### Palindrome

My instructor was showing us palindrome numbers today and instead if using string size method she used some division method where last number got extracted. Can anyone give me a hint in how to find if a certain number is palindrome? I thought about dividing the number by ten to extract the last digit, but I can clearly figure out how to compare without using string size.

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Are the numbers stored as integers or strings?

Here's one method to get the first and last digit of an integer:

 ``1234567891011121314151617181920212223242526272829`` ``````#include long long pow( int x , int n ) { return( n > 1 ? x * pow( x , n - 1 ) : 1 ); } int main() { int number = 12345; int power_ten = 0; int first_digit = 0; int last_digit = number; //using last_digit as a temporary variable right now while( last_digit > 0 ) { last_digit /= 10; ++power_ten; } //get first digit first_digit = number / pow( 10 , power_ten ); //get last digit last_digit = number % 10; std::cout << "first digit: " << first_digit << std::endl; std::cout << "last digit: " << last_digit << std::endl; }``````

This is probably how your teacher did it if she used the divide method.

 ``12`` `````` return( n > 1 ? x * pow( x , n - 1 ) : 1 ); ``````

Giblit, can you explain to me what this code does here. I know it's similar to if and else, but the expression I can't understand.
Its a c++ operator called the ternary operator.

Basically the pseudo code is like this

(return statement or a variable you are assigning a value to ) (condition to test) ? ( if true return this value ) : (otherwise return this value )

so lets look at my example

The thing I am using the value with is the return statement.
the condition I am using n > 1 //should of been n > 0 I had a typo because x^0 == 1
if n is greater than 1 I am using recursion ( return the value of x * pow( x , n - 1 )
if n is less than or equal to 1 then return 1 //again should have been less than 1 aka 0

Basically it can be rewritten as either of these two:

 ``123456789101112131415161718`` ``````long long pow( int x , int n ) { if( n > 0 ) return( x * pow( x , n - 1 ); return( 1 ); //if x is not greater than 1 } //or long long pow( int x , int n ) { int increment = x; if( n == 0 ) return( 1 ); for( int i = 1; i < n; ++i ) x *= increment; return( x ); }``````

Then it should have been `first_digit = number / pow( 10 , power_ten - 1 )` since 1000 = 10^3 not 10^4 I had a brain fart.

Here is somethign to read up on:
http://www.cplusplus.com/articles/1AUq5Di1/
http://www.learncpp.com/cpp-tutorial/34-sizeof-comma-and-arithmetic-if-operators/

*edit also you may use the pow function in the math library if you want. I just didn't feel like it.

**edit
The power_ten in my code is really the amount of digits in the number.
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