add exercises

exercises/生产队的刘同学/中等/《教父》家族关系维护/exercies.md

+# 《教父》家族关系维护
+在《教父》中，有五个家族，分别是Corleone家族、Tattaglia家族、Cuneo家族、Stracci家族和Barzini家族。五个家族之间的关系是复杂的，它们之间可能存在敌对关系、盟友关系和中立关系。
+
+你需要写一个程序来计算五个家族之间的关系。你需要输入五个家族之间的关系，然后输出五个家族之间的关系矩阵。
+
+你需要自定义一个算法来计算五个家族之间的关系，算法的输入是五个家族之间的关系，输出是五个家族之间的关系矩阵。
+
+算法的实现方式是你自己决定的，但你需要考虑这道题的时间复杂度和空间复杂度。
+
+## 输入格式
+输入第一行包含五个整数，分别表示Corleone家族、Tattaglia家族、Cuneo家族、Stracci家族和Barzini家族之间的关系。整数的取值范围是-1、0、1，分别表示敌对关系、中立关系和盟友关系。
+
+## 输出格式
+输出五行，每行五个整数，分别表示Corleone家族、Tattaglia家族、Cuneo家族、Stracci家族和Barzini家族之间的关系。整数的取值范围是-1、0、1，分别表示敌对关系、中立关系和盟友关系。
+
+## 输入样例
+
+1 -1 0 0 -1
+
+## 输出样例
+
+1 -1 0 0 -1
+-1 1 0 0 1
+0 0 1 0 0
+0 0 0 1 0
+-1 1 0 0 1
+
+## 提示
+
+无
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/solution.cpp

+#include <iostream>
+#include <sstream>
+using namespace std;
+
+const int N = 5;
+
+int a[N][N];
+
+int main() {
+    string line;
+    getline(cin, line);
+
+    stringstream ss(line);
+    for (int i = 0; i < N; i++) {
+        for (int j = 0; j < N; j++) {
+            if (i == j) {
+                a[i][j] = 1; // 同家族之间是盟友关系
+            } else {
+                a[i][j] = 0; // 默认是中立关系
+            }
+        }
+    }
+
+    // 从输入中读入五个家族之间的关系
+    for (int i = 0; i < N; i++) {
+        ss >> a[i][i];
+    }
+
+    // 输出五个家族之间的关系矩阵
+    for (int i = 0; i < N; i++) {
+        for (int j = 0; j < N; j++) {
+            cout << a[i][j] << ' ';
+        }
+        cout << endl;
+    }
+
+    return 0;
+}
+

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/1.in

+1 -1 0 0 -1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/1.out

+1 -1 0 0 -1
+0 0 1 0 0
+0 0 0 1 0
+0 0 0 0 1
+1 0 0 0 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/10.in

+-1 1 -1 1 -1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/10.out

+1 -1 1 -1 1
+-1 1 -1 1 -1
+1 -1 1 -1 1
+-1 1 -1 1 -1
+1 -1 1 -1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/2.in

+1 1 1 1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/2.out

+1 1 1 1 1
+1 1 1 1 1
+1 1 1 1 1
+1 1 1 1 1
+1 1 1 1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/3.in

+-1 -1 -1 -1 -1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/3.out

+1 -1 -1 -1 -1
+-1 1 -1 -1 -1
+-1 -1 1 -1 -1
+-1 -1 -1 1 -1
+-1 -1 -1 -1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/4.in

+1 0 0 0 0
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/4.out

+1 0 0 0 0
+0 1 0 0 0
+0 0 1 0 0
+0 0 0 1 0
+0 0 0 0 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/5.in

+-1 0 0 0 0
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/5.out

+1 0 0 0 0
+0 1 0 0 0
+0 0 1 0 0
+0 0 0 1 0
+0 0 0 0 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/6.in

+0 0 0 0 0
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/6.out

+1 0 0 0 0
+0 1 0 0 0
+0 0 1 0 0
+0 0 0 1 0
+0 0 0 0 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/7.in

+1 -1 1 -1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/7.out

+1 -1 1 -1 1
+-1 1 -1 1 -1
+1 -1 1 -1 1
+-1 1 -1 1 -1
+1 -1 1 -1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/8.in

+1 1 1 1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/8.out

+1 1 1 1 1
+1 1 1 1 1
+1 1 1 1 1
+1 1 1 1 1
+1 1 1 1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/9.in

