C++ Exercises: Check whether a number is Disarium or not

C++ Numbers: Exercise-18 with Solution

Write a program in C++ to check whether a number is Disarium or not.

Pictorial Presentation:

Sample Solution:

C++ Code :

 #include<bits/stdc++.h>
using namespace std;
 
int DigiCount(int n)
{
    int ctr_digi = 0;
    int tmpx = n;
    while (tmpx)
    {
        tmpx = tmpx/10;
 
        ctr_digi++;
    }
    return ctr_digi;
}
bool chkDisarum(int n)
{
    int ctr_digi = DigiCount(n);
    int s = 0; 
    int x = n;
    int pr;
    while (x)
    {
        pr = x % 10;
        s = s + pow(pr, ctr_digi--);
        x = x/10;
    }
    return (s == n);
}
int main()
{

int dino;
 cout << "\n\n Check whether a number is Disarium Number or not: \n";
 cout << " ---------------------------------------------------\n";
 cout << " Input a number: ";
 cin >> dino;	
	
    if( chkDisarum(dino))
        cout << " The given number is a Disarium Number."<<endl;
    else
        cout << " The given number is not a Disarium Number."<<endl;
    return 0;
}

Sample Output:

Check whether a number is Disarium Number or not:                                                   
 ---------------------------------------------------                                                 
 Input a number: 9                                                                                   
 The given number is a Disarium Number.

Flowchart:

C++ Code Editor:

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