PAT1024. Palindromic Number (25)

输入即为回文的情况要考虑

 #include <iostream>
 #include <algorithm> //reverse
 using namespace std;
 string n;
 int k;
 string strAdd(string a){
     string b;
     int num,cc=;
     int len=a.size();
     for(int i=;i<len;i++){
         num=a[i]+a[len-i-]-*''+cc;
         cc=num/;
         b.push_back(num%+'');
     }
     if(cc!=) b.push_back('');
     reverse(b.begin(),b.end());
     return b;
 }
 bool isPali(string &a){
     int len=a.size();
     for(int i=;i<len/;i++){
         if(a[i]==a[len--i]){continue;}
         else{return false;}
     }
     return true;
 }
 int main()
 {
     cin>>n>>k;
     int c=k;
     string result=n;
     if(isPali(result)) {cout<<result<<endl<<"";return ;}
     while(c--){
         result=strAdd(result);
         if(isPali(result))break;
     }
     if(c<=){
         cout<<result<<endl;
         cout<<k;
     }else{
         cout<<result<<endl;
         cout<<(k-c);
     }
     return ;
 }