Project Euler task 55

If we take 47, reverse and add, 47 + 74 = 121, which is palindromic.

Not all numbers produce palindromes so quickly. For example,

349 + 943 = 1292,
1292 + 2921 = 4213
4213 + 3124 = 7337

That is, 349 took three iterations to arrive at a palindrome.

Although no one has proved it yet, it is thought that some numbers, like 196, never produce a palindrome. A number that never forms a palindrome through the reverse and add process is called a Lychrel number. Due to the theoretical nature of these numbers, and for the purpose of this problem, we shall assume that a number is Lychrel until proven otherwise. In addition you are given that for every number below ten-thousand, it will either (i) become a palindrome in less than fifty iterations, or, (ii) no one, with all the computing power that exists, has managed so far to map it to a palindrome. In fact, 10677 is the first number to be shown to require over fifty iterations before producing a palindrome: 4668731596684224866951378664 (53 iterations, 28-digits).

Surprisingly, there are palindromic numbers that are themselves Lychrel numbers; the first example is 4994.

How many Lychrel numbers are there below ten-thousand?

NOTE: Wording was modified slightly on 24 April 2007 to emphasise the theoretical nature of Lychrel numbers.

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#include < iostream >

using namespace std;

typedef unsigned long long int uint64;

uint64 getReverse(uint64 _n) {

while (_n % 10 == 0)

_n /= 10;

uint64 iRes = 0;

while(_n) {

iRes *= 10;

iRes += (_n % 10);

_n /= 10;

}

return iRes;

}

bool isPalindromic(uint64 _n) {

uint64 _nMirr = getReverse(_n);

while (_nMirr) {

if (_nMirr % 10 != _n % 10)

return false;

_nMirr /= 10;

_n /= 10;

}

return true;

}

bool isLychrell(uint64 _n) {

uint64 iTemp = _n;

int j = 0;

for (; j < 50; j++) {

uint64 iTempNew = iTemp + getReverse(iTemp);

if (isPalindromic(iTempNew))

return false;

iTemp = iTempNew;

}

return true;

}

int main() {

int iLychrell = 0;

for (uint64 i = 1; i < 10000; i++)

iLychrell += isLychrell(i) ? 1 : 0;

cout << iLychrell << endl;

return 0;

}