实验6
task1
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include <string.h> 4 #define N 3 5 typedef struct student { 6 int id; 7 char name[20]; 8 char subject[20]; 9 double perf; 10 double mid; 11 double final; 12 double total; 13 char level[10]; 14 }STU; 15 void input(STU[], int); 16 void output(STU[], int); 17 void calc(STU[], int); 18 int fail(STU[], STU[], int); 19 void sort(STU[], int); 20 int main() { 21 STU st[N], fst[N]; 22 int k; 23 24 printf("录入学生成绩信息：\n"); 25 input(st, N); 26 27 printf("\n成绩处理...\n"); 28 calc(st, N); 29 30 k = fail(st, fst, N); 31 sort(st, N); 32 printf("\n学生成绩排名情况：\n"); 33 output(st, N); 34 printf("\n不及格学生信息：\n"); 35 output(fst, k); 36 37 return 0; 38 } 39 void input(STU s[], int n) { 40 for (int i = 0; i < n; i++) { 41 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); 42 } 43 }void output(STU s[], int n) { 44 int i; 45 printf("-------------------------\n"); 46 printf("学号 姓名 科目 平时 期中 期末 总评 等级\n"); 47 for (i = 0; i < n; i++) { 48 printf("%d %-6s %-4s %-4.0f %-4.0f %-4.0f %-4.lf %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); 49 } 50 } 51 void calc(STU s[], int n) { 52 for (int i = 0; i < n; i++) { 53 s[i].total = s[i].perf * 0.2 + s[i].mid * 0.2 + s[i].final * 0.6; 54 if (s[i].total >= 90) 55 strcpy(s[i].level, "优"); 56 else if (s[i].total >= 80) 57 strcpy(s[i].level, "良"); 58 else if (s[i].total >= 70) 59 strcpy(s[i].level, "中"); 60 else if (s[i].total >= 60) 61 strcpy(s[i].level, "及格"); 62 else 63 strcpy(s[i].level, "不及格"); 64 } 65 } 66 int fail(STU s[], STU t[], int n) { 67 int cnt = 0; 68 for (int i = 0; i < n; i++) 69 if (s[i].total < 60) 70 t[cnt++] = s[i]; 71 return cnt; 72 } 73 void sort(STU s[], int n) { 74 STU t; 75 for (int i = 0; i < n - 1; i++) { 76 for (int j = 0; j < n - 1 - i; j++) 77 if (s[j].total < s[j + 1].total) { 78 t = s[j]; 79 s[j] = s[j + 1]; 80 s[j + 1] = t; 81 } 82 } 83 }
运行结果
task2
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include<string.h> 4 #define N 10 5 #define M 80 6 typedef struct { 7 char name[M]; 8 char author[M]; 9 }Book; 10 int main() { 11 Book x[N] = { {"《一九八四》", "乔治.奥威尔"}, 12 {"《美丽新世界》", "赫胥黎"}, 13 {"《昨日的世界》", "斯蒂芬.茨威格"}, 14 {"《万历十五年》", "黄仁宇"}, 15 {"《一只特立独行的猪》", "王小波"}, 16 {"《百年孤独》", "马尔克斯"}, 17 {"《查令十字街84号》", "海莲.汉芙"}, 18 {"《只是孩子》", "帕蒂.史密斯"}, 19 {"《刀锋》", "毛姆"}, 20 {"《沉默的大多数》", "王小波"} }; 21 Book* ptr; 22 char author[M]; 23 int found; 24 printf("-------------------所有图书信息-------------------\n"); 25 for (ptr = x; ptr < x + N; ++ptr) 26 printf("%-30s%-30s\n", ptr->name, ptr->author); 27 printf("\n-------------------按作者查询图书-------------------\n"); 28 printf("输入作者名："); 29 gets(author); 30 found = 0; 31 for (ptr = x; ptr < x + N; ++ptr) { 32 if (strcmp(ptr->author, author) == 0) { 33 found = 1; 34 printf("%-30s%-30s\n", ptr->name, ptr->author); 35 } 36 } 37 if (!found) 38 printf("暂未收录该作者书籍！\n"); 39 return 0; 40 41 }
运行结果
task3_1
