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2 Commits
4f399c8649
...
2f8d74a33d
| Author | SHA1 | Date | |
|---|---|---|---|
| 2f8d74a33d | |||
| 33a316f373 |
30
.obsidian/workspace.json
vendored
30
.obsidian/workspace.json
vendored
@@ -50,6 +50,20 @@
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"title": "CQ2605"
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}
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},
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{
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||||
"id": "55c82e03f06c4fa8",
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||||
"type": "leaf",
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"state": {
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||||
"type": "webviewer",
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||||
"state": {
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||||
"url": "https://share.note.youdao.com/ynoteshare/index.html?id=a66c778ce8c6e4ee03608e493e59ba7a&type=notebook&_time=1768473840238#/WEB2d5da30532f421ebd7b7f91289780c02",
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"title": "CQ2605",
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||||
"mode": "webview"
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||||
},
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||||
"icon": "globe-2",
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||||
"title": "CQ2605"
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}
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},
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{
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"id": "5d8214f251dfc58d",
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"type": "leaf",
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@@ -133,13 +147,12 @@
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"state": {
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"type": "outline",
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"state": {
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"file": "YueQian/Homework/1-29.md",
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"followCursor": true,
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"showSearch": false,
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"searchQuery": ""
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},
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"icon": "lucide-list",
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||||
"title": "1-29 的大纲"
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"title": "大纲"
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}
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},
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{
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||||
@@ -291,10 +304,14 @@
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||||
"remotely-save:Remotely Save": false
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||||
}
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},
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"active": "5d8214f251dfc58d",
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"active": "55c82e03f06c4fa8",
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"lastOpenFiles": [
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"YueQian/相关网站.md",
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"YueQian/Homework/1-29.md",
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"YueQian/Homework/assets/1-29/file-20260129202925505.png",
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"YueQian/Homework/assets/1-29/file-20260129201415992.png",
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"YueQian/Homework/assets/1-29",
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"YueQian/Homework/assets",
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"YueQian/相关网站.md",
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"Diary/2026-1/2026-1-29 周四.md",
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"YueQian/Homework/~WRL0001.tmp",
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"YueQian/Homework/~WRD0000.tmp",
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@@ -304,8 +321,6 @@
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"YueQian/Homework/~WRL0003.tmp",
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"YueQian/Homework/~WRD0002.tmp",
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"YueQian/Homework/~WRL1582.tmp",
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"YueQian/Homework/~WRD1582.tmp",
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"YueQian/Homework/~WRL3055.tmp",
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"Diary/2026-1/2026-1-28 周三.md",
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"Collection/连~都忘记了的小知识.md",
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"Excalidraw/Drawing 20260122.md",
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@@ -328,7 +343,6 @@
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"Diary/2026-1/2026-1-19 周一.md",
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"Collection/PID控制算法.md",
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"Collection/linux命令.md",
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"README.md",
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"Excalidraw/Untitled.canvas",
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"Article/Studyroad/vivado_ip_img/IPCore10.png",
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"Article/Studyroad/vivado_ip_img/IPCore07.png",
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@@ -338,8 +352,6 @@
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"Article/Studyroad/vivado_ip_img/IPCore02.png",
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"Article/Studyroad/vivado_ip_img/IPCore03.png",
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"Article/Studyroad/vivado_ip_img/IPCore17.png",
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"Article/Studyroad/vivado_ip_img/IPCore04.png",
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"Article/Studyroad/vivado_ip_img/IPCore14.png",
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"YueQian/Homework/未命名.canvas",
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"Excalidraw/未命名.canvas",
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"未命名.canvas"
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@@ -5,36 +5,160 @@ tags:
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aliases: empty
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日期: 2026/1/29
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---
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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
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1. 判定该数组是否有序
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2. 若无序则将数组中的数据进行排序
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3. 找出数组中是否存在众数并输出结果
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```c
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```
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2. 定义一个二维数组存储一个3x3的矩阵，实现如下功能
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```c
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```
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3. 计算矩阵对角元素的和
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```c
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```
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4. 判断矩阵是否是对称矩阵
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```c
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```
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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：
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- 从某位员工开始顺时针报数
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- 报到数字M的倍数员工被淘汰出局
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- 从下一位员工继续报数，重复步骤2
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- 游戏继续直到只剩下K名员工（获胜者）
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```c
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```
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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：
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- 输入总人数N、报数间隔M和获胜人数K
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- 计算并输出安全的座位位置（最后剩下的K人）
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||||
```c
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||||
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||||
```
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||||
**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**
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||||
1. 判定该数组是否有序
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||||
2. 若无序则将数组中的数据进行排序
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||||
3. 找出数组中是否存在众数并输出结果
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```c
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#include <stdio.h>
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void bubble_sort(int arr[],int lenth)
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{
