Merge remote-tracking branch 'origin/drafts'

Author: Tramis

src/algorithm/砍树.md

@@ -0,0 +1,33 @@
+# 砍树
+
+现在有 $n$ 颗树，$a_i$ 表示树的初始高度，$b_i$ 表示树的每天生长速度，砍一棵树的代价是树的总高度
+
+## 1. 每天砍一棵，总代价最小
+
+应当每天砍一棵生长速度最快的
+
+## 2. 不砍树，问前 $n$ 天最高的树的高度分别是多少
+
+```
+生 1 3 5 7
+初 4 3 2 1
+```
+
+先按生长速度排序，再维护一个单调队列，保证队列中每棵树超过前一棵树的日期递增
+
+首先，显然队列中的初始高度递减，否则下面的第一棵树总是不会超过第二棵树
+
+```
+生 1 3
+初 4 5
+```
+
+计算超过的时间 $a_{i+1} + tb_{i+1} = a_i + t b_i \to t = \frac {a_{i+1} - a_i} {b_i - b_{i+1}}$
+
+可知队列中的 $t$ 总应该是正的
+
+## 3. 每天可以免费锯一棵树。必须先砍后锯
+
+
+
+## 4. 奇数天可以免费锯一棵树。必须先砍后锯

src/math/拉格朗日乘数法.md

@@ -108,3 +108,44 @@ $$
 
 于是得到拉格朗日乘数法的新约束
 
+## 一个例题
+
+求 $|z|$ 在
+$$
+\begin{cases}
+    x^2+9y^2 - z^2 = 0\\
+    x + 3y + 2z = 5
+\end{cases}
+$$
+
+约束下的最大值和最小值
+
+---
+
+此题有两个优化
+
+- $|z|$ 与 $z^2$ 的最值在同一点取到，可以去掉绝对值
+- 不必将 $z$ 看作 $z(x, y)$ 的函数，而是取 $f(x, y, z) = z^2$ 视为三元函数
+
+两个约束条件写为
+
+$$
+\begin{cases}
+    F(x, y, z) = 0\\
+    G(x, y, z) = 0
+\end{cases}
+$$
+
+于是可以取拉格朗日函数：
+
+$$
+L(x, y, z, \lambda_1, \lambda_2) = f - \lambda_1 F  - \lambda_2 G
+$$
+
+## 问
+
+拉格朗日函数 $L = f - \sum \lambda_i g$ 能不能写作 $L = f + \sum \lambda_i g$
+
+--- 
+
+可以，因为不改变切空间

src/math/矩阵乘法.ipynb

@@ -71,6 +71,61 @@
    "source": [
     "a * b"
    ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "ename": "KeyError",
+     "evalue": "'push1'",
+     "output_type": "error",
+     "traceback": [
+      "\u001b[1;31m---------------------------------------------------------------------------\u001b[0m",
+      "\u001b[1;31mKeyError\u001b[0m                                  Traceback (most recent call last)",
+      "Cell \u001b[1;32mIn[2], line 9\u001b[0m\n\u001b[0;32m      7\u001b[0m \u001b[38;5;28;01mfor\u001b[39;00m op \u001b[38;5;129;01min\u001b[39;00m ops:\n\u001b[0;32m      8\u001b[0m     \u001b[38;5;28;01mif\u001b[39;00m op\u001b[38;5;241m.\u001b[39mstartswith(\u001b[38;5;124m\"\u001b[39m\u001b[38;5;124mpush\u001b[39m\u001b[38;5;124m\"\u001b[39m):\n\u001b[1;32m----> 9\u001b[0m         \u001b[38;5;28;01mif\u001b[39;00m \u001b[43mnow\u001b[49m\u001b[43m[\u001b[49m\u001b[43mop\u001b[49m\u001b[43m]\u001b[49m \u001b[38;5;129;01mis\u001b[39;00m \u001b[38;5;28;01mNone\u001b[39;00m:\n\u001b[0;32m     10\u001b[0m             now[op] \u001b[38;5;241m=\u001b[39m {}\n\u001b[0;32m     11\u001b[0m         prev\u001b[38;5;241m.\u001b[39mpush(now)\n",
+      "\u001b[1;31mKeyError\u001b[0m: 'push1'"
+     ]
+    }
+   ],
+   "source": [
+    "ops = [\"push1\", \"push2\", \"pop\"]\n",
+    "\n",
+    "prev = []\n",
+    "root = {}\n",
+    "now = root\n",
+    "\n",
+    "for op in ops:\n",
+    "    if op.startswith(\"push\"):\n",
+    "        if op not in now:\n",
+    "            now[op] = {}\n",
+    "        prev.push(now)\n",
+    "        now = now[op]\n",
+    "    else:\n",
+    "        now = prev.back()\n",
+    "        prev.pop()\n",
+    "\n",
+    "total = 0\n",
+    "\n",
+    "def cnt(now):\n",
+    "    total += 1\n",
+    "    \n",
+    "    if now is None:\n",
+    "        return\n",
+    "    for child in now:\n",
+    "        cnt(child)\n",
+    "\n",
+    "cnt(root)\n",
+    "print(total)\n"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": []
   }
  ],
  "metadata": {