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Neon Number in C++
A neon number is a number where the sum of digits of the square of the number is equal to the number. Let's take an example.
n = 9
square = 81
sum of digits of square = 8 + 1 = 9
So, the number 9 is a neon number.
We need to check whether the given number is a neon number or not. If the given number is a neon number, then print Yes else print No.
Algorithm
- Initialise the number n.
- Find the square of the number.
- Find the sum of the digits of the square
- If the sum of digits of the square is equal to the given number then result true else false.
Implementation
Following is the implementation of the above algorithm in C++
#include <bits/stdc++.h>
using namespace std;
int isNeonNumber(int x) {
int square = x * x;
int digitsSum = 0;
while (square != 0) {
digitsSum += (square % 10);
square = square / 10;
}
return digitsSum == x;
}
int main(void) {
string result;
result = isNeonNumber(1) ? "Yes" : "No";
cout << 1 << "->" << result << endl;
result = isNeonNumber(3) ? "Yes" : "No";
cout << 3 << "->" << result << endl;
result = isNeonNumber(9) ? "Yes" : "No";
cout << 9 << "->" << result << endl;
}Output
If you run the above code, then you will get the following result.
1->Yes 3->No 9->Yes
Updated on: 23-Oct-2021
2K+ Views
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