commit 2163c32f563088d1eeadef7fc272ca0822d9ddc1
Author: zzwcreator <2267110628@qq.com>
Date:   Sat Nov 9 19:02:14 2024 +0800

    第一次提交

diff --git a/.vscode/settings.json b/.vscode/settings.json
new file mode 100644
index 0000000..4df6722
--- /dev/null
+++ b/.vscode/settings.json
@@ -0,0 +1,52 @@
+{
+    "files.associations": {
+        "vector": "cpp",
+        "algorithm": "cpp",
+        "atomic": "cpp",
+        "bit": "cpp",
+        "cctype": "cpp",
+        "clocale": "cpp",
+        "cmath": "cpp",
+        "compare": "cpp",
+        "concepts": "cpp",
+        "cstddef": "cpp",
+        "cstdint": "cpp",
+        "cstdio": "cpp",
+        "cstdlib": "cpp",
+        "cstring": "cpp",
+        "ctime": "cpp",
+        "cwchar": "cpp",
+        "exception": "cpp",
+        "initializer_list": "cpp",
+        "ios": "cpp",
+        "iosfwd": "cpp",
+        "iostream": "cpp",
+        "istream": "cpp",
+        "iterator": "cpp",
+        "limits": "cpp",
+        "memory": "cpp",
+        "new": "cpp",
+        "ostream": "cpp",
+        "stdexcept": "cpp",
+        "streambuf": "cpp",
+        "system_error": "cpp",
+        "tuple": "cpp",
+        "type_traits": "cpp",
+        "typeinfo": "cpp",
+        "utility": "cpp",
+        "xfacet": "cpp",
+        "xiosbase": "cpp",
+        "xlocale": "cpp",
+        "xlocinfo": "cpp",
+        "xlocnum": "cpp",
+        "xmemory": "cpp",
+        "xstddef": "cpp",
+        "xstring": "cpp",
+        "xtr1common": "cpp",
+        "xutility": "cpp",
+        "queue": "cpp",
+        "numeric": "cpp",
+        "unordered_set": "cpp",
+        "xhash": "cpp"
+    }
+}
\ No newline at end of file
diff --git a/DiDi/ques1.cpp b/DiDi/ques1.cpp
new file mode 100644
index 0000000..36d794e
--- /dev/null
+++ b/DiDi/ques1.cpp
@@ -0,0 +1,38 @@
+/*
+从数组中选择尽可能多的元素，只能连续选择，使得相加小于m，求最多可以选择几个数。输入为一个个数为n的数组和一个整数，输出为最多选择元素的个数。
+*/
+#include <iostream>
+#include <vector>
+using namespace std;
+int maxConsecutiveElementsUnderSum(const std::vector<int>& nums, int m) {
+    int n = nums.size();
+    int max_count = 0; // 最多选择的元素个数
+    int current_sum = 0; // 当前窗口内元素的和
+    int left = 0; // 左指针
+    // 遍历数组，用右指针表示窗口的右端
+    for (int right = 0; right < n; ++right) {
+        current_sum += nums[right]; // 增加当前右指针对应的元素
+        // 当当前窗口和大于 m 时，收缩左边界
+        while (current_sum > m && left <= right) {
+            current_sum -= nums[left]; // 减去左指针的元素
+            left++; // 左指针右移
+        }
+        // 更新最多选择的元素个数
+        max_count = std::max(max_count, right - left + 1);
+    }
+    return max_count;
+}
+
+int main() {
+    vector<int> nums;
+    int n,m;
+    cin>>n>>m;
+    int val;
+    for(int i = 0; i < n; i++){
+        cin >> val;
+        nums.push_back(val);
+    }
+    int result = maxConsecutiveElementsUnderSum(nums, m);
+    cout << result << endl;
+    return 0;
+}
diff --git a/DiDi/ques2.cpp b/DiDi/ques2.cpp
new file mode 100644