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include<stdlib.h> 4 #define N 80 5 typedef struct Filminfo { 6 char name[N]; 7 char director[N]; 8 char region[N]; 9 int year; 10 struct Filminfo* next; 11 }Film; 12 void output(Film* head); 13 Film* insert(Film* head, int n); 14 int main() { 15 int n; 16 Film* head; 17 head = NULL; 18 printf("输入影片数目："); 19 scanf("%d", &n); 20 head = insert(head, n); 21 printf("\n所有影片信息如下：\n"); 22 output(head); 23 return 0; 24 } 25 Film* insert(Film* head, int n) { 26 Film* p; 27 for (int i = 1; i <= n; i++) { 28 p = (Film*)malloc(sizeof(Film)); 29 printf("请输入第%d部影片信息：", i); 30 scanf("%s %s %s %d", p->name,p->director, p->region, &p->year); 31 p->next = head; 32 head = p; 33 } 34 return head; 35 } 36 void output(Film* head) { 37 Film* p; 38 p = head; 39 while (p != NULL) { 40 printf("%-20s %-20s %-20s %d\n", p->name, p->director, p->region, p->year); 41 p = p->next; 42 } 43 }
运行结果
task3_2
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include<stdlib.h> 4 #define N 80 5 typedef struct Filminfo { 6 char name[N]; 7 char director[N]; 8 char region[N]; 9 int year; 10 struct Filminfo* next; 11 }Film; 12 void output(Film* head); 13 Film* insert(Film* head, int n); 14 int main() { 15 int n; 16 Film* head; 17 Film* p; 18 p = (Film*)malloc(sizeof(Film)); 19 p->next = NULL; 20 head = p; 21 printf("输入影片数目："); 22 scanf("%d", &n); 23 head = insert(head, n); 24 printf("\n所有影片信息如下：\n"); 25 output(head); 26 return 0; 27 } 28 Film* insert(Film* head, int n) { 29 Film* p; 30 for (int i = 1; i <= n; i++) { 31 p = (Film*)malloc(sizeof(Film)); 32 printf("请输入第%d部影片信息：", i); 33 scanf("%s %s %s %d", p->name, p->director, p->region, &p->year); 34 p->next = head->next; 35 head->next = p; 36 } 37 return head; 38 } 39 void output(Film* head) { 40 Film* p; 41 p = head->next; 42 while (p != NULL) { 43 printf("%-20s %-20s %-20s %d\n", p->name, p->director, p->region, p->year); 44 p = p->next; 45 } 46 }
运行结果
task4
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #define N 10 4 typedef struct { 5 char isbn[20]; 6 char name[80]; 7 char author[80]; 8 double sales_price; 9 int sales_count; 10 }Book; 11 void output(Book x[], int n); 12 void sort(Book x[], int n); 13 double sales_amount(Book x[], int n); 14 int main() { 15 Book x[N] = { {"978-7-5327-6082-4", "门将之死", "罗纳德.伦", 42, 51}, 16 {"978-7-308-17047-5", "自由与爱之地：入以色列记", "云也退", 49 , 30}, 17 {"978-7-5404-9344-8", "伦敦人", "克莱格泰勒", 68, 27}, 18 {"978-7-5447-5246-6", "软件体的生命周期", "特德姜", 35, 90}, 19 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49}, 20 {"978-7-5133-5750-0", "主机战争", "布莱克.J.哈里斯", 128, 42}, 21 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}, 22 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42}, 23 {"978-7-1155-0509-5", "无穷的开始:世界进步的本源", "戴维·多伊奇", 37.5, 24 55}, 25 {"978-7-229-14156-1", "源泉", "安.兰德", 84, 59} }; 26 printf("图书销量排名（按销售册数）：\n"); 27 sort(x, N); 28 output(x, N); 29 printf("\n图书销售总额：%.2f\n", sales_amount(x, N)); 30 } 31 void output(Book x[], int n) { 32 printf("ISBN号\t\t 书名\t\t 作者\t\t 售价\t\t 销售册数\n"); 33 for (int i = 0; i < n; i++) { 34 printf("%-20s%-30s%-20s%-20lf%-20d\n", x[i].isbn, x[i].name, x[i].author, x[i].sales_price, x[i].sales_count); 35 } 36 } 37 void sort(Book x[], int n) { 38 Book temp; 39 for (int i = 0; i < n-1; i++) { 40 for (int j = 0; j < n-i-1; j++) { 41 if (x[j].sales_count < x[j + 1].sales_count) { 42 temp = x[j]; 43 x[j] = x[j + 1]; 44 x[j + 1] = temp; 45 } 46 } 47 } 48 49 } 50 double sales_amount(Book x[], int n) { 51 double total = 0; 52 for (int i = 0; i < n; i++) { 53 total += x[i].sales_price * x[i].sales_count; 54 } 55 return total; 56 57 }
