实验6

任务1

源代码

#define N 10        // 运行程序输入测试时，可以把N改小一些输入测试 

typedef struct student {
    int id;             // 学号 
    char name[20];         // 姓名 
    char subject[20];     // 考试科目
    double perf;         // 平时成绩 
    double mid;         // 期中成绩 
    double final;         // 期末成绩
    double total;         // 总评成绩 
    char level[10];     // 成绩等级
} STU;

void input(STU [], int);            // 录入学生信息
void output(STU [], int);            // 输出学生信息
void calc(STU [], int);                // 计算总评和等级 
int fail(STU [], STU [], int);        // 统计不及格学生信息
void sort(STU [], int);                // 排序 

int main() {
    STU st[N], fst[N];   // 数组st记录学生信息，fst记录不及格学生信息 
    int k;  // 用于记录不及格学生个数 
    
    printf("录入学生成绩信息:\n");
    input(st, N);
    
    printf("\n成绩处理...\n");
    calc(st, N);
    
    k = fail(st, fst, N);
    sort(st, N);
    printf("\n学生成绩排名情况:\n");
    output(st, N);
    
    printf("\n不及格学生信息:\n");
    output(fst, k);
    
    return 0;
} 

void input(STU s[], int n) {
    int i;
    
    for(i = 0; i < n; i++)
        scanf("%d %s %s %lf %lf %lf", &s[i].id, s[i].name, s[i].subject,
                                      &s[i].perf, &s[i].mid, &s[i].final);
}

void output(STU s[], int n) {
       int i;
   
      printf("-----------------\n");
      printf("学号   姓名     科目   平时   期中   期末   总评   等级\n");
       for(i = 0; i<n; i++)
          printf("%d   %-6s   %-4s   %-4.0f   %-4.0f   %-4.0f   %-4.1f   %s\n",s[i].id,s[i].name,s[i].subject,s[i].perf,s[i].mid,s[i].final,s[i].total,s[i].level);
}


void calc(STU s[],int n) {
    int i;

    for(i = 0; i < n; i++) {    
        s[i].total = s[i].perf * 0.2 + 
                     s[i].mid * 0.2 +
                     s[i].final * 0.6;
        
        if(s[i].total >= 90)
          strcpy(s[i].level, "优");
        else if(s[i].total >= 80)
          strcpy(s[i].level, "良");
        else if(s[i].total >= 70)
          strcpy(s[i].level, "中"); 
        else if(s[i].total >= 60)
          strcpy(s[i].level, "及格");
        else
          strcpy(s[i].level, "不及格");         
    }
}

int fail(STU s[], STU t[], int n) {
      int i, cnt = 0;
      
      for(i = 0; i < n; i++)
          if(s[i].total < 60)
            t[cnt++] = s[i];
            
    return cnt;
}

void sort(STU s[], int n) {
    int i, j;
    STU t;
    
    for(i = 0; i < n-1; i++)
      for(j = 0; j < n-1-i; j++)
        if(s[j].total < s[j+1].total) {
            t = s[j];
            s[j] = s[j+1];
            s[j+1] = t;
        }
}

运行结果

任务2

源代码

#include <stdio.h>
#include <string.h>
#define N 10
#define M 80

typedef struct {
    char name[M];       // 书名
    char author[M];     // 作者
} Book;

int main() {
    Book x[N] = { {"《一九八四》", "乔治.奥威尔"},
                  {"《美丽新世界》", "赫胥黎"},
                  {"《昨日的世界》", "斯蒂芬.茨威格"}, 
                  {"《万历十五年》", "黄仁宇"},
                  {"《一只特立独行的猪》", "王小波"},
                  {"《百年孤独》", "马尔克斯"},
                  {"《查令十字街84号》", "海莲.汉芙"},
                  {"《只是孩子》", "帕蒂.史密斯"}, 
                  {"《刀锋》", "毛姆"},
                  {"《沉默的大多数》", "王小波"} };
    Book *ptr;
    int i;
    char author[M];
    int found;