index 0000000..0c3fbbe
--- /dev/null
+++ b/DiDi/ques2.cpp
@@ -0,0 +1,77 @@
+/*
+小A正在玩游戏，在游戏中一共有n个不同星球，星球间共有m条双向航道，小A的任务是摧毁这些星球。若有多个星球间两两可达，则我们称它们属于同一个联盟。特别的，若一个星球与其他星球间都没有航道，则也称其为一个联盟。小A将按照星球的编号从小到大依次摧毁各个星球，当一个星球被摧毁后，与之相连的航道也将相继被摧毁，现在小A想知道在每个星球被摧毁时还剩下多少个联盟。不同星球间可能有多条航道，但每条航道连接的两个星球必然不同。上述题意可以被简化为，给定n个点，m条边的无向图，按照编号大小依次删去各个节点，请问在每个节点被删去时，还剩下多少个连通块。保证给定图无自环，但可能有重边，
+
+输入描述
+第一行两个正整数n,m，表示星球数与航道数。接下来m行，每行2个正整数u,v，表示两星球间有一条航道。1<n,m<5e4
+输出描述
+输出一行n个正整数，表示答案。
+
+样例输入
+5 6
+1 2
+2 3
+3 1
+4 5
+5 1
+2 4
+样例输出
+1 2 1 1 0
+*/
+#include <stdio.h>
+#define MAXNUM 1000
+int Group[MAXNUM];			//记录图中节点是否被访问过
+int dist[MAXNUM][MAXNUM];	//通过邻接矩阵保存图
+void DFS(int i, int n);
+int main()
+{
+	int n, m, p, q, k = 0;
+	for (int i = 0; i < MAXNUM; i++)
+	{
+		for (int j = 0; j < MAXNUM; j++)
+			dist[i][j] = 0;			//初始化邻接矩阵，初始值为0
+		Group[i] = 0;				//同上
+	}
+	scanf("%d%d", &n, &m);			//获取图的节点数n，和边数m
+	for (int i = 0; i < m; i++)
+	{
+		scanf("%d%d", &p, &q);
+		dist[p-1][q-1] = 1;			//如果节点1和节3相连 则另dist[1][3]和dist[3][1]的值为1
+		dist[q-1][p-1] = 1;
+	}
+
+    for(int zz = 1; zz <= n; zz++){
+        for(int j = 1; j <= n; j++){    // 删除节点zz相连的边
+            p = zz; 
+            q = j;
+            dist[p-1][q-1] = 0;		
+		    dist[q-1][p-1] = 0;
+        }
+
+        for (int i = 0; i < MAXNUM; i++)    // 清零
+        {
+            Group[i] = 0;					
+        }
+
+        for (int i = 0; i < n; i++)
+        {
+            if (Group[i] != 1)
+            {
+                DFS(i, n);
+                k++;				    //执行一次DFS就是k的值自增1,
+            }
+        }
+        printf("%d", k - zz);       // k为删除边但没有删除节点的连通数 zz表示每次删除一个节点
+        k = 0;
+    }
+	return 0;
+}
+void DFS(int i,int n)
+{
+	Group[i] = 1;
+	for (int j = 0; j < n; j++)
+	{
+		if (dist[i][j] == 1)
+			if (Group[j] != 1)
+				DFS(j, n);
+	}
+}
diff --git a/HuaWei/output/ques.exe b/HuaWei/output/ques.exe
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diff --git a/HuaWei/output/ques2.exe b/HuaWei/output/ques2.exe
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diff --git a/HuaWei/ques.cpp b/HuaWei/ques.cpp
new file mode 100644
index 0000000..7d7fa35
--- /dev/null
+++ b/HuaWei/ques.cpp
@@ -0,0 +1,104 @@
+/*
+c++实现以下问题：
+有一个 N x N 大小的迷宫。初始状态下，配送员位于迷宫的左上角，他希望前往迷宫的右下角。配送员只能沿若上下左右四个方向移动，从每个格子移动到相邻格子所需要的时间是 1个单位，他必须用最多 K 个(也可以少于 K 个)单位时间到达右下角格子。迷宫的每个格子都有辐射值，配送员必须穿着防护能力不低于相应辐射值的防护服，才能通过该格子。他希望知道，防护服的防护能力最少要达到多少，他才能顺利完成任务。注意:配送员需要通过迷宫的左上角和右下角，因此防护服的防护能力必须大于等于这两个格子的辐射值。
+解答要求:
+时间限制: C/C++1000ms,其他语言:2000ms内存限制: C/C++256MB,其他语言:512MB
+输入:
+前两行各包含一个正整数，分别对应 N 和 K后 N 行各包含 N 整数，以空格分隔，表示地图上每个位置的辐射值,2≤N≤100。 K≥2N-2，以保证题目有解。所有辐射值都是非负整数，绝对值不超过 1e4
+输出:
+一个整数，表示配送员穿着防护服的最低防护能力。
+
+样例1
+输入:
+2
+2
+1 3
+2 1
+输出:2
+解释:配送员可以选择通过左下角(辐射值为2)的路线，耗费2单位时间。
+
+样例2
+输入:
+5
+12
+0 0 0 0 0
+9 9 3 9 0
+0 0 0 0 0
+0 9 5 9 9
+0 0 0 0 0
+输出:3
+解释:最优路线:往右2格，往下2格，往左2格，往下2格，往右4格，耗费12单位时间，经过格子的最大辐射值为3。
+另外，在地图不变的情况下，如果K=16，输出为0;如果K=8，输出为5。
+
+*/
+#include <iostream>
+#include <vector>
