add introduction of the algorithms for A B C D G H

A.ipynb

@@ -93,11 +93,11 @@
    ]
   },
   {
-   "cell_type": "code",
-   "execution_count": null,
+   "cell_type": "markdown",
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "source": [
+    "简单模拟即可"
+   ]
   }
  ],
  "metadata": {

B.ipynb

@@ -111,11 +111,11 @@
    ]
   },
   {
-   "cell_type": "code",
-   "execution_count": null,
+   "cell_type": "markdown",
    "metadata": {},
-   "outputs": [],
-   "source": []
+   "source": [
+    "找到规律，发现环保数为奇数或者4的倍数（4本身除外），然后用二分搜索进行查找即可"
+   ]
   }
  ],
  "metadata": {

C.ipynb

@@ -1 +1 @@
-{"cells":[{"cell_type":"code","execution_count":2,"metadata":{},"outputs":[{"name":"stdout","output_type":"stream","text":["5.000000\n"]}],"source":["import math\n","\n","# 定义函数，计算最小值\n","def calculate_min_value(a, b):\n","    # 根据给出的表达式 x = (sqrt(a) * b) / (sqrt(a) + 1)\n","    x = (math.sqrt(a) * b) / (math.sqrt(a) + 1)\n","    \n","    # 根据公式 f(x) = sqrt(x^2 + a) 和 g(x) = sqrt((b - x)^2 + 1)\n","    f_x = math.sqrt(x**2 + a)\n","    g_x = math.sqrt((b - x)**2 + 1)\n","    \n","    # 返回 f(x) + g(x)\n","    return f_x + g_x\n","\n","# 读取输入\n","a, b = map(float, input().split())\n","\n","# 计算并输出结果，保留小数点后 6 位\n","result = calculate_min_value(a, b)\n","print(f\"{result:.6f}\")"]},{"cell_type":"code","execution_count":3,"metadata":{},"outputs":[{"name":"stdout","output_type":"stream","text":["5.000000\n"]}],"source":["from math import sqrt\n","\n","# 定义函数，计算最小值\n","def calculate_min_value(a, b):\n","    ret = sqrt(b**2 + a + 2*sqrt(a) + 1)\n","    return ret\n","\n","# 读取输入\n","a, b = map(float, input().split())\n","\n","# 计算并输出结果，保留小数点后 6 位\n","result = calculate_min_value(a, b)\n","print(f\"{result:.6f}\")"]},{"cell_type":"code","execution_count":null,"metadata":{},"outputs":[],"source":[]}],"metadata":{"kernelspec":{"display_name":"base","language":"python","name":"python3"},"language_info":{"codemirror_mode":{"name":"ipython","version":3},"file_extension":".py","mimetype":"text/x-python","name":"python","nbconvert_exporter":"python","pygments_lexer":"ipython3","version":"3.11.5"}},"nbformat":4,"nbformat_minor":2}
+{"cells":[{"cell_type":"code","execution_count":2,"metadata":{},"outputs":[{"name":"stdout","output_type":"stream","text":["5.000000\n"]}],"source":["import math\n","\n","# 定义函数，计算最小值\n","def calculate_min_value(a, b):\n","    # 根据给出的表达式 x = (sqrt(a) * b) / (sqrt(a) + 1)\n","    x = (math.sqrt(a) * b) / (math.sqrt(a) + 1)\n","    \n","    # 根据公式 f(x) = sqrt(x^2 + a) 和 g(x) = sqrt((b - x)^2 + 1)\n","    f_x = math.sqrt(x**2 + a)\n","    g_x = math.sqrt((b - x)**2 + 1)\n","    \n","    # 返回 f(x) + g(x)\n","    return f_x + g_x\n","\n","# 读取输入\n","a, b = map(float, input().split())\n","\n","# 计算并输出结果，保留小数点后 6 位\n","result = calculate_min_value(a, b)\n","print(f\"{result:.6f}\")"]},{"cell_type":"code","execution_count":3,"metadata":{},"outputs":[{"name":"stdout","output_type":"stream","text":["5.000000\n"]}],"source":["from math import sqrt\n","\n","# 定义函数，计算最小值\n","def calculate_min_value(a, b):\n","    ret = sqrt(b**2 + a + 2*sqrt(a) + 1)\n","    return ret\n","\n","# 读取输入\n","a, b = map(float, input().split())\n","\n","# 计算并输出结果，保留小数点后 6 位\n","result = calculate_min_value(a, b)\n","print(f\"{result:.6f}\")"]},{"cell_type":"markdown","metadata":{},"source":["将军饮马问题"]}],"metadata":{"kernelspec":{"display_name":"base","language":"python","name":"python3"},"language_info":{"codemirror_mode":{"name":"ipython","version":3},"file_extension":".py","mimetype":"text/x-python","name":"python","nbconvert_exporter":"python","pygments_lexer":"ipython3","version":"3.11.5"}},"nbformat":4,"nbformat_minor":2}