运行结果
task5
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 4 typedef struct { 5 int year; 6 int month; 7 int day; 8 } Date; 9 10 // 函数声明 11 void input(Date* pd); // 输入日期给pd指向的Date变量 12 int day_of_year(Date d); // 返回日期d是这一年的第多少天 13 int compare_dates(Date d1, Date d2); // 比较两个日期: 14 // 如果d1在d2之前，返回-1; 15 // 如果d1在d2之后，返回1 16 // 如果d1和d2相同，返回0 17 18 void test1() { 19 Date d; 20 int i; 21 22 printf("输入日期:(以形如2025-12-19这样的形式输入)\n"); 23 for (i = 0; i < 3; ++i) { 24 input(&d); 25 printf("%d-%02d-%02d是这一年中第%d天\n\n", d.year, d.month, d.day, day_of_year(d)); 26 } 27 } 28 29 void test2() { 30 Date Alice_birth, Bob_birth; 31 int i; 32 int ans; 33 34 printf("输入Alice和Bob出生日期:(以形如2025-12-19这样的形式输入)\n"); 35 for (i = 0; i < 3; ++i) { 36 input(&Alice_birth); 37 input(&Bob_birth); 38 ans = compare_dates(Alice_birth, Bob_birth); 39 40 if (ans == 0) 41 printf("Alice和Bob一样大\n\n"); 42 else if (ans == -1) 43 printf("Alice比Bob大\n\n"); 44 else 45 printf("Alice比Bob小\n\n"); 46 } 47 } 48 49 int main() { 50 printf("测试1: 输入日期, 打印输出这是一年中第多少天\n"); 51 test1(); 52 53 printf("\n测试2: 两个人年龄大小关系\n"); 54 test2(); 55 } 56 57 // 补足函数input实现 58 // 功能: 输入日期给pd指向的Date变量 59 void input(Date* pd) { 60 scanf("%d-%d-%d", &(pd->year), &(pd->month), &(pd->day)); 61 62 // 清除本行剩余字符（包括换行符），避免影响下一次输入 63 int c; 64 while ((c = getchar()) != '\n' && c != EOF); 65 66 67 } 68 int runnian(int year) { 69 if (year % 100 != 0 && year % 4 == 0) 70 return 1; 71 else if (year % 400 == 0) 72 return 1; 73 else 74 return 0; 75 } 76 77 // 补足函数day_of_year实现 78 // 功能：返回日期d是这一年的第多少天 79 int day_of_year(Date d) { 80 int a; 81 int sum=0; 82 if (runnian(d.year)) 83 a = 29; 84 else 85 a = 28; 86 int day_num[13] = { 0,31,a,31,30,31,30,31,31,30,31,30,31 }; 87 for (int i = 1; i < d.month; i++) { 88 sum += day_num[i]; 89 } 90 sum += d.day; 91 return sum; 92 } 93 94 // 补足函数compare_dates实现 95 // 功能：比较两个日期: 96 // 如果d1在d2之前，返回-1; 97 // 如果d1在d2之后，返回1 98 // 如果d1和d2相同，返回0 99 int compare_dates(Date d1, Date d2) { 100 if (d1.year < d2.year) 101 return -1; 102 else if (d1.year > d2.year) 103 return 1; 104 else { 105 if (d1.month < d2.month) 106 return -1; 107 else if (d1.month > d2.month) 108 return 1; 109 else { 110 if (d1.day < d2.day) 111 return -1; 112 else if (d1.day > d2.day) 113 return 1; 114 else 115 return 0; 116 117 } 118 } 119 120 }
运行结果
task6