+#include <queue>
+#include <algorithm>
+using namespace std;
+const int INF = 1e9; // 定义一个足够大的值表示不可达
+int N, K;
+vector<vector<int>> radiation;
+// 四个方向的移动
+int dx[] = {1, -1, 0, 0};
+int dy[] = {0, 0, 1, -1};
+// 判断是否可以在指定的防护能力下从 (0,0) 到 (N-1,N-1)
+bool canReach(int limit) {
+    if (radiation[0][0] > limit || radiation[N-1][N-1] > limit) {
+        return false;
+    }
+    vector<vector<bool>> visited(N, vector<bool>(N, false));
+    queue<pair<int, int>> q;
+    q.push({0, 0});
+    visited[0][0] = true;
+    int time = 0;
+    while (!q.empty()) {
+        int qsize = q.size();
+        if (time > K) return false; // 时间超过 K
+        while (qsize--) {
+            int x = q.front().first;
+            int y = q.front().second;
+            q.pop();
+            
+            if (x == N-1 && y == N-1) {
+                return true; // 成功到达终点
+            }
+            for (int i = 0; i < 4; ++i) {
+                int nx = x + dx[i];
+                int ny = y + dy[i];
+                if (nx >= 0 && nx < N && ny >= 0 && ny < N && !visited[nx][ny] && radiation[nx][ny] <= limit) {
+                    visited[nx][ny] = true;
+                    q.push({nx, ny});
+                }
+            }
+        }
+        ++time; // 每次扩展一层，时间增加
+    }
+    return false; // 没有在 K 时间内到达终点
+}
+
+int main() {
+    // 读入 N 和 K
+    cin >> N >> K;
+    radiation.resize(N, vector<int>(N));
+    // 读入迷宫的辐射值
+    for (int i = 0; i < N; ++i) {
+        for (int j = 0; j < N; ++j) {
+            cin >> radiation[i][j];
+        }
+    }
+    // 二分查找防护能力
+    int low = 0, high = 1e4, ans = INF;
+    while (low <= high) {
+        int mid = (low + high) / 2;
+        if (canReach(mid)) {
+            ans = mid;
+            high = mid - 1;
+        } else {
+            low = mid + 1;
+        }
+    }
+    // 输出结果
+    cout << ans << endl;
+    return 0;
+}
diff --git a/HuaWei/ques2.cpp b/HuaWei/ques2.cpp
new file mode 100644
index 0000000..9fc7af4
--- /dev/null
+++ b/HuaWei/ques2.cpp
@@ -0,0 +1,72 @@
+#include <iostream>
+#include <vector>
+#include <algorithm>
+using namespace std;
+// 日志文件结构体
+struct LogFile {
+    int process;  // 进程序号
+    int size;     // 文件大小
+    int downloads; // 下载次数
+};
+int main() {
+    int N, M, C;
+    cin >> N >> M >> C;
+    vector<vector<LogFile>> logs(N); // 每个进程的日志文件
+    // 读取日志文件信息
+    for (int i = 0; i < N * M; ++i) {
+        int process, size, downloads;
+        cin >> process >> size >> downloads;
+        logs[process].push_back({process, size, downloads});
+    }
+    int minCapacity = 0; // 最少需要的容量
+    vector<int> mustHaveDownloads; // 每个进程必须要保存的日志文件的下载次数
+    vector<pair<int, int>> optionalFiles; // 额外的日志文件，保存 {文件大小，下载次数}
+    // 选择每个进程的最小日志文件，并记录额外的文件
+    for (int i = 0; i < N; ++i) {
+        // 找到当前进程中最小的文件
+        int minSize = 21, minDownloads = 0;
+        for (const auto& log : logs[i]) {
+            if (log.size < minSize) {
