DISCO Presents Discovery Channel Code Contest 2020 Qual Task E. Majority of Balls

Not able to solve this problem during the contest (virtual participation).

The first observation is that if we can identify \(N-1\) balls of which half is blue and the other half is red, then with these \(N - 1\) balls we can identify the color of the rest \(N+1\) balls.

It's not hard to note that if there are more blue balls than red balls among balls numbered from \(l\) to \(l + N - 1\) but less among balls numbered from \(l + 1\) to \(l + N\), we then know that

1. The \(l\)-th ball and the \(l+N\)-th ball must be of different colors.
2. The \(l\)-th ball must be blue and the \(l+N\)-th ball must be red.
3. There are equal number of blue and red balls among balls numbered from \(l + 1\) to \(l+N - 1\).

The problem is whether such \(l\) even exists? The answer is ,fortunately, YES.

Here comes the second interesting observation:
Let's assume without loss of generality, there are more blue balls than red balls among the first \(N\) balls, then there must be more red balls among the last \(N\) balls. So such \(l\) as described above must exist, and we can find one using binary search which costs at most \(\log_2 N + 1\) queries. When the binary search finishes, we get \(l\) and the color of the \(l\)-th and \(l+N\)-th balls.

When \(l\) is found, for each ball numbered from \(1\) to \(l - 1\) or from \(l + N + 1\) to \(2N\), we can know its color with a query. Note that exactly half of these \(N - 1\) known balls are blue, so we can use these balls to identify color of balls numbered from \(l + 1\) to \(l + N -1\) in a similar way.

code

int main() {
  int n;
  cin >> n;
  auto ask = [&](int l) {
    cout << '?';
    for (int i = 0; i < n; i++) {
      cout << ' ' << l + i;
    }
    cout << endl;
    string res;
    cin >> res;
    return res.front();
  };
  char ml = ask(1);
  int l = 1, r = n + 1;
while (r - l > 1) {

int mid = l + (r - l) / 2;

if (ask(mid) == ml) {

l = mid;

} else {

r = mid;

}

}

vector ans(2 * n + 1);

ans[l] = ml;

ans[l + n] = ml == 'R' ? 'B' : 'R';
auto ask2 = [&](int pos) {

cout << '?';

for (int i = 1; i < n; i++) {

cout << ' ' << l + i;

}

cout << ' ' << pos << endl;

string res;

cin >> res;

return res.front();

};

// [l + 1, l + n)

rng (i, 1, 2 * n + 1) {

if (i < l || i > l + n) {

ans[i] = ask2(i);

}

}

auto ask3 = [&](int pos) {

cout << '?';

rng (i, 1, 2 * n + 1) {

if (i < l || i > l + n) {

cout << ' ' << i;

}

}

cout << ' ' << pos << endl;

string res;

cin >> res;

return res.front();

};

rng (i, l + 1, l + n) {

ans[i] = ask3(i);

}
cout << "! ";

for (int i = 1; i <= 2 * n; i++) {

cout << ans[i];

}

cout << '\n';

return 0;

}