    // 使用指针遍历结构体数组
    printf("-------------------所有图书信息-------------------\n");
    for(ptr = x; ptr < x + N; ++ptr)
        printf("%-30s%-30s\n", ptr->name, ptr->author);

    // 查找指定作者的图书
    printf("\n-------------------按作者查询图书-------------------\n");
    printf("输入作者名: ");
    gets(author);
    found = 0;
    for(ptr = x; ptr < x + N; ++ptr)
        if(strcmp(ptr->author, author) == 0) {
            found = 1;
            printf("%-30s%-30s\n", ptr->name, ptr->author);
        }
    
    if(!found) 
        printf("暂未收录该作者书籍!\n");

    return 0;
}

运行结果

任务3

3.1源代码

#include <stdio.h>
#include <stdlib.h>
#define N 80

typedef struct FilmInfo {
    char name[N];
    char director[N];
    char region[N];
    int year;
    struct FilmInfo *next;
} Film;


void output(Film *head);   // 遍历输出链表信息
Film *insert(Film *head, int n);   // 向链表中插入n个结点，返回头指针


int main() {
    int n;          // 结点数
    Film *head;     // 头指针变量，存放链表中第一个节点的地址

    head = NULL;
    printf("输入影片数目: ");
    scanf("%d", &n);

    // 向链表中插入n部影片信息
    head = insert(head, n);

    // 遍历输出链表中所有影片信息
    printf("\n所有影片信息如下: \n");
    output(head);

    return 0;
}

// 向链表中插入n个结点，从表头插入，返回头指针变量
Film *insert(Film *head, int n) {
    int i;
    Film *p;

    for(i = 1; i <= n; ++i) {
        p = (Film *)malloc(sizeof(Film));
        printf("请输入第%d部影片信息: ", i);
        scanf("%s %s %s %d", p->name, p->director, p->region, &p->year);
        
        // 把结点从表头插入到链表中
        p->next = head;
        head = p;   // 更新头指针变量
    }

    return head;
}

// 遍历输出链表信息
void output(Film *head) {
    Film *p;

    p = head;
    while(p != NULL) {
        printf("%-20s %-20s %-20s %d\n", p->name, p->director, p->region, p->year);
        p = p -> next;
    }
}

运行结果

3.2 源代码

#include <stdio.h>
#include <stdlib.h>
#define N 80

typedef struct FilmInfo {
    char name[N];
    char director[N];
    char region[N];
    int year;
    struct FilmInfo *next;
} Film;


void output(Film *head);   // 遍历输出链表信息
Film *insert(Film *head, int n);   // 向链表中插入n个结点，返回头指针


int main() {
    int n;          // 结点数
    Film *head;     // 头指针变量，存放链表中第一个节点的地址
    Film *p;        // 存放新申请的Film节点内存空间地址

    // 创建头结点
    p = (Film *)malloc(sizeof(Film));
    p->next = NULL;
    head = p;       // 头指针变量存放头节点的地址

    printf("输入影片数目: ");
    scanf("%d", &n);

    // 向链表中插入n部影片信息
    head = insert(head, n);

    // 遍历输出链表中所有影片信息
    printf("\n所有影片信息如下: \n");
    output(head);

    return 0;
}

// 向链表中插入n个结点，从表头插入，返回头指针变量
Film *insert(Film *head, int n) {
    int i;
    Film *p;

    for(i = 1; i <= n; ++i) {
        p = (Film *)malloc(sizeof(Film));
        printf("请输入第%d部影片信息: ", i);
        scanf("%s %s %s %d", p->name, p->director, p->region, &p->year);
        
        // 把结点从表头插入到链表中
        p->next = head->next;
        head->next = p;
    }

    return head;
}

// 遍历输出链表信息
void output(Film *head) {
    Film *p;

    p = head->next;
    while(p != NULL) {
        printf("%-20s %-20s %-20s %d\n", p->name, p->director, p->region, p->year);
        p = p -> next;
    }
}

