 1) {
cout << " The given number is a Keith Number.\n"; // Displaying output
}
if (kk == 0) {
cout << " The given number is not a Keith Number.\n"; // Displaying output
}
}
Sample Output:
Check whether a number is Keith or not: Sample Keith numbers: 197, 742, 1104, 1537, 2208, 2580, 3684, 4788, 7385.. ----------------------------------------------------------------------- Input a number : 197 1 + 9 + 7 = 17 9 + 7 + 17 = 33 7 + 17 + 33 = 57 17 + 33 + 57 = 107 33 + 57 + 107 = 197 The given number is a Keith Number.
Flowchart:
C++ Code Editor:
Contribute your code and comments through Disqus.
Previous: Find Strong Numbers within a range of numbers.
Next: Create the first twenty Hamming numbers.
What is the difficulty level of this exercise?
It will be nice if you may share this link in any developer community or anywhere else, from where other developers may find this content. Thanks.
https://w3resource.com/cpp-exercises/numbers/cpp-numbers-exercise-41.php
- Weekly Trends and Language Statistics
- Weekly Trends and Language Statistics