CF1088D Ehab and another another xor problem

思路：

根据异或的性质一位一位来搞。参考了https://blog.lucien.ink/archives/362/

实现：

 #include <bits/stdc++.h>
 using namespace std;
 void out(char x, int a, int b)
 {
     cout << x << " " << a << " " << b << endl;
     fflush(stdout);
 }
 int main()
 {
     int a, b;
     char big = 'a';
     out('?', , );
     cin >> a;
     if (a == -) big = 'b';
     int x = , y = ;
     for (int i = ; i >= ; i--)
     {
         int t =  << i;
         x |= t; y &= ~t;
         out('?', x, y);
         cin >> a;
         x ^= t; y ^= t;
         out('?', x, y);
         cin >> b;
         if (a == - && b == )
         {
             x |= t; y |= t;
         }
         else if (a ==  && b == -)
         {
             x &= ~t; y &= ~t;
         }
         else if (a == b)
         {
             if (big == 'a')
             {
                 x |= t; y &= ~t;
                 if (a == -) big = 'b';
             }
             else
             {
                 x &= ~t; y |= t;
                 if (a == ) big = 'a';
             }
         }
     }
     out('!', x, y);
     return ;
 }