用c语言结合线性结构特点,将某班级的学生成绩单设计为线性表,使用顺序表或链表结构,表中存储每位学生的信息包括学号、姓名、班级、数据结构成绩、Java成绩、数据库成绩六个部分,并设计算法统计及格人数和最高分。
时间: 2024-11-13 10:22:48 浏览: 16
在C语言中,我们可以使用顺序表或链表作为基础数据结构来表示班级的成绩单。这里我们选择链表作为例子,因为它提供了更好的动态插入和删除元素的能力。
首先,我们需要创建一个`Student`结构体,包含学号、姓名、班级、数据结构成绩、Java成绩和数据库成绩等字段:
```c
typedef struct {
int student_id;
char name[50];
char class_name[50];
float ds_score;
float java_score;
float db_score;
} StudentNode;
```
接下来,我们可以定义一个`StudentList`链表结构,用于存储`StudentNode`:
```c
typedef struct {
StudentNode* head;
int size;
} StudentList;
```
然后,我们可以定义函数来初始化链表,添加学生,以及计算及格人数和最高分:
1. 初始化链表:
```c
void initStudentList(StudentList* list) {
list->head = NULL;
list->size = 0;
}
```
2. 添加学生到链表:
```c
void addStudent(StudentList* list, const StudentNode* new_student) {
StudentNode* newNode = (StudentNode*)malloc(sizeof(StudentNode));
if (newNode != NULL) {
newNode->student_id = new_student->student_id;
strcpy(newNode->name, new_student->name);
strcpy(newNode->class_name, new_student->class_name);
newNode->ds_score = new_student->ds_score;
newNode->java_score = new_student->java_score;
newNode->db_score = new_student->db_score;
newNode->next = list->head;
list->head = newNode;
list->size++;
}
}
```
3. 计算及格人数(假设60分为及格线):
```c
int countPassingStudents(StudentList* list) {
int passing_count = 0;
StudentNode* current = list->head;
while (current != NULL) {
if (current->ds_score >= 60 && current->java_score >= 60 && current->db_score >= 60) {
passing_count++;
}
current = current->next;
}
return passing_count;
}
```
4. 找出最高分(可以针对每个科目分别找出):
```c
float findHighestScore(StudentList* list, char subject) {
float highest_score = 0.0;
StudentNode* current = list->head;
while (current != NULL) {
if (strcmp(subject, "ds") == 0 && current->ds_score > highest_score) {
highest_score = current->ds_score;
} else if (strcmp(subject, "java") == 0 && current->java_score > highest_score) {
highest_score = current->java_score;
} else if (strcmp(subject, "db") == 0 && current->db_score > highest_score) {
highest_score = current->db_score;
}
current = current->next;
}
return highest_score;
}
```
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