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;}