Question

How to check whether a given number is a disarium number in C++

Answer 1

// 1^1 + 3^2 + 5^3 = 1 + 9 + 125 = 135

#include <iostream>
#include <cmath>
 
int main() {
    int num = 135, remainder = 0;
    int len = log10(num) + 1;
    float sum = 0.0f;
 
    int temp = num;
    while (temp > 0) {
        remainder = temp % 10;
        sum = sum + pow(remainder, len);
        temp = temp / 10;
        len--;
    }
 
    if (num == (int)sum)
        std::cout << num << " is a disarium number";
    else
        std::cout << num << " is not a disarium number";
 
    return 0;
}
 
 

 
 
/*
run:
 
135 is a disarium number
 
*/

 

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