测试点3,4,5是对输入N是否为回文数的判断

测试点6存在大数运算，不能单纯进行整数的加减，需要进行一位一位的处理

C++

 1 #include <iostream>
 2 #include <string>
 3 #include <algorithm>
 4 
 5 using namespace std;
 6 
 7 bool isPalindromeNumber(string &s) {                            //判断回文数的函数
 8     for (int i = 0; i < s.size() / 2; ++i) {
 9         if(s[i] != s[s.size() - 1 - i]) return 0;
10     }
11     return 1;
12 }
13 
14 int main() {
15     string N, rN;
16     string sum;
17     cin >> N;
18     bool N_judge = isPalindromeNumber(N);
19     if (N_judge) {                                              //测试点3,4,5，首先对N进行判断是否为回文数
20         cout << N << " is a palindromic number." << endl;
21         return 0;
22     }
23     for(int i = 0; i < 10; ++i) {
24         rN = N;
25         reverse(N.begin(), N.end());
26         int carry_bit = 0;
27         for (int j = N.size() - 1; j >= 0; --j) {               //测试点6会进行大数运算，所以不能普通的整数加减，需要进行进位加法处理
28             int tmp = (N[j] - '0') + (rN[j] - '0') + carry_bit;
29             carry_bit = 0;
30             if (tmp >= 10) {
31                 carry_bit = 1;
32                 tmp -= 10;
33             }
34             sum = to_string(tmp) + sum;
35         }
36         if(carry_bit) sum = to_string(carry_bit) + sum;         //判断加到最后是否存在进位
37         bool judge = isPalindromeNumber(sum);
38         cout << rN << " + " << N << " = " << sum << endl;
39         if(judge) {
40             cout << sum << " is a palindromic number." << endl;
41             return 0;
42         }
43         N = sum;
44         sum.clear();                                            //记得对加和的字符串进行清除，方便下一轮判断重复利用
45     }
46     cout << "Not found in 10 iterations." << endl;
47     return 0;
48 }