D.ipynb

@@ -1 +1,352 @@
 {
+ "cells": [
+  {
+   "cell_type": "code",
+   "execution_count": 1,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "6\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 输入处理\n",
+    "n, k = map(int, input().split())\n",
+    "citizens = list(map(int, input().split()))\n",
+    "\n",
+    "# 初始滑动窗口的和\n",
+    "current_sum = sum(citizens[:k])\n",
+    "max_sum = current_sum\n",
+    "\n",
+    "# 滑动窗口遍历\n",
+    "for i in range(1, len(citizens) - k + 1):\n",
+    "    current_sum = current_sum - citizens[i - 1] + citizens[i + k - 1]\n",
+    "    max_sum = max(max_sum, current_sum)\n",
+    "\n",
+    "# 输出结果\n",
+    "print(max_sum)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 2,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "5\n"
+     ]
+    }
+   ],
+   "source": [
+    "# 输入处理\n",
+    "n, k = map(int, input().split())\n",
+    "\n",
+    "# 构建上三角矩阵\n",
+    "citizens = []\n",
+    "for i in range(n - 1):\n",
+    "    citizens.append(list(map(int, input().split())))\n",
+    "\n",
+    "# 计算最大总和\n",
+    "max_sum = 0\n",
+    "\n",
+    "# 遍历所有可能的起始路口\n",
+    "for i in range(n - k):\n",
+    "    current_sum = 0\n",
+    "    for j in range(k):\n",
+    "        current_sum += citizens[i][j]  # 累加相邻的市民数量\n",
+    "    \n",
+    "    max_sum = max(max_sum, current_sum)\n",
+    "\n",
+    "# 输出结果\n",
+    "print(max_sum)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 3,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "6\n"
+     ]
+    }
+   ],
+   "source": [
+    "def max_citizens(n, k, citizens_matrix):\n",
+    "    # 将所有相邻路口之间的市民数量拉平成一个列表\n",
+    "    citizens = []\n",
+    "    for i in range(n - 1):\n",
+    "        citizens += citizens_matrix[i]\n",
+    "    \n",
+    "    # 使用滑动窗口，计算所有长度为 k 的连续子区间的和\n",
+    "    max_sum = 0\n",
+    "    current_sum = sum(citizens[:k])\n",
+    "    max_sum = current_sum\n",
+    "    \n",
+    "    for i in range(1, len(citizens) - k + 1):\n",
+    "        current_sum = current_sum - citizens[i - 1] + citizens[i + k - 1]\n",
+    "        max_sum = max(max_sum, current_sum)\n",
+    "    \n",
+    "    return max_sum\n",
+    "\n",
+    "# 读取输入\n",
+    "n, k = map(int, input().split())\n",
+    "citizens_matrix = []\n",
+    "for i in range(n - 1):\n",
+    "    citizens_matrix.append(list(map(int, input().split())))\n",
+    "\n",
+    "# 计算并输出结果\n",
+    "result = max_citizens(n, k, citizens_matrix)\n",
+    "print(result)"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "metadata": {},
+   "outputs": [],
+   "source": [
+    "def max_citizens(n, k, C_input):\n",
+    "    # Initialize C[s][t] as a 2D list\n",
+    "    C = [[0] * (n + 1) for _ in range(n + 1)]\n",
+    "    idx = 0\n",
+    "    for s in range(1, n):\n",
+    "        for t in range(s + 1, n + 1):\n",
+    "            C[s][t] = C_input[idx]\n",
+    "            idx += 1\n",
+    "\n",
+    "    # Compute pre_sum[s][x] = sum of C[s][x] to C[s][n]\n",
+    "    pre_sum = [[0] * (n + 2) for _ in range(n + 1)]\n",