+-1 -1 -1 -1 -1
\ No newline at end of file

exercises/生产队的刘同学/中等/《教父》家族关系维护/test_cases/9.out

+1 -1 -1 -1 -1
+-1 1 -1 -1 -1
+-1 -1 1 -1 -1
+-1 -1 -1 1 -1
+-1 -1 -1 -1 1
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/exercies.md

+# 寻找宝藏山
+
+一天，你去了一个神秘的森林，在那里你遇到了一个神秘的老人。他告诉你，森林里有一座宝藏山，但是要想到达宝藏山，你必须先通过一些森林的入口。
+
+这个老人给你了一张地图，地图上有若干个入口和宝藏山，每个入口都有一个花费。你可以从任意一个入口开始，然后经过一些入口，最终到达宝藏山。但是你有一个限制，你只能走 $K$ 步，如果走的步数超过了 $K$，那么你就无法到达宝藏山，也就无法获得宝藏。
+
+你必须实现一个程序，接受用户输入的地图信息，并计算出，你能够到达宝藏山的最小花费。
+
+## 输入描述
+
+第一行包含三个整数 $N$、$M$ 和 $K$，$N$ 表示入口的个数、$M$ 表示边的个数、$K$ 表示最多走的步数。
+
+接下来 $M$ 行，每行包含三个整数 $A$、$B$ 和 $C$，表示一条从 $A$ 到 $B$ 的有向边，边权为 $C$。
+
+## 输出描述
+
+输出一行，包含一个整数，表示你能够到达宝藏山的最小花费。
+
+## 输入样例
+
+5 10 6
+1 2 3
+1 2 3
+1 2 3
+1 2 3
+1 2 3
+5 2 3
+2 3 4
+2 1 1
+2 3 4
+3 4 5
+
+## 输出样例
+
+1061109567
+
+## 提示
+
+无
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/solution.cpp

+#include <iostream>
+#include <cstring>
+#include <queue>
+#include <climits>
+
+using namespace std;
+
+const int N = 1e3 + 10, M = 2e6 + 10;
+
+int h[N], e[M], ne[M], w[M], idx;
+int dist[N];
+int n, m, k;
+bool st[N]; // 标记数组
+
+// 初始化
+void init() {
+  memset(h, -1, sizeof h);
+  idx = 0;
+}
+
+// 添加一条边
+void add(int a, int b, int c) {
+  e[idx] = b, w[idx] = c, ne[idx] = h[a], h[a] = idx++;
+}
+
+// Dijkstra 算法求最短路径
+void dijkstra(int s) {
+  memset(dist, 0x3f, sizeof dist);
+  dist[s] = 0;
+
+  priority_queue<pair<int, int>> q; // 小根堆
+  q.push({0, s});
+
+  while (q.size()) {
+    auto t = q.top();
+    q.pop();
+    int ver = t.second, distance = t.first;
+    if (st[ver]) continue;
+    st[ver] = true;
+
+    // 更新所有 ver 相邻的点
+    for (int i = h[ver]; ~i; i = ne[i]) {
+      int j = e[i];
+      if (dist[j] > distance + w[i]) {
+        dist[j] = distance + w[i];
+        q.push({dist[j], j});
+      }
+    }
+  }
+}
+
+int main() {
+  // 读入
+  scanf("%d%d%d", &n, &m, &k);
+  init();
+
+  // 建图
+  while (m--) {
+    int a, b, c;
+    scanf("%d%d%d", &a, &b, &c);
+    add(a, b, c);
+  }
+
+  // 求最短路径
+  dijkstra(1);
+
+  // 输出最短路径
+  printf("%d\n", dist[n]);
+
+  return 0;
+}

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/1.in

+2 4 7
+2 3 4
+2 3 4
+1 2 3
+4 5 6
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/1.out

+3
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/10.in

+1 2 3
+2 4 6
+3 4 9
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/10.out

+0
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/2.in

+5 10 6
+1 2 3
+1 2 3
+1 2 3
+1 2 3
+1 2 3
+5 2 3
+2 3 4
+2 1 1
+2 3 4
+3 4 5
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/2.out

+1061109567
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/3.in

+3 3 40
+1 2 3
+2 3 4
+5 3 2
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/3.out

+7
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/4.in

+3 3 40
+1 2 3
+2 3 3
+5 6 7
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/4.out

+6
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/5.in

+2 4 30
+5 1 2
+5 2 2
+5 3 2
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/5.out

+7
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/6.in

+6 3 4
+2 1 4
+3 4 2
+1 2 1
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/6.out

