diff --git a/content/post/CF_997C.md b/content/post/CF_997C.md
index 3d101e5..27bdb32 100644
--- a/content/post/CF_997C.md
+++ b/content/post/CF_997C.md
@@ -1,5 +1,5 @@
 +++
-title = 'CF 997C Sky Full of Stars(二项式反演，DP）'
+title = 'CF 997C Sky Full of Stars（二项式反演，DP）'
 date = 2021-03-18T10:47:44+08:00
 categories = ['题解']
 tags = ['二项式反演', 'DP']
diff --git a/content/post/Luogu_P2824.md b/content/post/Luogu_P2824.md
new file mode 100644
index 0000000..4039a95
--- /dev/null
+++ b/content/post/Luogu_P2824.md
@@ -0,0 +1,177 @@
++++
+title = 'Luogu P2824 [HEOI2016/TJOI2016]排序（线段树，二分答案）'
+date = 2021-03-23T10:50:37+08:00
+categories = ['题解']
+tags = ['线段树', '二分答案']
++++
+
+[题目链接](https://www.luogu.com.cn/problem/P2824)
+
+{{% question %}}
+
+给出一个 $1$ 到 $n$ 的排列，现在对这个排列序列进行 $m$ 次局部排序，排序分为两种：
+
+- `0 L R` 表示将区间 $[L, R]$ 的数字升序排序
+- `1 L R` 表示将区间 $[L, R]$ 的数字降序排序
+
+注意，这里是对下标在区间 $[L, R]$ 内的数排序
+
+最后询问第 $q$ 位置上的数字
+
+$ n, m \leq 10^5$
+
+{{% /question %}}
+
+<!--more-->
+
+## 简要做法
+
+考虑二分答案，对于 $mid$，将原排列中 $\geq mid$ 的数标记为 $1$，$< mid$ 的数标记为 $0$
+
+问题转化为二分一个最大的 $mid$，使得排序后 $mid$ 位置上的数字是 $1$
+
+考虑处理排序
+
+对于区间 $[L, R]$，记区间 $[L, R]$ 内 $1$ 的个数为 $cnt$
+
+若为升序排序，则 
+
+$$
+\begin{aligned}
+\begin{cases}
+	&[L, R - cnt] \leftarrow 0\\\\
+	\\\\
+	&[R - cnt + 1, R] \leftarrow 1
+\end{cases}
+\end{aligned}
+$$
+
+若为降序排序，则 
+
+$$
+\begin{aligned}
+\begin{cases}
+	&[L + cnt, R] \leftarrow 0\\\\
+	\\\\
+	&[L, L + cnt - 1] \leftarrow 1
+\end{cases}
+\end{aligned}
+$$
+
+发现需要一种同时支持快速 区间查询 区间修改 单点查询 的数据结构
+
+使用线段树即可，复杂度 $O(m \log^2n)$
+
+## 参考代码
+
+```cpp
+#include <stdio.h>
+#include <algorithm>
+#include <memory.h>
+
+int read(int x = 0, int f = 0, char ch = getchar())
+{
+    while ('0' > ch or ch > '9')
+        f = ch == '-', ch = getchar();
+    while ('0' <= ch and ch <= '9')
+        x = x * 10 + (ch ^ 48), ch = getchar();
+    return f ? -x : x;
+}
+
+const int N = 1e6 + 5;
+
+int n, m, p, ans;
+int a[N];
+int opt[N], L[N], R[N];
+
+struct Seg_Tree
+{
+#define ls (p << 1)
+#define rs (p << 1 | 1)
+#define mid ((l + r) >> 1)
+    int tag[N << 2], sum[N << 2];
+    void push_up(int p) { sum[p] = sum[ls] + sum[rs]; }
+    void push_down(int p, int l, int r)
+    {
+        if (!tag[p])
+            return;
+        tag[ls] = tag[rs] = tag[p];
+        if (tag[p] == 1)
+            sum[ls] = mid - l + 1, sum[rs] = r - mid;
+        else
+            sum[ls] = sum[rs] = 0;
+        tag[p] = 0;
+    }
+    void build(int p, int l, int r, int x)
+    {
+        tag[p] = 0;
+        if (l == r)
+            return sum[p] = a[l] >= x, tag[p] = 0, void();
