Palindromes And Project Euler C++ Version -Tips - c++

Question

Can Anyone Tell Me What's Wrong In My Code?
Thank you :)
// A palindromic number reads the same both ways.
// The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 × 99.
// Find the largest palindrome made from the product of two 3-digit numbers.
#include <iostream>
using namespace std;
int main()
{
int x = 2;
int product;
int n, digit, rev = 0;
int greatest = 0;
for(int i = 2;i<100;i++){
product = x * i;
n = product;
cout << x << " * " << i << " = " << product << endl;
do
{
digit = product % 10;
rev = (rev * 10) + digit;
product = product / 10;
} while(product != 0);
cout << " The reverse of the number is: " << rev << endl;
if(n == rev){
cout << "Therefore Palindrome" << endl;
if(rev > greatest){
cout << "REV Greater Than Greatest Palindrome" << endl;
greatest = rev;
}
}
if(i == 99){
if(x < 99){
x++;
i = 1;
cout << "Go For The Next Loop" << endl;
}
}
}
cout << "The Greatest Palindrome Number Is " << greatest << endl;
return 0;
}

Answer 1

Several things:
You should be counting down from 999 thru 100.
You should be using two loops for the two factors of your product
rev should be initially zero before each entrance into your inner do-while loop
The results would look something like this, with all but the detected palindromes and final result being output (the amount of worthless noise in your output is mind-bending)
#include <iostream>
using namespace std;
int main()
{
int product;
int n, digit, rev = 0;
int greatest = 0;
for(int i = 999;i>=100; --i)
{
for (int j =999; j>=100; --j)
{
product = j * i;
n = product;
rev = 0;
do
{
rev = (rev * 10) + (product % 10);
product /= 10;
} while(product != 0);
if(n == rev)
{
cout << "Palindrome : " << i << " * " << j << " = " << n << endl;
if(rev > greatest)
greatest = rev;
}
}
}
cout << "The Greatest Palindrome Number Is : " << greatest << endl;
return 0;
}
This will conclude the correct answer you seek.
Faster
Optionally, you can squeeze better performance out of this by understanding a few additional details:
The task is to find the largest palindrome, not just the longest palindrome. Therefore, once you find one, any product, palindrome or not, that is a smaller magnitude than the current largest palindrome is pointless to even check and the inner loop can be terminated (thus the reason you're counting down).
A little number crunching will allow you to conclude that at least one of the factors must be divisible by 11. I won't cover why, but do the math, it's true. Therefore, you can make one of your loops a count-down from the largest 3-digit multiple of 11 (990) to 100, in steps of (-11).
The result looks like this:
#include <iostream>
using namespace std;
int main()
{
int greatest = 0;
for(int i = 999;i>=100; --i)
{
for (int j =990; j>=100; j-=11)
{
int product = j * i;
if (product < greatest)
break;
int n = product;
int rev = 0;
do
{
rev = (rev * 10) + (product % 10);
product /= 10;
} while(product != 0);
if(n == rev)
{
cout << "Palindrome : " << i << " * " << j << " = " << n << endl;
if(rev > greatest)
greatest = rev;
}
}
}
cout << "The Greatest Palindrome Number Is : " << greatest << endl;
return 0;
}
Output
Palindrome : 995 * 583 = 580085
Palindrome : 993 * 913 = 906609
The Greatest Palindrome Number Is : 906609