Project Euler：Problem 55 Lychrel numbers

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.

求10000以内的不可按以上方法迭代得出回文的数的个数。

#include <iostream>
#include <string>
using namespace std;

string num2str(int n)
{
	string ans = "";
	while (n)
	{
		int a = n % 10;
		char b = a + '0';
		ans = b + ans;
		n /= 10;
	}
	return ans;
}

string strplus(string a, string b)
{
	int len = a.length();

	int flag = 0;
	string ans = "";
	for (int i = len - 1; i >= 0; i--)
	{
		int tmp = a[i] + b[i] - '0' - '0' + flag;
		flag = tmp / 10;
		tmp = tmp % 10;
		char p = tmp + '0';
		ans = p + ans;
	}
	if (flag == 1)
		ans = '1' + ans;
	return ans;
}

bool pali(string a)
{
	for (int i = 0; i < a.length() / 2; i++)
	{
		if (a[i] != a[a.length() - 1 - i])
			return false;
	}
	return true;
}

bool isLychrel(int n)
{
	string a, b;
	a = num2str(n);
	b.assign(a.rbegin(), a.rend());
	for (int i = 1; i <= 50; i++)
	{
		a = strplus(a, b);
		if (pali(a))
			return false;
		b.assign(a.rbegin(), a.rend());
	}
	return true;
}

int main()
{

	int count = 0;
	for (int i = 1; i <= 10000; i++)
	{
		if (isLychrel(i))
			count++;
	}
	cout << count << endl;
	system("pause");
	return 0;
}