+    "    for s in range(1, n):\n",
+    "        pre_sum[s][n] = C[s][n]\n",
+    "        for t in range(n - 1, s, -1):\n",
+    "            pre_sum[s][t] = C[s][t] + pre_sum[s][t + 1]\n",
+    "\n",
+    "    max_total = 0\n",
+    "    # Iterate over all possible windows\n",
+    "    for l in range(1, n - k + 1):\n",
+    "        window_end = l + k - 1\n",
+    "        current_sum = 0\n",
+    "        for s in range(1, window_end + 1):\n",
+    "            current_sum += pre_sum[s][l + 1]\n",
+    "        if current_sum > max_total:\n",
+    "            max_total = current_sum\n",
+    "\n",
+    "    return max_total\n",
+    "\n",
+    "# 读取输入\n",
+    "import sys\n",
+    "\n",
+    "def main():\n",
+    "    import sys\n",
+    "    input = sys.stdin.read\n",
+    "    data = input().split()\n",
+    "    n = int(data[0])\n",
+    "    k = int(data[1])\n",
+    "    C_input = list(map(int, data[2:]))\n",
+    "    result = max_citizens(n, k, C_input)\n",
+    "    print(result)\n",
+    "\n",
+    "if __name__ == \"__main__\":\n",
+    "    main()"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 6,
+   "metadata": {},
+   "outputs": [
+    {
+     "name": "stdout",
+     "output_type": "stream",
+     "text": [
+      "14\n"
+     ]
+    }
+   ],
+   "source": [
+    "def max_citizens(n, k, C_input):\n",
+    "    # 初始化二维列表 C[s][t]\n",
+    "    C = [[0] * (n + 1) for _ in range(n + 1)]\n",
+    "    idx = 0\n",
+    "    for s in range(1, n):\n",
+    "        for t in range(s + 1, n + 1):\n",
+    "            C[s][t] = C_input[idx]\n",
+    "            idx += 1\n",
+    "\n",
+    "    # 计算 pre_sum[s][t] = 从 t 到 n 的 C[s][t] 之和\n",
+    "    pre_sum = [[0] * (n + 2) for _ in range(n + 1)]\n",
+    "    for s in range(1, n):\n",
+    "        pre_sum[s][n] = C[s][n]\n",
+    "        for t in range(n - 1, s, -1):\n",
+    "            pre_sum[s][t] = C[s][t] + pre_sum[s][t + 1]\n",
+    "\n",
+    "    max_total = 0\n",
+    "    # 遍历所有可能的窗口\n",
+    "    for l in range(1, n - k + 1):\n",
+    "        window_end = l + k - 1\n",
+    "        current_sum = 0\n",
+    "        for s in range(1, window_end + 1):\n",
+    "            current_sum += pre_sum[s][l + 1]\n",
+    "        if current_sum > max_total:\n",
+    "            max_total = current_sum\n",
+    "\n",
+    "    return max_total\n",
+    "\n",
+    "def main():\n",
+    "    # 读取第一行：n 和 k\n",
+    "    first_line = input().strip().split()\n",
+    "    n = int(first_line[0])\n",
+    "    k = int(first_line[1])\n",
+    "\n",
+    "    # 读取接下来的 n-1 行市民数量数据\n",
+    "    C_input = []\n",
+    "    for i in range(n - 1):\n",
+    "        line = input().strip()\n",
+    "        if line:\n",
+    "            nums = list(map(int, line.split()))\n",
+    "            C_input.extend(nums)\n",
+    "\n",
+    "    # 计算结果并输出\n",
+    "    result = max_citizens(n, k, C_input)\n",
+    "    print(result)\n",
+    "\n",
+    "if __name__ == \"__main__\":\n",
