小波矩阵树：高效静态区间第 K 大查询
手写 bitset 压位，以 \(\mathcal O(N \log N)\) 的时间复杂度和 \(\mathcal O(N + \frac{N \log N}{64})\) 的空间建树后，实现单次 \(\mathcal O(\log N)\) 复杂度的区间第 \(k\) 大值询问。建议使用 \(\texttt{0-idx}\) 计数法，但是经测试 \(\texttt{1-idx}\) 也有效，但需要更多的检验。
#define __count(x) __builtin_popcountll(x)
struct Wavelet {vector<int> val, sum;vector<u64> bit;int t, n;int getSum(int i) {return sum[i >> 6] + __count(bit[i >> 6] & ((1ULL << (i & 63)) - 1));}Wavelet(vector<int> v) : val(v), n(v.size()) {sort(val.begin(), val.end());val.erase(unique(val.begin(), val.end()), val.end());int n_ = val.size();t = __lg(2 * n_ - 1);bit.resize((t * n + 64) >> 6);sum.resize(bit.size());vector<int> cnt(n_ + 1);for (int &x : v) {x = lower_bound(val.begin(), val.end(), x) - val.begin();cnt[x + 1]++;}for (int i = 1; i < n_; ++i) {cnt[i] += cnt[i - 1];}for (int j = 0; j < t; ++j) {for (int i : v) {int tmp = i >> (t - 1 - j);int pos = (tmp >> 1) << (t - j);auto setBit = [&](int i, u64 v) {bit[i >> 6] |= (v << (i & 63));};setBit(j * n + cnt[pos], tmp & 1);cnt[pos]++;}for (int i : v) {cnt[(i >> (t - j)) << (t - j)]--;}}for (int i = 1; i < sum.size(); ++i) {sum[i] = sum[i - 1] + __count(bit[i - 1]);}}int small(int l, int r, int k) {r++;for (int j = 0, x = 0, y = n, res = 0;; ++j) {if (j == t) return val[res];int A = getSum(n * j + x), B = getSum(n * j + l);int C = getSum(n * j + r), D = getSum(n * j + y);int ab_zeros = r - l - C + B;if (ab_zeros > k) {res = res << 1;y -= D - A;l -= B - A;r -= C - A;} else {res = (res << 1) | 1;k -= ab_zeros;x += y - x - D + A;l += y - l - D + B;r += y - r - D + C;}}}int large(int l, int r, int k) {return small(l, r, r - l - k);}
};
