update

_config.fluid.yml

@@ -918,18 +918,24 @@ links:
         link: "https://www.cnblogs.com/chinakevin",
         avatar: "/img/chinakevin.jpg",
       }
-    - {
-        title: "montreal",
-        intro: "天下第一寝室长",
-        link: "https://blog.moontreal.cf",
-        avatar: "/img/montreal.jpg",
-      }
     - {
         title: "lllllan",
         intro: "苏老师太强啦",
         link: "https://blog.lllllan.cn",
         avatar: "/img/lllllan.png",
       }
+    - {
+        title: "tangent",
+        intro: "谭老板 Orz",
+        link: "https://tangent1231.com/",
+        avatar: "/img/tangent.svg",
+      }
+    - {
+        title: "light",
+        intro: "信院第一人",
+        link: "https://lllighter.com",
+        avatar: "/img/light.jpeg",
+      }
 
   # 当成员头像加载失败时，替换为指定图片
   # When the member avatar fails to load, replace the specified image

source/_posts/oi/notes-on-math-theory.md

@@ -14,7 +14,7 @@ date: 2017-07-04 10:00:35
 
 ### Lucas
 
-$n=(a_ka_{k-1}a_{k-2}\ldots a_0)_p, \ m=(b_kb_{k-1}b_{k-2}\ldots b_0)_p, \ {n \choose m} \equiv \prod_{i=0}^k {a_i \choose b_i} \pmod p$.
+$n=(a_ka_{k-1}a_{k-2}\ldots a_0)\_p, \ m=(b_kb_{k-1}b_{k-2}\ldots b_0)\_p, \ {n \choose m} \equiv \prod_{i=0}^k {a_i \choose b_i} \pmod p$.
 
 $n!$中$p$的次数$=\sum_{i=1}^{\infty} [{n \over p^i}]={n-S_p(n) \over p-1}$
 
@@ -238,7 +238,7 @@ $({2 \over p})=(-1)^{p^2-1 \over 8}=(-1)^{[{p \over 2}]-[{p \over 4}]}$.
 
 ### Summary
 
-$n=(a_ka_{k-1}a_{k-2}\ldots a_0)_p, m=(b_kb_{k-1}b_{k-2}\ldots b_0)_p, {n \choose m} \equiv \prod_{i=0}^k {a_i \choose b_i} \pmod p$.
+$n=(a_ka_{k-1}a_{k-2}\ldots a_0)\_p, m=(b_kb_{k-1}b_{k-2}\ldots b_0)\_p, {n \choose m} \equiv \prod_{i=0}^k {a_i \choose b_i} \pmod p$.
 
 $x^{\varphi(n)} \equiv 1 \pmod n, (x, n)=1$.
 

source/_posts/oi/tsr-and-variance.md

@@ -15,21 +15,21 @@ Tsr is a cute boy with handsome moustache.
 
 You are given a sequence with length $n$. Tsr wants you to calculate the sum of variance of each successive subsequence. Note: The variance in this problem should't be divided by length.
 
-Recall $\overline a_{l, r}={1 \over r-l+1} \sum_{i=l}^r a_i$. Then you are supposed to calculate $\sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k-\overline a_{i, j})^2$.
+Recall $\overline{a}\_{l, r}={1 \over r-l+1} \sum_{i=l}^r a_i$. Then you are supposed to calculate $\sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k-\overline{a}\_{i, j})^2$.
 
 ## 题意概述
 
-给定一个长度为$n$的序列，求$\sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k-\overline a_{i, j})^2$。有$T$组数据。
+给定一个长度为$n$的序列，求$\sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k-\overline{a}\_{i, j})^2$。有$T$组数据。
 
 数据范围：$1 \le T \le 20, \ 1 \le n \le 10000, \ 1 \le a_i \le 10$。
 
 ## 算法分析
 
-$$\begin{align} \sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k-\overline a_{i, j})^2 &= \sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k^2-2a_k \overline a_{i, j}+\overline a_{i, j}^2) \\\\ &= \sum_{k=1}^n \sum_{i=1}^k \sum_{j=k}^n a_k^2-2\sum_{i=1}^n \sum_{j=i}^n \overline a_{i, j} \sum_{k=i}^j a_k+ \sum_{i=1}^n \sum_{j=i}^n (j-i+1) \overline a_{i, j}^2 \\\\ &= \sum_{k=1}^n k(n-k+1)a_k^2-\sum_{i=1}^n \sum_{j=i}^n (j-i+1) \overline a_{i, j}^2 \end{align}$$
+$$\begin{align} \sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k-\overline{a}\_{i, j})^2 &= \sum_{i=1}^n \sum_{j=i}^n \sum_{k=i}^j (a_k^2-2a_k \overline{a}\_{i, j}+\overline{a}\_{i, j}^2) \\\\ &= \sum_{k=1}^n \sum_{i=1}^k \sum_{j=k}^n a_k^2-2\sum_{i=1}^n \sum_{j=i}^n \overline{a}\_{i, j} \sum_{k=i}^j a_k+ \sum_{i=1}^n \sum_{j=i}^n (j-i+1) \overline{a}\_{i, j}^2 \\\\ &= \sum_{k=1}^n k(n-k+1)a_k^2-\sum_{i=1}^n \sum_{j=i}^n (j-i+1) \overline{a}\_{i, j}^2 \end{align}$$
 
 前一项可以在$O(n)$时间内求出，只需考虑如何求后一项。令$s_i=\sum_{j=1}^i a_j$。
 
-$$\begin{align} \sum_{i=1}^n \sum_{j=i}^n (j-i+1) \overline a_{i, j}^2  &= \sum_{j=1}^n \sum_{i=1}^j {(s_j-s_{i-1})^2 \over j-i+1} \\\\ &= \sum_{j=1}^n \sum_{i=1}^j {s_j^2-2s_js_{i-1}+s_{i-1}^2 \over j-i+1} \\\\ &= \sum_{j=1}^n s_j^2 \sum_{i=1}^j {1 \over i}-2\sum_{j=1}^n s_j \sum_{i=1}^j {s_{i-1} \over j-(i-1)}+\sum_{j=1}^n \sum_{i=1}^j {s_{i-1}^2 \over j-(i-1)} \end{align}$$
+$$\begin{align} \sum_{i=1}^n \sum_{j=i}^n (j-i+1) \overline{a}\_{i, j}^2  &= \sum_{j=1}^n \sum_{i=1}^j {(s_j-s_{i-1})^2 \over j-i+1} \\\\ &= \sum_{j=1}^n \sum_{i=1}^j {s_j^2-2s_js_{i-1}+s_{i-1}^2 \over j-i+1} \\\\ &= \sum_{j=1}^n s_j^2 \sum_{i=1}^j {1 \over i}-2\sum_{j=1}^n s_j \sum_{i=1}^j {s_{i-1} \over j-(i-1)}+\sum_{j=1}^n \sum_{i=1}^j {s_{i-1}^2 \over j-(i-1)} \end{align}$$
 
 第一项也可以在$O(n)$时间内求出，而后面两项都是卷积的形式，可以用 FFT 在$O(n\log n)$时间内求出。