+    "    main()"
+   ]
+  },
+  {
+   "cell_type": "markdown",
+   "metadata": {},
+   "source": [
+    "当然，可以将程序调整为逐行从键盘读取输入。以下是调整后的完整 Python 代码，它逐行读取输入并计算受到宣传的最大市民数。\n",
+    "\n",
+    "### 问题回顾\n",
+    "\n",
+    "**目标**：在 **n** 个相邻路口中选择 **k** 个连续的路段安装宣传装置，以最大化受到宣传的市民总数。\n",
+    "\n",
+    "**输入格式**：\n",
+    "1. 第一行：两个整数 `n` 和 `k`，表示路口数和可安装装置的数量。\n",
+    "2. 接下来的 `n-1` 行：第 `i` 行包含 `n-i` 个整数，第 `j` 个数表示从路口 `i` 到路口 `i+j` 的每天通行市民数量。\n",
+    "\n",
+    "**输出格式**：\n",
+    "- 输出一个整数，表示能够受到宣传的最大市民总数。\n",
+    "\n",
+    "### 解题思路\n",
+    "\n",
+    "1. **输入处理**：\n",
+    "   - 逐行读取输入，首先获取 `n` 和 `k`。\n",
+    "   - 接着读取 `n-1` 行的市民数量数据，并将其存储在一个一维列表中。\n",
+    "\n",
+    "2. **数据结构构建**：\n",
+    "   - 使用二维列表 `C[s][t]` 来表示从路口 `s` 到路口 `t` 的市民数量，其中 `s < t`。\n",
+    "\n",
+    "3. **前缀和计算**：\n",
+    "   - 为了高效计算每个窗口内的市民总数，预先计算每个起点 `s` 从 `t` 到 `n` 的前缀和 `pre_sum[s][t]`。\n",
+    "\n",
+    "4. **滑动窗口遍历**：\n",
+    "   - 遍历所有可能的 `k` 个连续路段窗口。\n",
+    "   - 对于每个窗口，计算所有经过这些路段的市民总数。\n",
+    "   - 更新并记录最大的市民总数。\n",
+    "\n",
+    "5. **输出结果**：\n",
+    "   - 输出计算得到的最大市民总数。\n",
+    "\n",
+    "### 代码说明\n",
+    "\n",
+    "1. **函数 `max_citizens`**：\n",
+    "   - **输入参数**：\n",
+    "     - `n`：路口数量。\n",
+    "     - `k`：可安装装置的数量。\n",
+    "     - `C_input`：包含所有路段市民数量的一维列表。\n",
+    "   - **功能**：\n",
+    "     - 将 `C_input` 转换为二维列表 `C[s][t]`。\n",
+    "     - 计算每个起点 `s` 从 `t` 到 `n` 的前缀和 `pre_sum[s][t]`。\n",
+    "     - 使用滑动窗口方法，遍历所有可能的 `k` 个连续路段，计算每个窗口内的市民总数，并记录最大值。\n",
+    "   - **返回值**：\n",
+    "     - 最大的市民总数。\n",
+    "\n",
+    "2. **函数 `main`**：\n",
+    "   - **功能**：\n",
+    "     - 逐行读取输入数据。\n",
+    "     - 调用 `max_citizens` 函数计算结果。\n",
+    "     - 输出结果。\n",
+    "\n",
+    "### 示例验证\n",
+    "\n",
+    "让我们验证一下提供的示例：\n",
+    "\n",
+    "**输入**：\n",
+    "```\n",
+    "4 1\n",
+    "5 0 6\n",
+    "5 3\n",
+    "5\n",
+    "```\n",
+    "\n",
+    "**输出**：\n",
+    "```\n",
+    "14\n",
+    "```\n",
+    "\n",
+    "**解释**：\n",
+    "- 当 `k=1` 时，选择安装在第 3-4 个路段，可以覆盖的市民数量为 `6 + 3 + 5 = 14`，这与输出一致。\n",
+    "\n",
+    "### 运行方式\n",
+    "\n",
+    "你可以将上述代码复制到你的 Python 环境中，运行后按照输入格式逐行输入数据。例如：\n",
+    "\n",
+    "```\n",
+    "4 1\n",
+    "5 0 6\n",
+    "5 3\n",
+    "5\n",
+    "```\n",
+    "\n",
+    "然后程序会输出：\n",
+    "\n",
+    "```\n",
+    "14\n",
+    "```\n",
+    "\n",
+    "### 注意事项\n",
+    "\n",
+    "- **输入数据的格式**：确保每行输入的数据与题目描述相符，且没有多余的空格或空行。\n",
+    "- **数据范围**：根据题目描述，`n` 的范围为 `2 ≤ n ≤ 500`，`k` 的范围为 `1 ≤ k ≤ n-1`，每对路口之间的市民数量不超过 `100`。\n",
+    "\n",
+    "这样调整后的代码能够正确地从键盘逐行读取输入，并计算出最大受宣传的市民总数。"
+   ]
+  }
+ ],
+ "metadata": {
+  "kernelspec": {
+   "display_name": "base",
+   "language": "python",
+   "name": "python3"
+  },
+  "language_info": {
+   "codemirror_mode": {
+    "name": "ipython",
+    "version": 3
+   },
+   "file_extension": ".py",
+   "mimetype": "text/x-python",
+   "name": "python",
+   "nbconvert_exporter": "python",
+   "pygments_lexer": "ipython3",
+   "version": "3.11.5"
+  }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 2
+}