实验六

task 4

#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] = {{"978-7-229-14156-1", "源泉", "安.兰德", 84, 59},
                 {"978-7-5133-5261-1", "李白来到旧金山", "谭夏阳", 48, 16},
                 {"978-7-5617-4347-8", "陌生人日记", "周怡芳", 72.6, 27},
                 {"978-7-5722-5475-8", "芯片简史", "汪波", 74.9, 49},
                 {"978-7-5046-9568-0", "数据化决策", "道格拉斯·W·哈伯德", 49, 42},
                 {"978-7-5133-4388-6", "美好时代的背后", "凯瑟琳.布", 34.5, 39},
                 {"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},
                 {"978-7-5321-5691-7", "何为良好生活", "陈嘉映", 29.5 , 31},
                 {"978-7-5133-5109-6", "你好外星人", "英国未来出版集团", 118, 42},
                 {"978-7-2011-4617-1", "世界尽头的咖啡馆", "约翰·史崔勒基", 22.5, 44}};
    
    printf("图书销量排名: \n");
    sort(x, N);
    output(x, N);

    printf("\n图书销售总额: %.2f\n", sales_amount(x, N));
    
    return 0;
}


void output(Book x[], int n){
    int i;
    printf("ISBN  书名  作者  售价  销售册数\n");
    for(i=0;i<n;i++){
    printf("%-13s\t %-30s\t %-28s\t %-10.1lf\t %-4d\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-1; i++){
      for(j = 0; j < n-1-i; j++){
        if(x[j].sales_count < x[j+1].sales_count) {
            t = x[j];
            x[j] = x[j+1];
            x[j+1]= t;}}}
}


double sales_amount(Book x[], int n){
        int i;
        double sum=0.0;
        for(i=0;i<n;i++){
            sum=sum+x[i].sales_count*x[i].sales_price;}
        return sum;
        }

task 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();                      
void test2();   


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 test1() {
    Date d;
    int i;

    printf("输入日期:(以形如2024-06-01这样的形式输入)\n");
    for(i = 0; i < 3; ++i) {
        input(&d);
        printf("%04d-%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Alice和Bob出生日期:(以形如2005-08-11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\n");
        else if(ans == -1)
            printf("Alice比Bob大\n\n");
        else
            printf("Alice比Bob小\n\n");
    }
}

void input(Date *pd) {    
    scanf("%d-%d-%d", &pd->year, &pd->month, &pd->day);
}

int day_of_year(Date d) {
    int i,dath=0;
    for(i=1;i<d.month;i++){
        if(i==1||i==3||i==5||i==7||i==8||i==10||i==12){
        dath+=31;
        }
        else if(i==2)
            {if((d.year%4==0&&d.year%100!=0)||d.year%400==0)
            dath+=29;
            else
            dath+=28;
        }
        else
            dath+=30;
    }
   dath+=d.day;
   return dath;
}

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

task 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("*");
        }
        printf("    \t");
        switch (x[i].type) {
            case admin:
                printf("admin\n");
                break;
            case student:
                printf("student\n");
                break;
            case teacher:
                printf("teacher\n");
                break;
        }
    }
      
}

 

posted @ 2024-06-09 22:01  刘梦佳  阅读(5)  评论(0编辑  收藏  举报