B 题非常有意思，C 题的难度感觉不应该放在C题，D 题构造头皮都抠破了

A - Palindromic Indices

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    string s;
    cin >> n >> s;
    int r = (n + 1) / 2, ans = 1;
    while (s[r] == s[n / 2] && r < n) {
      ans += 2;
      r++;
    }
    cout << ans - (n % 2 == 0) << '\n';      
  }
  return 0;
}

B - AND Sorting

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    int ans = (int)(1ll << 32) - 1;
    for (int i = 0; i < n; i++) {
      if (a[i] != i) {
        ans &= (a[i] & i);
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

C - LIS or Reverse LIS?

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    map<int, int> mp1, mp2;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      if (!mp1.count(x) && !mp2.count(x)) {
        mp1[x]++;
      } else {
        mp2[x]++;
        mp1.erase(x);
      }
    }
    cout << (int) (mp1.size() + 1) / 2 + (int) mp2.size() << '\n';
  }  
  return 0;
}

D - Circular Spanning Tree

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    string s;
    cin >> n >> s;
    int root = -1;
    for (int i = 0; i < n; i++) {
      if (s[i] == '1') {
        root = (i + 1) % n;
        break;
      }
    }
    if (root == -1) {
      cout << "NO\n";
      continue;
    }
    vector<bool> adj(n, true);
    for (int i = root + 1; i < n; i++) {
      if (s[i] == '0') {
        adj[(i + 1) % n] = false;
      }
    }
    for (int i = 0; i < root - 1; i++) {
      if (s[i] == '0') {
        adj[i + 1] = false;
      }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      if (i == root) {
        continue;
      }
      ans += (int) adj[i];
    }
    if (s[root] - '0' != (ans & 1)) {
      cout << "NO\n";
      continue;
    }
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      if (i == root) {
        continue;
      }
      if (adj[i]) {
        cout << i + 1 << ' ' << root + 1 << '\n';
      } else {
        cout << (i - 1 + n) % n + 1 << ' ' << i + 1 << '\n';
      }
    }
  }
  return 0;
}