+                minSize = log.size;
+                minDownloads = log.downloads;
+            }
+        }
+        // 计算最少需要的容量
+        minCapacity += minSize;
+        mustHaveDownloads.push_back(minDownloads);
+        
+        // 记录下当前进程中非最小的日志文件，作为可选的日志文件
+        for (const auto& log : logs[i]) {
+            if (log.size != minSize || log.downloads != minDownloads) {
+                optionalFiles.push_back({log.size, log.downloads});
+            }
+        }
+    }
+    // 如果最小容量已经超出了U盘容量，返回-1
+    if (minCapacity > C) {
+        cout << -1 << endl;
+        return 0;
+    }
+    // 计算每个进程必须保存的文件的下载次数之和
+    int totalDownloads = 0;
+    for (int d : mustHaveDownloads) {
+        totalDownloads += d;
+    }
+    // 剩余的U盘容量
+    int remainingCapacity = C - minCapacity;
+    // 动态规划数组，dp[j] 表示容量为 j 时的最大下载次数
+    vector<int> dp(remainingCapacity + 1, 0);
+    // 0-1 背包：选择可选的日志文件
+    for (const auto& file : optionalFiles) {
+        int size = file.first;
+        int downloads = file.second;
+        for (int j = remainingCapacity; j >= size; --j) {
+            dp[j] = max(dp[j], dp[j - size] + downloads);
+        }
+    }
+    // 最大的下载次数等于已选择的最小文件的下载次数 + 背包中选出的最大下载次数
+    int maxDownloads = totalDownloads + dp[remainingCapacity];
+    // 输出结果
+    cout << maxDownloads << endl;
+    return 0;
+}
diff --git a/LiXiang/TEST.PY b/LiXiang/TEST.PY
new file mode 100644
index 0000000..4b27755
--- /dev/null
+++ b/LiXiang/TEST.PY
@@ -0,0 +1,34 @@
+import math
+
+# 定义向量x和y
+x = (0, 1, 0, 1)
+y = (1, 0, 1, 0)
+
+# 计算余弦相似度
+def cosine(x, y):
+    numerator = sum([x[i] * y[i] for i in range(len(x))])
+    denominator = math.sqrt(sum([x[i] ** 2 for i in range(len(x))])) * math.sqrt(sum([y[i] ** 2 for i in range(len(y))]))
+    return numerator / denominator
+
+# 计算相关系数
+def correlation(x, y):
+    n = len(x)
+    sum_x = sum(x)
+    sum_y = sum(y)
+    sum_xy = sum([x[i] * y[i] for i in range(n)])
+    sum_x2 = sum([x[i] ** 2 for i in range(n)])
+    sum_y2 = sum([y[i] ** 2 for i in range(n)])
+    numerator = n * sum_xy - sum_x * sum_y
+    denominator = math.sqrt((n * sum_x2 - sum_x ** 2) * (n * sum_y2 - sum_y ** 2))
+    return numerator / denominator
+
+# 计算Jaccard系数
+def jaccard(x, y):
+    intersection = sum([x[i] and y[i] for i in range(len(x))])
+    union = sum([x[i] or y[i] for i in range(len(x))])
+    return intersection / union
+
+# 输出结果
+print('余弦相似度:', cosine(x, y))
+print('相关系数:', correlation(x, y))
+print('Jaccard系数:', jaccard(x, y))
\ No newline at end of file
diff --git a/LiXiang/output/ques.exe b/LiXiang/output/ques.exe
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diff --git a/LiXiang/ques.cpp b/LiXiang/ques.cpp
new file mode 100644
index 0000000..fac6fee
--- /dev/null
+++ b/LiXiang/ques.cpp
@@ -0,0 +1,32 @@
+#include <iostream>
+#include <iomanip>
+using namespace std;
+
+/*杨辉三角最终显示效果是一个等腰三角形，两个最外边都是1
+  杨辉三角的根本在于，每个数等于它上方两数之和
+*/
+int main(){
+  int n;//变量n在此处起到了限制输出行数的作用，可优化成用户输入
