vault backup: 2026-01-29 20:32:42

.obsidian/workspace.json

@@ -50,6 +50,20 @@
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             }
           },
+          {
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+            "type": "leaf",
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+                "title": "CQ2605",
+                "mode": "webview"
+              },
+              "icon": "globe-2",
+              "title": "CQ2605"
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@@ -133,13 +147,12 @@
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-              "title": "1-29 的大纲"
+              "title": "大纲"
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@@ -291,10 +304,14 @@
       "remotely-save:Remotely Save": false
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+    "YueQian/Homework/assets/1-29/file-20260129201415992.png",
+    "YueQian/Homework/assets/1-29",
+    "YueQian/Homework/assets",
+    "YueQian/相关网站.md",
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     "Collection/PID控制算法.md",
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YueQian/Homework/1-29.md

@@ -5,26 +5,150 @@ tags:
 aliases: empty
 日期: 2026/1/29
 ---
-1. 定义一个长度为10的数组并从键盘输入10个整数对数组进行赋值，并实现下列功能
+**1.定义一个长度为10的数组并从键盘输入10个整数对数组进行赋值，并实现下列功能**
 1. 判定该数组是否有序
 2. 若无序则将数组中的数据进行排序
 3. 找出数组中是否存在众数并输出结果
 ```c
+#include <stdio.h>
 
+void bubble_sort(int arr[],int lenth)
+{
+	for(int i = 0;i < lenth;i++){
+		for(int j = i+1;j < lenth;j++){
+			if(arr[i] > arr[j]){
+				arr[i] ^= arr[j];
+				arr[j] ^= arr[i];
+				arr[i] ^= arr[j];
+			}
+		}
+	}
+}
+
+int sort_check(int arr[],int lenth)
+{
+	int count = 0,fcount=0;
+	for(int i=0,j=1;i<lenth-1;i++,j++){
+		if(arr[i] >= arr[j])
+			count++;
+		if(arr[i] <= arr[j])
+			fcount++;
+	}
+	if(count == lenth-1 || fcount == lenth-1)
+		return 1;
+	else
+		return 0;
+}
+
+void find_mode(int arr[], int length)
+{
+	int count[10] = {0};
+	int max_freq = 1;
+
+	for (int i = 0; i < length; i++) {
+		if (count[i] != -1) {
+			int freq = 1;
+			for (int j = i + 1; j < length; j++) {
+				if (arr[i] == arr[j]) {
+					freq++;
+					count[j] = -1;
+				}
+			}
+			count[i] = freq;
+			if (freq > max_freq) {
+				max_freq = freq;
+			}
+		}
+	}
+	printf("数组众数结果：");
+	int has_mode = 0;
+	for (int i = 0; i < length; i++) {
+		if (count[i] == max_freq) {
+			if (has_mode) printf("、");
+			printf("%d（出现%d次）", arr[i], max_freq);
+			has_mode = 1;
+		}
+	}
+	if (max_freq == 1) {
+		printf("所有元素均仅出现1次，无有效众数\n");
+	} else {
+		printf("（最大频次：%d）\n", max_freq);
+	}
+}
+int main(int argc, char const *argv[])
+{
+	int arr[10];
+	int mode;
+	printf("输入10个数:");
+	for(int i=0;i<10;i++)
+		scanf("%d",&arr[i]);
+	if(sort_check(arr,10))
+		printf("已排序\n");
+	else{
+		printf("未排序\n");
+		bubble_sort(arr,10);
+		printf("排序后为:");
+		for(int i=0;i<10;i++)
+			printf("%d ",arr[i]);
+		printf("\n");
+	}
+	find_mode(arr,10);
+	return 0;
+}
 ```
-2. 定义一个二维数组存储一个3x3的矩阵，实现如下功能
+![700](assets/1-29/file-20260129201415992.png)
+**2.定义一个二维数组存储一个3x3的矩阵，实现如下功能**
+1. 计算矩阵对角元素的和
+2. 判断矩阵是否是对称矩阵
 ```c
+#include <stdio.h>
 
-	```
+#define ROW 3
-3. 计算矩阵对角元素的和
+#define COL 3
-	```c
 
-	```
+int calcDiagSum(int matrix[ROW][COL])
-4. 判断矩阵是否是对称矩阵
+{
-	```c
+    int sum = 0;
+    for (int i = 0; i < ROW; i++) {
+        sum += matrix[i][i];
+    }
+    return sum;
+}
 
+int isSymmetricMatrix(int matrix[ROW][COL])
+{
+    for (int i = 0; i < ROW; i++) {
+        for (int j = 0; j < COL; j++) {
+            if (matrix[i][j] != matrix[j][i]) {
+                return 0;
+            }
+        }
+    }
+    return 1;
+}
+
+int main(int argc, char const *argv[]) 
+{
+    int matrix[ROW][COL];
+    printf("请输入3x3矩阵的9个整数：\n");
+    for (int i = 0; i < ROW; i++) {
+        for (int j = 0; j < COL; j++) {
+            scanf("%d", &matrix[i][j]);
+        }
+    }
+
+    int diagSum = calcDiagSum(matrix);
+    printf("\n矩阵主对角线元素的和为：%d\n", diagSum);
+    if (isSymmetricMatrix(matrix))
+        printf("该3x3矩阵是对称矩阵\n");
+    else
+        printf("该3x3矩阵不是对称矩阵\n");
+
+    return 0;
+}
 ```
-5. 小张作为公司的HR设计了一个年会游戏，来观察员工的团队协作能力、测试员工在压力下的决策能力，寻找潜在的项目领导者，现在有N名员工参与并围坐一圈，游戏规则如下：
+![700](assets/1-29/file-20260129202925505.png)
+**3.小张作为公司的HR设计了一个年会游戏，来观察员工的团队协作能力、测试员工在压力下的决策能力，寻找潜在的项目领导者，现在有N名员工参与并围坐一圈，游戏规则如下：**
 - 从某位员工开始顺时针报数
 - 报到数字M的倍数员工被淘汰出局
 - 从下一位员工继续报数，重复步骤2
@@ -32,7 +156,7 @@ aliases: empty
 ```c
 
 ```
-6. 小刘作为公司的技术骨干编写了一个程序，帮助他每次都能处在胜利者中并最终获得晋升资格，请你使用C语言编程复现小刘的程序实现以下功能：
+**4.小刘作为公司的技术骨干编写了一个程序，帮助他每次都能处在胜利者中并最终获得晋升资格，请你使用C语言编程复现小刘的程序实现以下功能：**
 - 输入总人数N、报数间隔M和获胜人数K
 - 计算并输出安全的座位位置（最后剩下的K人）
 ```c