实验 6

任务 1

#include <stdio.h>
#include <string.h> 
#define N 3     

typedef struct student {
    int id;         
    char name[20];         
    char subject[20];     
    double perf;         
    double mid;         
    double final;         
    double total;     
    char level[10]; 
} STU;

void input(STU [], int);            
void output(STU [], int);           
void calc(STU [], int);                
int fail(STU [], STU [], int);        
void sort(STU [], int);                

int main() {
    STU st[N], fst[N];    
    int k;  
    
    printf("录入学生成绩信息:\n");
    input(st, N);
    
    printf("\n loading ...\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("---------我是分割线--------\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;
        }
}

 

 

 

任务 2

#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] = { {"《十日终焉》", "杀虫队队长"},
                  {"《美丽新世界》", "赫胥黎"},
                  {"《飘》", "米切尔"}, 
                  {"《小王子》", "安托万"},
                  {"《天天睡大觉》", "沃兹基硕德"},
                  {"《百年孤独》", "马尔克斯"},
                  {"《撒野》", "巫哲"},
                  {"《悲惨世界》", "雨果"}, 
                  {"《鼠疫》", "加缪"},
                  {"《高等数学》", "王顺凤"} };
    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;
}

 

 

 

任务 3

#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);   
Film *insert(Film *head, int n);   


int main() {
    int n;          
    Film *head;     

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

   
    head = insert(head, n);

    
    printf("\n所有影片信息如下: \n");
    output(head);

    return 0;
}


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);
        
        
        p->next = head;
        head = p;   
    }

    return head;
}


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;
    }
}

 

 

 

任务 4

#include<stdio.h>
#define N 10

typedef struct{
    char isbn[20];
    char name[80];
    char author[80];
    double sales_price;
    int sales_count;
} Book ;

void output(Book x[], int n);
void sort(Book x[], int n);
double salse_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("图书馆销量排名（按销售册数）：\n");
sort(x,N);
output(x,N);

printf("\n图书销售总额：%.2f\n",salse_amount(x,N));

return 0;

}


void output(Book x[], int n){
    printf("ISBN号            书名                      作者           售价   销售册数\n");
    int i;
    for(i=0;i<n;i++){
        printf("%-15s %-25s %-15s %-5g %-5d\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 t;;
    for(i=0;i<n;i++){
        for(j=0;j<n-1-i;j++){
            if(x[j+1].sales_count>x[j].sales_count){
                t=x[j];
                x[j]=x[j+1];
                x[j+1]=t;
            }
        }
    }
    
    
}




double salse_amount(Book x[], int n){
    int i;
    double sum=0;
    for(i=0;i<n;i++){
        sum=sum+(x[i].sales_count*x[i].sales_price);
    }
    
    return sum;
}

 

 

 

任务 5

#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：\n");
    for(i=0;i<3;++i){
        input(&d);
        printf("%d-%02d-%02d是这一年中第%d天\n\n",d.year,d.month,d.day,day_of_year(d));
    }
}




void test2(){
    Date Alice_birth,Bob_birth;
    int i;
    int ans;
    
    printf("输入出生日期:\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\n");
        else if(ans==-1)   printf("Alice比Bob大\n\n");
        else    printf("Alice比Bob小\n\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("%d-%d-%d", &(pd->year), &(pd->month), &(pd->day));
}


int day_of_year(Date d){
    int i;
    int sum=0;
    if(d.year%4==0&&d.year%100!=0||d.year%400==0){
        for(i=1;i<d.month;i++){
            if(i==1||i==3||i==5||i==7||i==8||i==10)    sum=sum+31;
            if(i==2)  sum=sum+29;
            if(i==4||i==6||i==9||i==11)    sum=sum+30;
            
        }
        sum=sum+d.day;
    }
    else{
        for(i=1;i<d.month;i++){
            if(i==1||i==3||i==5||i==7||i==8||i==10)    sum=sum+31;
            if(i==2)  sum=sum+28;
            if(i==4||i==6||i==9||i==11)    sum=sum+30;
            
        }
        sum=sum+d.day;
    } 
    
    return sum;
}



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

 

 

 

任务 6

#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){
    int i,j;
    for(i=0;i<n;i++){
        printf("%s\t",x[i].username) ;
        for(j=0;j<strlen(x[i].password);j++)  printf("*");
        if(i==0)   printf("    ");
        printf("\t");
        switch (x[i].type) {
            case admin:
                printf("admin\n");
                break;
            case student:
                printf("student\n");
                break;
            case teacher:
                printf("teacher\n");
                break;
            }
    }
    
}

 

 

 

任务 7

#include <stdio.h>
#include <string.h>
#include <stdlib.h>


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

// 函数声明
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:(按姓名字典序升序排列,紧急联系人最先显示)\n");
    display(list, N);

    return 0;
}

// 补足函数set_vip_contact实现
// 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）
void set_vip_contact(Contact x[], int n, char name[]) {
    int i;
    for(i = 0; i < n; i++) {
        if (!strcmp(name, x[i].name)) {
            x[i].vip = 1;
        
        }
    }
}

// 补足函数display实现
// 功能: 显示联系人数组x中的联系人信息
//      按姓名字典序升序显示, 紧急联系人显示在最前面
void display(Contact x[], int n) {
    Contact temp ;
    int i, j;
    
    for (i = 0; i < n - 1; i++) {
        for (j = 0; j < n - i - 1; j++) {
            if (x[j].vip < x[j + 1].vip) {
                
                
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            } else if (x[j].vip == x[j + 1].vip && strcmp(x[j].name, x[j + 1].name) > 0) {
                
                
                temp = x[j];
                x[j] = x[j + 1];
                x[j + 1] = temp;
            }
        }
    }
    
    
    for (i = 0; i < n; i++) {
        printf("%-10s%-15s", x[i].name, x[i].phone);
        if (x[i].vip)
            printf("%5s", "*");
        printf("\n");
    }
    
}


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");
    }
}