There only 10 situations!!!!!
//1. in
//2. input
//3. inputon
//4. inputone

//5. out
//6. output
//7. outputon
//8. outputone

//9. puton
//10. one

Check their in inverse order.

   bool check(char const* s){
        while(*s != '\n'){
               if (s == "one") s += 3;
              // there s == "one" only pseudocode, you may check
             //  as memcmp(s, "one",3) == 0
               else  // ......

               //.............

               // 10 times else
               else if (s == "in") s +=2;
               else
                  return false;
         }
        return true;
   }

Is it DFA?

thnx a lot  c_pp
atleast i learn something