2^x-2^y其实可以枚举,O(30*30),可以过

 for (ll j = 0; j < 30; j++)
        {
            for (ll k = 0; k < 30; k++)
            {
                if (dif == (1 << j) - (1 << k))
                {
                    ge[j]++;
                    sh[k]++;
                    flag = 1;
                    break;
                }
            }
            if (flag)
                break;
        }
        if (flag == 0)
        {
            cout << "NO" << endl;
            return;
        }