+  cin>>n;
+  int i,j,a[n][n];
+
+  //使第一列和对角线元素的值为1
+  for (i=0;i<n;i++){            //前两行全为1，拿出来单独处理
+     a[i][i]=1;//使最右侧边全为1
+     a[i][0]=1;//使最左侧边全为1
+  }
+
+  //从第三行开始处理
+  for (i=2;i<n;i++)                //三行开始出现变化
+    for (j=1;j<=i-1;j++) //j始终慢i一步
+       a[i][j]=a[i-1][j-1]+a[i-1][j];//每个数等于它上方两数之和，如a32=a21+a22
+
+  //输出数组各元素的值
+  for (i=0;i<n;i++){            //从第一行开始
+     for (j=0;j<=i;j++)            //利用j将每一行的数据全部输出
+       cout<<a[i][j]<<" ";    //在C++中，setw(int n)用来控制输出间隔，这里是指前元素末尾与后元素末尾之间有个5空格位
+     cout<<endl;
+  }
+  //cout<<endl;
+  return 0;
+}
\ No newline at end of file
diff --git a/XiaoHongShu/output/ques.exe b/XiaoHongShu/output/ques.exe
new file mode 100644
index 0000000..a6c35a2
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diff --git a/XiaoHongShu/ques.cpp b/XiaoHongShu/ques.cpp
new file mode 100644
index 0000000..295a5a4
--- /dev/null
+++ b/XiaoHongShu/ques.cpp
@@ -0,0 +1,92 @@
+/*
+小红书最近决定举办一场盛大的红蓝PK!为了公平的分出红蓝两个阵营，每个人都获得了一个数字ai代表目己的战斗力，随后，由小红书“抽奖组件”这个程序来自动划分阵营，使得红方队员的总战斗力等于蓝方的总战斗力
+然而，并不是每一局游戏都能找到这样完美的分配方案，已有的程序显然已经不够用了。让我们来新增一个总战斗力。前置程序吧，你只需要计算出每一局双方战斗力差值的最小绝对值。
+*/
+#include <iostream>
+#include <vector>
+#include <cmath>
+#include <numeric>
+using namespace std;
+
+int findClosestSum(vector<int>& nums, int target) {
+    int sum = accumulate(nums.begin(), nums.end(), 0);  // 数组元素的总和
+    int maxSum = min(sum, target + abs(sum - target));  // 限制最大的可能和
+    vector<bool> dp(maxSum + 1, false);  // dp数组，初始为 false
+    dp[0] = true;  // 和为 0 是可能的
+    // 动态规划填充 dp 数组
+    for (int num : nums) {
+        for (int i = maxSum; i >= num; --i) {
+            dp[i] = dp[i] || dp[i - num];
+        }
+    }
+    // 从 dp 数组中找到最接近目标值的和
+    int closest = 0;
+    for (int i = 0; i <= maxSum; ++i) {
+        if (dp[i] && abs(i - target) < abs(closest - target)) {
+            closest = i;
+        }
+    }
+    return closest;
+}
+
+int main() {
+    int n;
+    vector<int> arr;
+    int val, sum = 0;
+    cin >> n;
+    for(int i = 0; i < n; i++){
+        cin >> val;
+        sum += val;
+        arr.push_back(val);
+    }
+    int arv = sum / 2;
+    int tmp = findClosestSum(arr, arv);
+    cout << tmp << endl;
+    cout << arv - tmp + sum - tmp - arv << endl;
+    return 0;
+}
+
+
+#include <iostream>
+#include <vector>
+#include <cmath>
+#include <numeric>
+
+using namespace std;
+
+int findClosestSum(vector<int>& nums, int target) {
+    int n = nums.size();
+    int sum = accumulate(nums.begin(), nums.end(), 0);  // 数组元素的总和
+
+    int maxSum = min(sum, target + abs(sum - target));  // 限制最大的可能和
+    vector<bool> dp(maxSum + 1, false);  // dp数组，初始为 false
+    dp[0] = true;  // 和为 0 是可能的
+
+    // 动态规划填充 dp 数组
+    for (int num : nums) {
+        for (int i = maxSum; i >= num; --i) {
+            dp[i] = dp[i] || dp[i - num];
+        }
+    }
+
+    // 从 dp 数组中找到最接近目标值的和
+    int closest = 0;
+    for (int i = 0; i <= maxSum; ++i) {
