【1136】A Delayed Palindrome (20分)

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>  
#include<map>
#include<vector>
#include<queue> 
#include<string>
/*
/*注意1：字符串存数字时低位存数字的高位
   注意2："1"+s是在字符串s首加了1
*/

using namespace std;  
string rev(string s){
   reverse(s.begin(),s.end());
   return s;
}
string add(string s1,string s2){//大整数加法
   string s=s1;
   int carry=0;
   for(int i=s1.size()-1;i>=0;i--){
      s[i]=(s1[i]-'0'+s2[i]-'0'+carry)%10+'0';
      carry=(s1[i]-'0'+s2[i]-'0'+carry)/10;
   }
   if(carry>0)  s="1"+s;
   return s;
}
       
int main(){   
   string s,sum;
   int n=10;
   cin>>s;
   if(s==rev(s)){
      cout<<s<<" is a palindromic number.\n";
      return 0;
   }
   while(n--){
      sum=add(s,rev(s));
      cout<<s<<" + "<<rev(s)<<" = "<<sum<<endl;
      if(sum==rev(sum)){
         cout<<sum<<" is a palindromic number.\n";
         return 0;
      }
      s=sum;
   }
   cout <<"Not found in 10 iterations.\n";
   //system("pause");
    return 0;   
}