Consider a positive integer N written in standard notation with k+1 digits a?i?? as a?k???a?1??a?0?? with 0≤a?i??<10 for all i and a?k??>0. Then N is palindromic if and only if a?i??=a?k?i?? for all i. Zero is written 0 and is also palindromic by definition.
Non-palindromic numbers can be paired with palindromic ones via a series of operations. First, the non-palindromic number is reversed and the result is added to the original number. If the result is not a palindromic number, this is repeated until it gives a palindromic number. Such number is called a delayed palindrome. (Quoted from https://en.wikipedia.org/wiki/Palindromic_number )
Given any positive integer, you are supposed to find its paired palindromic number.
Input Specification:
Each input file contains one test case which gives a positive integer no more than 1000 digits.
Output Specification:
For each test case, print line by line the process of finding the palindromic number. The format of each line is the following:
A + B = C
where A is the original number, B is the reversed A, and C is their sum. A starts being the input number, and this process ends until C becomes a palindromic number -- in this case we print in the last line C is a palindromic number.; or if a palindromic number cannot be found in 10 iterations, print Not found in 10 iterations. instead.
Sample Input 1:
97152
Sample Output 1:
97152 + 25179 = 122331
122331 + 133221 = 255552
255552 is a palindromic number.
Sample Input 2:
196
Sample Output 2:
196 + 691 = 887
887 + 788 = 1675
1675 + 5761 = 7436
7436 + 6347 = 13783
13783 + 38731 = 52514
52514 + 41525 = 94039
94039 + 93049 = 187088
187088 + 880781 = 1067869
1067869 + 9687601 = 10755470
10755470 + 07455701 = 18211171
Not found in 10 iterations.
思路：因为他的数字个数不超过1000个，所以用字符串来保存数据；然后判断是不是回文串，反过来比较是否相同即可；
其中用到了reverse（）函数，把一个字符串反过来，大数相加，存进位；因为是加法，所以进位最大也只可能是1；
注意事项：将单个字符存入字符串：ans += char(num + ‘0‘);将个位数转换为char类型，要 + ‘0‘；不能 - ‘0’
代码如下： 1 #include<bits/stdc++.h>
2 using namespace std;
3 #define ll long long
4
5 string add(string a)
6 {
7 string ans;
8 int l = a.length();
9 int co = 0;
10 for(int i = 0;i < l;i++)
11 {
12 int num = (a[i] - ‘0‘) + (a[l - 1 - i] - ‘0‘) + co;
13 //cout << num << "---" << endl;
14 co = 0;
15 if(num > 9)
16 {
17 co = 1;
18 num -= 10;
19 }
20 ans += char(num + ‘0‘);
21 }
22 if(co == 1)
23 ans += ‘1‘;
24 reverse(ans.begin(),ans.end());
25 //cout << ans << endl;
26 return ans;
27 }
28
29
30 string a,b;
31 int main()
32 {
33 cin >> a;
34 int i,len;
35 for(i = 0;i < 10;i++)
36 {
37 b = a;
38 reverse(b.begin(),b.end());
39 if(a == b)
40 break;
41 cout << a << " + " << b << " = " << add(a) << endl;
42 a = add(a);
43 }
44 if(i == 10)
45 cout << "Not found in 10 iterations." << endl;
46 else
47 cout << a << " is a palindromic number." << endl;
48 return 0;
49 }
原文地址：https://www.cnblogs.com/lu1nacy/p/10088765.html