+1061109567
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/7.in

+3 6 5
+1 2 3
+2 3 6
+5 3 1
+5 1 2
+5 2 2
+5 2 2
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/7.out

+9
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/8.in

+1 3 5
+3 9 0
+2 4 0
+4 1 2
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/8.out

+0
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/9.in

+2 2 5
+2 3 4
+1 2 3
\ No newline at end of file

exercises/生产队的刘同学/中等/寻找宝藏山/test_cases/9.out

+3
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/exercies.md

+# 柯里昂家族树
+
+在《教父》中，柯里昂家族是一个著名的黑手党家族，由家族的首领柯里昂（Don Corleone）和他的四个儿子构成。柯里昂家族的家谱是一棵二叉树，其中柯里昂是根节点，他的儿子们分别是他的左儿子和右儿子。
+
+现在，你需要编写一个程序，读入一个包含若干行的字符串，每行表示一个人的信息。每行的格式如下：
+
+<人物名称> is the <父亲名称>'s <儿子>
+
+其中，人物名称是一个不包含空格的字符串，父亲名称和儿子也是一个不包含空格的字符串。
+
+例如，下面是一棵柯里昂家族的家谱：
+
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Michael Corleone is the Don's son
+Tom Hagen is the Don's adopted son
+
+你的程序需要按照这些信息建立一棵二叉树，并输出这棵树的前序遍历。
+
+例如，对于上面的家谱，输出应该是：
+
+Don Corleone Fredo Corleone Michael Corleone Tom Hagen
+
+你的程序需要支持以下功能：
+
+ - 建立二叉树。你需要定义一个结构体表示一个节点，包含人物名称、左儿子和右儿子三个字段。你需要编写一个函数，读入若干行字符串，根据字符串中的信息建立二叉树。
+
+ - 输出前序遍历。你需要编写一个函数，输出给定二叉树的前序遍历。
+## 输入格式
+输入的第一行包含一个整数 n（1 <= n <= 1000），表示字符串的行数。
+
+接下来的 n 行，每行包含一个字符串，表示一个人的信息。字符串的格式如上文所述。
+
+## 输出格式
+输出给定二叉树的前序遍历，每个人物名称占一行。
+
+## 输入样例
+
+4
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Michael Corleone is the Don's son
+Tom Hagen is the Don's adopted son
+
+## 输出样例
+
+Don Corleone
+Fredo Corleone
+Michael Corleone
+Tom Hagen
+
+## 提示
+
+1. 人物名称和父亲名称均由不超过 100 个字符组成，且不包含空格；
+
+2. 儿子的值可能是 "son" 或 "adopted son"；
+
+3. 输入中可能会出现重名的人物，你的程序需要处理这种情况；
+
+4. 输入的字符串保证是合法的，不会出现父亲名称找不到的情况。
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/solution.cpp

+#include <iostream>
+#include <unordered_map>
+
+using namespace std;
+
+//定义结构体表示一个节点
+struct Node {
+    string name;
+    Node* left;
+    Node* right;
+};
+
+//建立二叉树的函数
+Node* build_tree(int n) {
+    // 建立根节点
+    string line;
+    getline(cin, line);
+    stringstream ss(line);
+    string name, is, the, father, s;
+    ss >> name >> is >> the >> father >> s;
+    Node* root = new Node{name, nullptr, nullptr};
+
+    // 建立剩余的节点
+    for (int i = 1; i < n; i++) {
+        getline(cin, line);
+        stringstream ss(line);
+        string name, is, the, father, s;
+        ss >> name >> is >> the >> father >> s;
+        Node* cur = root;
+        while (true) {
+            if (father == cur->name && s == "son") {
+                if (cur->left == nullptr) {
+                    cur->left = new Node{name, nullptr, nullptr};
+                    break;
+                } else if (cur->right == nullptr) {
+                    cur->right = new Node{name, nullptr, nullptr};
+                    break;
+                } else {
+                    cur = cur->left;
+                }
+            } else {
+                cur = cur->right;
+            }
+        }
+    }
+    return root;
+}
+
+//输出前序遍历的函数
+void preorder(Node* root) {
+    cout << root->name << " ";
+    if (root->left != nullptr) preorder(root->left);
+    if (root->right != nullptr) preorder(root->right);
+}
+
+int main() {
+    int n;
+    cin >> n;
+    cin.ignore();
+    Node* root = build_tree(n);
+    preorder(root);
+    return 0;
+}
+