+        build(ls, l, mid, x), build(rs, mid + 1, r, x), push_up(p);
+    }
+    int query(int p, int l, int r, int x, int y)
+    {
+        if (x == l and r == y)
+            return sum[p];
+        push_down(p, l, r);
+        int res = 0;
+        if (x <= mid)
+            res += query(ls, l, mid, x, std::min(mid, y));
+        if (mid < y)
+            res += query(rs, mid + 1, r, std::max(mid + 1, x), y);
+        return res;
+    }
+    void modify(int p, int l, int r, int x, int y, int val)
+    {
+        if (x > r or y < l)
+            return;
+        if (x == l and r == y)
+            return sum[p] = val * (r - l + 1), tag[p] = val ? 1 : -1, void();
+        push_down(p, l, r);
+        if (x <= mid)
+            modify(ls, l, mid, x, std::min(mid, y), val);
+        if (mid < y)
+            modify(rs, mid + 1, r, std::max(mid + 1, x), y, val);
+        push_up(p);
+    }
+#undef ls
+#undef rs
+#undef mid
+} T;
+
+bool judge(int x)
+{
+    T.build(1, 1, n, x);
+    for (int i = 1; i <= m; i++)
+    {
+        int cnt = T.query(1, 1, n, L[i], R[i]);
+        if (opt[i] == 0)
+        {
+            T.modify(1, 1, n, R[i] - cnt + 1, R[i], 1);
+            T.modify(1, 1, n, L[i], R[i] - cnt, 0);
+        }
+        else
+        {
+            T.modify(1, 1, n, L[i], L[i] + cnt - 1, 1);
+            T.modify(1, 1, n, L[i] + cnt, R[i], 0);
+        }
+    }
+    return T.query(1, 1, n, p, p);
+}
+
+int main()
+{
+    n = read(), m = read();
+    for (int i = 1; i <= n; i++)
+        a[i] = read();
+    for (int i = 1; i <= m; i++)
+        opt[i] = read(), L[i] = read(), R[i] = read();
+    p = read();
+    for (int L = 1, R = n, mid; L <= R;)
+        if (judge(mid = L + R >> 1))
+            ans = mid, L = mid + 1;
+        else
+            R = mid - 1;
+    return printf("%d", ans), 0;
+}
+```
diff --git a/content/post/Luogu_P5505.md b/content/post/Luogu_P5505.md
index 89e0590..95f5f59 100644
--- a/content/post/Luogu_P5505.md
+++ b/content/post/Luogu_P5505.md
@@ -1,5 +1,5 @@
 +++
-title = 'Luogu P5505 [JSOI2011]分特产(二项式反演，DP）'
+title = 'Luogu P5505 [JSOI2011]分特产（二项式反演，DP）'
 date = 2021-03-15T20:39:45+08:00
 categories = ['题解']
 tags = ['二项式反演', 'DP']
diff --git a/content/post/Luogu_P6478.md b/content/post/Luogu_P6478.md
index 0b8ac2b..2d6ae98 100644
--- a/content/post/Luogu_P6478.md
+++ b/content/post/Luogu_P6478.md
@@ -1,5 +1,5 @@
 +++
-title = 'Luogu P6478 [NOI Online #2 提高组] 游戏(二项式反演，树形DP）'
+title = 'Luogu P6478 [NOI Online #2 提高组] 游戏（二项式反演，树形DP）'
 date = 2021-03-16T18:19:45+08:00
 categories = ['题解']
 tags = ['二项式反演', 'DP', '树形DP']