UVa Problem 10018 Reverse and Add （反转相加）

// Reverse and Add （反转相加）// PC/UVa IDs: 110502/10018, Popularity: A, Success rate: low Level: 1// Verdict: Accepted// Submission Date: 2011-05-27// UVa Run Time: 0.028s//// 版权所有（C）2011，邱秋。metaphysis # yeah dot net//// 利用动态数组来保存数字的每一位，逆序存放。#include <iostream>#include <vector>using namespace std;// 判断一个数是否是回文数。该数的每一位逆序存放在动态数组中。bool is_palindrome_number(vector < int > digits){if (digits.size() == 1 || digits.size() == 0)return true;if (digits[0] != digits[digits.size() - 1])return false;digits.erase(digits.begin());digits.erase(digits.end() - 1);return is_palindrome_number(digits);}// 反向相加，判断是否是回文数，否则继续该操作。void reverse_add(vector < int > &digits){int step = 0;do{step++;vector < int > sum;int carry = 0;for (int i = 0, j = digits.size() - 1; i < digits.size(); i++, j--){int tmp = digits[i] + digits[j] + carry;carry = ((tmp - 10) >= 0 ? 1 : 0);sum.push_back(tmp % 10);}if(carry)sum.push_back(carry);digits = sum;} while (!is_palindrome_number(digits));cout << step << " ";for (int i = digits.size() - 1; i >= 0; i--)cout << digits[i];cout << endl;}int main(int ac, char *av[]){int cases;vector < int > digits;cin >> cases;while (cases--){int number;cin >> number;digits.clear();while (number){digits.push_back(number % 10);number /= 10;}reverse_add(digits);}return 0;}