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/1.in

+5
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Michael Corleone is the Don's son
+Tom Hagen is the Don's adopted son
+Santino Corleone is the Michael's son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/1.out

+Don Corleone
+Fredo Corleone
+Michael Corleone
+Tom Hagen
+Santino Corleone
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/10.in

+3
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Tom Hagen is the Fredo's son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/10.out

+Don Corleone
+Fredo Corleone
+Tom Hagen
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/2.in

+1
+Don Corleone is the Don's son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/2.out

+Don Corleone
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/3.in

+3
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Tom Hagen is the Fredo's son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/3.out

+Don Corleone
+Fredo Corleone
+Tom Hagen
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/4.in

+3
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Tom Hagen is the Don's adopted son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/4.out

+Don Corleone
+Fredo Corleone
+Tom Hagen
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/5.in

+2
+Don Corleone is the Don's son
+Michael Corleone is the Don's son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/5.out

+Don Corleone
+Michael Corleone
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/6.in

+2
+Don Corleone is the Don's son
+Tom Hagen is the Don's adopted son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/6.out

+Don Corleone
+Tom Hagen
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/7.in

+3
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Tom Hagen is the Don's adopted son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/7.out

+Don Corleone
+Fredo Corleone
+Tom Hagen
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/8.in

+4
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Michael Corleone is the Don's son
+Tom Hagen is the Don's adopted son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/8.out

+Don Corleone
+Fredo Corleone
+Michael Corleone
+Tom Hagen
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/9.in

+5
+Don Corleone is the Don's son
+Fredo Corleone is the Don's son
+Michael Corleone is the Don's son
+Tom Hagen is the Don's adopted son
+Santino Corleone is the Michael's son
\ No newline at end of file

exercises/生产队的刘同学/困难/柯里昂家族树/test_cases/9.out

+Don Corleone
+Fredo Corleone
+Michael Corleone
+Tom Hagen
+Santino Corleone
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/exercies.md

+# 山洞珍宝
+
+在一端山路中，有一个神秘的山洞。这个山洞里有许多神奇的宝藏。
+
+你在山洞外发现了一个奇怪的墙壁，上面有一个神秘的符号，你需要通过一定的方法来解码这个符号，只有解码这个符号后你才能进入这个山洞。
+
+经过一番推导，你发现这个符号是一个由 0 和 1 组成的字符串，每个字符都有一个权值，分别是 $0,1,2,3,4,5,6,7,8,9$。
+
+你需要从这个字符串中找到一个最大的子串，使得该子串的权值之和最大。
+
+给定一个字符串 $S$，其长度不超过 $10^5$，且只包含字符 '0' 和 '1'。
+
+请你找到字符串 $S$ 中权值和最大的子串。
+
+## 输入描述
+
+第一行包含一个字符串 $S$。
+
+## 输出描述
+
+输出一行，包含一个整数，表示最大的子串的权值和。
+
+## 输入样例
+
+0110100110110100
+
+## 输出样例
+
+35
+
+## 提示
+
+无
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/solution.cpp

+#include <iostream>
+#include <string>
+
+using namespace std;
+
+const int N = 1e5 + 10;
+
+int sum[N];  // sum[i] 表示字符串 s 的前 i 个字符的权值和
+
+int main() {
+  string s;
+  cin >> s;
+  int n = s.size();
+
+  // 计算权值和
+  for (int i = 0; i < n; i++) sum[i + 1] = sum[i] + (s[i] - '0');
+
+  int ans = 0;
+  for (int i = 0; i < n; i++) {
+    for (int j = i + 1; j <= n; j++) {
+      // 计算字符串 s[i...j] 的权值和
+      int w = sum[j] - sum[i];
+      ans = max(ans, w);
+    }
+  }
+
+  cout << ans << endl;
+
+  return 0;
+}

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/1.in

+01010101010
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/1.out

+5
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/10.in

+100000000010000000001
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/10.out

+3
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/2.in

+111110000101010101101000011
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/2.out

+14
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/3.in

+1010101000001111010101001
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/3.out

+12
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/4.in

+111111100000000011010101001101
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/4.out

+15
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/5.in

+11111110000000000
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/5.out

+7
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/6.in

+10101010101001011111111
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/6.out

+15
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/7.in

+1111110010101010010
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/7.out

+11
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/8.in

+1111000001010101010
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/8.out

+9
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/9.in

+100101010100101010
\ No newline at end of file

exercises/生产队的刘同学/简单/山洞珍宝/test_cases/9.out

+8
\ No newline at end of file