【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;   
}