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for(int i = 0;i < lenth;i++){
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for(int j = i+1;j < lenth;j++){
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if(arr[i] > arr[j]){
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arr[i] ^= arr[j];
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arr[j] ^= arr[i];
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arr[i] ^= arr[j];
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}
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}
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}
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}
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int sort_check(int arr[],int lenth)
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{
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int count = 0,fcount=0;
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for(int i=0,j=1;i<lenth-1;i++,j++){
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if(arr[i] >= arr[j])
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count++;
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if(arr[i] <= arr[j])
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fcount++;
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}
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if(count == lenth-1 || fcount == lenth-1)
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return 1;
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else
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return 0;
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}
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void find_mode(int arr[], int length)
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{
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int count[10] = {0};
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int max_freq = 1;
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for (int i = 0; i < length; i++) {
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if (count[i] != -1) {
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int freq = 1;
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for (int j = i + 1; j < length; j++) {
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if (arr[i] == arr[j]) {
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freq++;
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count[j] = -1;
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}
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}
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count[i] = freq;
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if (freq > max_freq) {
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max_freq = freq;
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}
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}
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}
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printf("数组众数结果：");
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int has_mode = 0;
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for (int i = 0; i < length; i++) {
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if (count[i] == max_freq) {
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if (has_mode) printf("、");
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printf("%d（出现%d次）", arr[i], max_freq);
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has_mode = 1;
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}
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}
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if (max_freq == 1) {
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printf("所有元素均仅出现1次，无有效众数\n");
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} else {
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printf("（最大频次：%d）\n", max_freq);
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}
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}
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int main(int argc, char const *argv[])
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{
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int arr[10];
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int mode;
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printf("输入10个数:");
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for(int i=0;i<10;i++)
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scanf("%d",&arr[i]);
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if(sort_check(arr,10))
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printf("已排序\n");
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else{
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printf("未排序\n");
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bubble_sort(arr,10);
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printf("排序后为:");
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for(int i=0;i<10;i++)
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printf("%d ",arr[i]);
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printf("\n");
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}
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find_mode(arr,10);
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return 0;
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}
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```
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**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
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2. 判断矩阵是否是对称矩阵
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```c
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#include <stdio.h>
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#define ROW 3
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#define COL 3
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int calcDiagSum(int matrix[ROW][COL])
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{
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int sum = 0;
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for (int i = 0; i < ROW; i++) {
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sum += matrix[i][i];
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}
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return sum;
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}
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int isSymmetricMatrix(int matrix[ROW][COL])
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{
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for (int i = 0; i < ROW; i++) {
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for (int j = 0; j < COL; j++) {
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if (matrix[i][j] != matrix[j][i]) {
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return 0;
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}
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}
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}
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return 1;
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}
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int main(int argc, char const *argv[])
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{
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int matrix[ROW][COL];
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printf("请输入3x3矩阵的9个整数：\n");
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for (int i = 0; i < ROW; i++) {
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for (int j = 0; j < COL; j++) {
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scanf("%d", &matrix[i][j]);
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}
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}
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int diagSum = calcDiagSum(matrix);
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printf("\n矩阵主对角线元素的和为：%d\n", diagSum);
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if (isSymmetricMatrix(matrix))
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printf("该3x3矩阵是对称矩阵\n");
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else
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printf("该3x3矩阵不是对称矩阵\n");
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return 0;
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}
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```
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**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
|
||||
- 游戏继续直到只剩下K名员工（获胜者）
|
||||
```c
|
||||
|
||||
```
|
||||
**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
|
||||
|
||||
```
|
||||
BIN
YueQian/Homework/assets/1-29/file-20260129201415992.png
Normal file
BIN
YueQian/Homework/assets/1-29/file-20260129201415992.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 53 KiB |
BIN
YueQian/Homework/assets/1-29/file-20260129202925505.png
Normal file
BIN
YueQian/Homework/assets/1-29/file-20260129202925505.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 43 KiB |
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