Finding Armstrong Number using C++ - c++

Question

I'm writing a program for finding whether a given number is an Armstrong Number:
int main()
{
int n,rem,sum=0;
cout<<"Enter the Number for checking"<<endl;
cin>>n;
while(n!=0)
{
rem=n%10;
sum=sum+(rem*rem*rem);
n=n/10;
}
if(sum==n)
{
cout<<"Armstrong Number"<<endl;
}
else
{
cout<<"It's not a armstrong number";
}
return 0;
}
When I run it, it always reports "It's not a armstrong number", regardless of input.
I changed the code as follows, and got the correct result. But I don't understand why I need to assign input to n1 and do the operation - why can't I directly do the operation with n?
int main()
{
int n,rem,sum=0,n1;
cout<<"Enter the Number for checking"<<endl;
cin>>n;
n1=n;
while(n1!=0)
{
rem=n1%10;
sum=sum+(rem*rem*rem);
n1=n1/10;
}
if(sum==n)
{
cout<<"Armstrong Number"<<endl;
}
else
{
cout<<"It's not a armstrong number";
}
return 0;
}

Answer 1

In the line if (sum==n) your program compares sum and n. In the second program n is initial number entered by user. But in the first program n==0 (see the loop above it).
So, in the first program the check if (sum==n) works as if (sum==0). But value of sum is never 0 (except user entered 0). So, first program always returns "It's not a armstrong number".
And about style: It is much better to use functions instead of putting the whole logic into one main() function. For instance, you can create a function for calculation of the intermediate sum for cheching of Armstrong Number:
int getSumOfCubesOfDigits(int n)
{
int sum = 0;
while (n)
{
const int rem = n % 10;
sum += rem * rem * rem;
n = n / 10;
}
}
In this case your program will be much simpler and it will be hard to make the mistake you have in the first program of your question:
int main()
{
int n;
cout << "Enter the Number for checking" << endl;
cin >> n;
if(getSumOfCubesOfDigits(n) == n)
cout<<"Armstrong Number"<<endl;
else
cout<<"It's not a armstrong number";
return 0;
}

Answer 2

In the first program, the original number is entered into 'n'. The only problem in your logic is, you forgot that by the time you exit from the while loop, 'n' will no longer be your original number since you are repeatedly doing n=n/10, and hence 'sum==n' never satisfies even for an Armstrong number.
So before you enter the while loop, save the original number into another variable, say n1 (as done in the second program you provided), and only use n1 for operations, ie, n1=n1/10. Leave n alone so that, in the end 'n' will still contain the original number, which you can finally compare with 'sum' to find your answer.

Answer 3

Which number do you compare ? , in first program in while loop , n value is changed ( in this variable you get the input) and finally check with sum == n , so it always get condition fail.
So temp (n1) variable is required , to compare the final result

Answer 4

Your code is different in the second code block, you are still testing if sum=n.
In the second code block, if you tested if(sum=n1), I would suspect it would work the same.

Answer 5

I got this solution for finding an Armstrong Numbers
int main() {
for (int i=10; i<=9999; i++) {
int k = i,z = 1, s = 0, n = i;
while ((k/=10) > 0) z++;
for (int t = z; t; t--, n/=10) {
s += pow(n % 10, z);
}
if (i == s) cout << i << endl;
}
return 0;
}