+        if (dp[i] && abs(i - target) < abs(closest - target)) {
+            closest = i;
+        }
+    }
+
+    return closest;
+}
+
+int main() {
+    vector<int> nums = {7,3,5,3};  // 输入数组
+    int target = 10;  // 目标值
+
+    int result = findClosestSum(nums, target);
+
+    cout << result << endl;
+
+    return 0;
+}
diff --git a/output/test copy.exe b/output/test copy.exe
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index 0000000..90d50ee
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diff --git a/output/test.exe b/output/test.exe
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diff --git a/test.cpp b/test.cpp
new file mode 100644
index 0000000..e61987f
--- /dev/null
+++ b/test.cpp
@@ -0,0 +1,59 @@
+#include <iostream>
+#include <vector>
+#include <set>
+#include <numeric>
+#include <algorithm>
+#include <cmath>
+
+using namespace std;
+
+// 计算一个数的质因数的最小公倍数
+long long lcm(long long a, long long b) {
+    return a / __gcd(a, b) * b;
+}
+
+// 计算一个城堡的完美度
+long long calculateCastlePerfectness(const vector<pair<int, int>>& blocks) {
+    long long product = 1;
+    for (const auto& block : blocks) {
+        product *= pow(block.first, block.second);
+    }
+    return product;
+}
+
+int main() {
+    int n;
+    cin >> n;
+
+    vector<long long> perfectness(n);
+    vector<vector<pair<int, int>>> castles(n);
+
+    for (int i = 0; i < n; ++i) {
+        int ti;
+        cin >> ti;
+        castles[i].resize(ti);
+        for (int j = 0; j < ti; ++j) {
+            int Pij, Cij;
+            cin >> Pij >> Cij;
+            castles[i][j] = {Pij, Cij};
+        }
+        perfectness[i] = calculateCastlePerfectness(castles[i]);
+    }
+
+    // 计算总完美度
+    set<long long> uniquePerfectness;
+    for (int i = 0; i < n; ++i) {
+        long long currentLcm = 1;
+        for (int j = 0; j < n; ++j) {
+            if (j != i) {
+                currentLcm = lcm(currentLcm, perfectness[j]);
+            }
+        }
+        uniquePerfectness.insert(currentLcm);
+    }
+
+    // 输出不同的总完美度的种数
+    cout << uniquePerfectness.size() << endl;
+
+    return 0;
+}
\ No newline at end of file
diff --git a/test.py b/test.py
new file mode 100644
index 0000000..c7338af
--- /dev/null
+++ b/test.py
@@ -0,0 +1,35 @@
+def outer(fun):
+    def wrapper(s):
+        # 获取字符串的前三个字符
+        first_three = s[1:4]
+        #倒序前三个字符
+        reversed_three = first_three[::-1]
+        # 返回拼接后的字符串
+        return first_three + reversed_three
+    return wrapper
+@outer
+def fun(s):
+    return s
+
+
+str = input()
+print(fun(str))
+
+
+# def process_list(li):
+#     li[:] = li[:6]
+#     li.sort(reverse=True)
+#     l1 = li[1::2]
+#     l2 = li[0::2]
+#     l2.reverse()
+#     li[:]= l2 + l1
+#     return li
+
+# # str="[2,1,6,4,5,3,7,8]"
+# str=input()
+# str=str[1:-1]
+
+# after5 = str.split(',')
+# res = process_list(after5)
+# res2 = [int(i) for i in res]
+# print(res2)
\ No newline at end of file