任务4

源代码

#include <stdio.h>
#define N 10

typedef struct {
    char isbn[20];          // isbn号
    char name[80];          // 书名
    char author[80];        // 作者
    double sales_price;     // 售价
    int  sales_count;       // 销售册数
} Book;

void output(Book x[], int n);
void sort(Book x[], int n);
double sales_amount(Book x[], int n);

int main() {
     Book x[N] = {{"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51},
                  {"978-7-308-17047-5", "自由与爱之地：入以色列记", "云也退", 49 , 30},
                  {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27},
                  {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90}, 
                  {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
                  {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42},
                  {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44},
                  {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42},
                  {"978-7-1155-0509-5", "无穷的开始:世界进步的本源", "戴维·多伊奇", 37.5, 55},
                  {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59}};
    
    printf("图书销量排名(按销售册数): \n");
    sort(x, N);
    output(x, N);
     double total=sales_amount(x,N);
    printf("\n图书销售总额:%.2lf 元\n",total);
    return 0;
}
void output(Book x[], int n){
    printf("%-22s%-25s%-15s%8s%8s\n","isbn","书名","作者","售价","销售册数");
    int i;
    for(i=0;i<n;i++){
        printf("%-22s%-25s%-15s%8.2lf%8d\n",x[i].isbn,x[i].name,x[i].author,x[i].sales_price,x[i].sales_count);    
    }
}

void sort(Book x[], int n){
    int i,j;
    Book temp;
    for(i=0;i<n-1;i++){
        for(j=0;j<n-1;j++){
            if(x[j].sales_count<x[j+1].sales_count){
                temp=x[j];
                x[j]=x[j+1];
                x[j+1]=temp;
            }
        }
    }
}
double sales_amount(Book x[], int n){
    double sum=0;
    for(int i=0;i<n;i++){
        sum+=x[i].sales_price*x[i].sales_count;
    return sum;}
    
}

运行结果

任务5

源代码

  #include <stdio.h>
  typedef struct {
     int year;
      int month;
      int day;
 } Date;
 
 void input(Date* pd);
  int day_of_year(Date d);
 int compare_dates(Date d1, Date d2);

 void test1() {
    Date d;
    int i;
     printf("输入日期:(以形如2025-12-19这样的形式输入)\n");
     for (i = 0; i < 3; ++i) {
        input(&d);
         printf("%d-%02d-%02d是这一年中第%d天\n", d.year, d.month, d.day, day_of_year(d));
     }
 }
 
 void test2() {
     Date Alice_birth, Bob_birth;
     int i;
     int ans;
     printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n");
    for (i = 0; i < 3; ++i) {
        input(&Alice_birth);
         input(&Bob_birth);
         ans = compare_dates(Alice_birth, Bob_birth);
         if (ans == 0)
             printf("Alice和Bob一样大\n");
         else if (ans == -1)
             printf("Alice比Bob大\n");
         else
             printf("Alice比Bob小\n");
     }
 }
 
 int main() {
     printf("测试1: 输入日期，打印输出这是一年中第多少天\n");
     test1();
     printf("\n测试2: 两个人年龄大小关系\n");
     test2();
 }
 
 void input(Date* pd) {
     scanf_s("%d-%d-%d", &pd->year, &pd->month, &pd->day);
 }
 
 int day_of_year(Date d) {
     int mon[12] = { 31,28,31,30,31,30,31,31,30,31,30,31 };
     int sum = d.day;
     for (int i = 0; i < d.month - 1; i++) {
         sum += mon[i];
     }
     if ((d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0)) {
         if (d.month > 2) sum++;
    }
     return sum;
 }

 int compare_dates(Date d1, Date d2) {
     if (d1.year > d2.year) return 1;
    if (d1.year < d2.year) return -1;
    if (d1.month > d2.month) return 1;
     if (d1.month < d2.month) return -1;
    if (d1.day > d2.day) return 1;
     if (d1.day < d2.day) return -1;
     return 0;
 }

