题解 | #数字反转#

#include<bits/stdc++.h>
using namespace std;


int diandao(int c){
    if(c/100000!=0)
    return c%10*100000+c%100/10*10000+c/100000+c%10000/1000*100+c%1000/100*1000+c%10000/1000*10;
    else if(c/10000!=0)
    return c%10*10000+c%100/10*1000+c/10000+c%10000/1000*10+c%1000/100*100;
    else if(c/1000!=0)
    return c/1000+c%1000/100*10+c%100/10*100+c%10*1000;
    else if(c/100!=0)
    return c/100+c%100/10*10+c%10*100;
     else if(c/10!=0)
    return c/10+c%10*10;
    else return c;
}
int main() {
    int a=0;
    while(cin>>a)
    {
         int b=0;
    cin>>b;

   int c=a+b;
   if(diandao(a)+diandao(b)==diandao(c))
   cout<<c<<endl;
   else cout<<"NO"<<endl;
    }
   
   
    return 0;
}