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=0;
    int len=a.size();
    for(int i=0;i<len;i++){
        num=a[i]+a[len-i-1]-2*'0'+cc;
        cc=num/10;
        b.push_back(num%10+'0');
    }
    if(cc!=0) b.push_back('1');
    reverse(b.begin(),b.end());
    return b;
}
bool isPali(string &a){
    int len=a.size();
    for(int i=0;i<len/2;i++){
        if(a[i]==a[len-1-i]){continue;}
        else{return false;}
    }
    return true;
}
int main()
{
    cin>>n>>k;
    int c=k;
    string result=n;
    if(isPali(result)) {cout<<result<<endl<<"0";return 0;}
    while(c--){
        result=strAdd(result);
        if(isPali(result))break;
    }
    if(c<=0){
        cout<<result<<endl;
        cout<<k;
    }else{
        cout<<result<<endl;
        cout<<(k-c);
    }
    return 0;
}