PAT A1136 A Delayed Palindrome （20 分）——回文，大整数

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.

 #include <stdio.h>
 #include <string>
 #include <iostream>
 #include <algorithm>
 #include <vector>
 #include <string.h>
 using namespace std;
 string add(string s1,string s2){
     string res="";
     int carry=;
     for(int i=s1.length()-;i>=;i--){
         int tmp=s2[i]-''+s1[i]-''+carry;
         res+=(tmp%+'');
         carry=tmp/;
     }
     if(carry!=)res += (carry+'');
     reverse(res.begin(),res.end());
     return res;
 }
 bool ispali(string s){
     string s2;
     s2=s;
     reverse(s2.begin(),s2.end());
     if(s==s2) return true;
     else return false;
 }
 int main(){
       string s1,s2,s3;
       cin>>s1;
       int i;
       if(ispali(s1)){
           cout<<s1<<" is a palindromic number."<<endl;
               return ;
       }
       for(i=;i<;i++){
           s2=s1;
           reverse(s2.begin(),s2.end());
           s3=add(s1,s2);
           cout<<s1<<" + "<<s2<<" = "<<s3<<endl;
           if(ispali(s3)){
               cout<<s3<<" is a palindromic number."<<endl;
               return ;
           }
           s1=s3;
       }
       printf("Not found in 10 iterations.\n");
 }

注意点：判断回文用string还是方便，string只能string+char，不能char+string，所以大数相加还是要最后反转一下。