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include <string.h> 4 5 enum Role { admin, student, teacher }; 6 7 typedef struct { 8 char username[20]; // 用户名 9 char password[20]; // 密码 10 enum Role type; // 账户类型 11 } Account; 12 13 14 // 函数声明 15 void output(Account x[], int n); // 输出账户数组x中n个账户信息，其中，密码用*替代显示 16 17 int main() { 18 Account x[] = { {"A1001", "123456", student}, 19 {"A1002", "123abcdef", student}, 20 {"A1009", "xyz12121", student}, 21 {"X1009", "9213071x", admin}, 22 {"C11553", "129dfg32k", teacher}, 23 {"X3005", "921kfmg917", student} }; 24 int n; 25 n = sizeof(x) / sizeof(Account); 26 output(x, n); 27 28 return 0; 29 } 30 31 // 待补足的函数output()实现 32 // 功能：遍历输出账户数组x中n个账户信息 33 // 显示时，密码字段以与原密码相同字段长度的*替代显示 34 void output(Account x[], int n) { 35 for (int i = 0; i < n; i++) { 36 int m = strlen(x[i].password); 37 for (int j=0; j < m; j++) { 38 x[i].password[j] = '*'; 39 } 40 } 41 for (int i = 0; i < n; i++) { 42 char str[100]; 43 switch (x[i].type) { 44 case admin:strcpy(str,"admin"); break; 45 case student:strcpy(str,"student"); break; 46 case teacher:strcpy(str,"teacher"); break; 47 } 48 printf("%-20s%-20s%-20s\n", x[i].username, x[i].password, str); 49 50 } 51 }
运行结果
task7
源代码
1 #define _CRT_SECURE_NO_WARNINGS 2 #include <stdio.h> 3 #include <string.h> 4 5 typedef struct { 6 char name[20]; // 姓名 7 char phone[12]; // 手机号 8 int vip; // 是否为紧急联系人，是取1；否则取0 9 } Contact; 10 11 12 // 函数声明 13 void set_vip_contact(Contact x[], int n, char name[]); // 设置紧急联系人 14 void output(Contact x[], int n); // 输出x中联系人信息 15 void display(Contact x[], int n); // 按联系人姓名字典序升序显示信息，紧急联系人最先显示 16 17 18 #define N 10 19 int main() { 20 Contact list[N] = { {"刘一", "15510846604", 0}, 21 {"陈二", "18038747351", 0}, 22 {"张三", "18853253914", 0}, 23 {"李四", "13230584477", 0}, 24 {"王五", "15547571923", 0}, 25 {"赵六", "18856659351", 0}, 26 {"周七", "17705843215", 0}, 27 {"孙八", "15552933732", 0}, 28 {"吴九", "18077702405", 0}, 29 {"郑十", "18820725036", 0} }; 30 int vip_cnt, i; 31 char name[20]; 32 33 printf("显示原始通讯录信息: \n"); 34 output(list, N); 35 36 printf("\n输入要设置的紧急联系人个数: "); 37 scanf("%d", &vip_cnt); 38 39 printf("输入%d个紧急联系人姓名:\n", vip_cnt); 40 for (i = 0; i < vip_cnt; ++i) { 41 scanf("%s", name); 42 set_vip_contact(list, N, name); 43 } 44 45 printf("\n显示通讯录列表:(按姓名字典序升序排列,紧急联系人最先显示)\n"); 46 display(list, N); 47 48 return 0; 49 } 50 51 // 补足函数set_vip_contact实现 52 // 功能：将联系人数组x中，联系人姓名与name一样的人，设置为紧急联系人（即成员vip值设为1） 53 void set_vip_contact(Contact x[], int n, char name[]) { 54 for (int i = 0; i < n; i++) { 55 if (!strcmp(name, x[i].name)) 56 x[i].vip = 1; 57 } 58 } 59 60 // 补足函数display实现 61 // 功能: 显示联系人数组x中的联系人信息 62 // 按姓名字典序升序显示, 紧急联系人显示在最前面 63 void display(Contact x[], int n) { 64 Contact temp; 65 for (int i = 0; i < n - 1; i++) { 66 for (int j = 0; j < n - 1 - i; j++) { 67 int h = 0; 68 if (x[j].vip < x[j + 1].vip) { 69 h = 1; 70 } 71 else if (x[j].vip == x[j + 1].vip) { 72 if (strcmp(x[j].name, x[j + 1].name) < 0) 73 h = 1; 74 } 75 if (h) { 76 temp = x[j]; 77 x[j] = x[j + 1]; 78 x[j + 1] = temp; 79 } 80 } 81 } 82 for (int i = 0; i < n; i++) { 83 printf("%-20s%-20s%-20d\n", x[i].name, x[i].phone, x[i].vip); 84 } 85 86 } 87 88 void output(Contact x[], int n) { 89 int i; 90 91 for (i = 0; i < n; ++i) { 92 printf("%-10s%-15s", x[i].name, x[i].phone); 93 if (x[i].vip) 94 printf("%5s", "*"); 95 printf("\n"); 96 } 97 }
运行结果
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