运行结果

任务6

源代码

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>

enum Role { admin, student, teacher };

typedef struct {
    char username[20];
    char password[20];
    enum Role type;
} Account;

void output(Account x[], int n);

int main() {
    Account x[] = {
        {"A1001", "123456", student},
        {"A1002", "123abcdef", student},
        {"A1009", "xyz12121", student},
        {"X1009", "9213071x", admin},
        {"C11553", "129dfg32k", teacher},
        {"x3005", "921kfmg917", student}
    };
    int n;
    n = sizeof(x) / sizeof(Account);
    output(x, n);
    return 0;
}

void output(Account x[], int n) {
    for (int i = 0; i < n; i++) {
        char pwd_mask[20] = { 0 };
        int len = strlen(x[i].password);
        for (int j = 0; j < len; j++) {
            pwd_mask[j] = '*';
        }
        char role_str[20];
        if (x[i].type == admin) strcpy(role_str, "admin");
        else if (x[i].type == student) strcpy(role_str, "student");
        else strcpy(role_str, "teacher");
        printf("用户名：%s  密码：%s  角色：%s\n", x[i].username, pwd_mask, role_str);
    }
}

运行结果

任务7

源代码

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string.h>

typedef struct {
    char name[20];
    char phone[12];
    int vip;
} Contact;

void set_vip_contact(Contact x[], int n, char name[]);
void output(Contact x[], int n);
void display(Contact x[], int n);

#define N 10
int main() {
    Contact list[N] = {
        {"刘一", "15510846604", 0},
        {"陈二", "18038747351", 0},
        {"张三", "18853253914", 0},
        {"李四", "13230584477", 0},
        {"王五", "15547571923", 0},
        {"赵六", "18856659351", 0},
        {"周七", "17705843215", 0},
        {"孙八", "15552933732", 0},
        {"吴九", "18077702405", 0},
        {"郑十", "18820725036", 0}
    };
    int vip_cnt, i;
    char name[20];

    printf("显示原始通讯录信息：\n");
    output(list, N);

    printf("\n输入要设置的紧急联系人个数：");
    scanf("%d", &vip_cnt);
    printf("输入%d个紧急联系人姓名:\n", vip_cnt);
    for (i = 0; i < vip_cnt; ++i) {
        scanf("%s", name);
        set_vip_contact(list, N, name);
    }

    printf("\n显示通讯录列表:(按姓名字典升序排列,紧急联系人最先显示)\n");
    display(list, N);

    return 0;
}

void set_vip_contact(Contact x[], int n, char name[]) {
    for (int i = 0; i < n; i++) {
        if (strcmp(x[i].name, name) == 0) {
            x[i].vip = 1;
            break;
        }
    }
}

void output(Contact x[], int n) {
    int i;
    for (i = 0; i < n; ++i) {
        printf("%-10s%-15s", x[i].name, x[i].phone);
        if (x[i].vip)
            printf("%5s", "*");
        printf("\n");
    }
}

void display(Contact x[], int n) {
    Contact temp[N];
    int vip_num = 0;
    for (int i = 0; i < n; i++) {
        if (x[i].vip == 1) {
            temp[vip_num++] = x[i];
        }
    }
    for (int i = 0; i < n; i++) {
        if (x[i].vip == 0) {
            temp[vip_num++] = x[i];
        }
    }
    for (int i = 0; i < n - 1; i++) {
        for (int j = 0; j < n - 1 - i; j++) {
            if (temp[j].vip == 0 && temp[j + 1].vip == 1) {
                Contact t = temp[j];
                temp[j] = temp[j + 1];
                temp[j + 1] = t;
            }
            else if (temp[j].vip == temp[j + 1].vip && strcmp(temp[j].name, temp[j + 1].name) > 0) {
                Contact t = temp[j];
                temp[j] = temp[j + 1];
                temp[j + 1] = t;
            }
        }
    }
    output(temp, n);
}

运行结果

posted @ 2026-06-16 21:54 FineLe 阅读(9) 评论(0) 收藏举报

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