实验6
task4
task4代码：
#include <stdio.h>
#include <stdlib.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);
printf("\n图书销售总额: %.2f\n", sales_amount(x, N));
system("pause");
return 0;
}
void output(Book x[], int n){
int i;
printf("%-20s %-30s %-20s %-10s %-10s\n",
"ISBN", "书名", "作者", "售价", "销量");
for (i = 0; i < n; i++) {
printf("%-20s %-30s %-20s %-10.1f %-10d\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;
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) {
Book temp = x[j];
x[j] = x[j + 1];
x[j + 1] = temp;
}
}
}
}
double sales_amount(Book x[], int n){
int i;
double s=0;
for(i = 0;i<n;i++)
{
s+=x[i].sales_count*x[i].sales_price;
}
return s;
}
task4截图：
task5
task5代码：
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdlib.h>
typedef struct {
int year;
int month;
int day;
} Date;
// 函数声明
void input(Date *pd); // 输入日期给pd指向的Date变量
int day_of_year(Date d); // 返回日期d是这一年的第多少天
int compare_dates(Date d1, Date d2); // 比较两个日期:
// 如果d1在d2之前，返回-1;
// 如果d1在d2之后，返回1
// 如果d1和d2相同，返回0
void test1() {
Date d;
int i;
printf("输入日期:(以形如2025-06-01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\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2025-06-01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");
}
}
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();
system("pause");
}
void input(Date *pd)
{
scanf("%d-%d-%d", &pd->year, &pd->month, &pd->day);
}
int day_of_year(Date d) {
int days_in_month[] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
int sum = 0;
int i;
int is_leap = (d.year % 4 == 0 && d.year % 100 != 0) || (d.year % 400 == 0);
if(is_leap) days_in_month[2] = 29;
for(i = 1; i < d.month; ++i)
{
sum += days_in_month[i];
}
sum+=d.day;
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;
}
task5截图：
task6
task6代码：
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
enum Role {admin, student, teacher};
typedef struct {
char username[20]; // 用户名
char password[20]; // 密码
enum Role type; // 账户类型
} Account;
void output(Account x[], int n);
int main() {
Account x[] = {{"A1001", "123456", student},
{"A1002", "123abcdef", student},
{"A1009", "xyz12121", student},
{"X1009", "9213071x", admin},
{"C11553", "12947632k", teacher},
{"X3005", "921Kfag917", student}};
int n = sizeof(x)/sizeof(Account);
output(x, n);
system("pause");
return 0;
}
void output(Account x[], int n) {
const char* role_names[] = {"admin", "student", "teacher"};
int i,j,pass_len;
for (i = 0; i < n; i++) {
printf("%-8s ", x[i].username);
pass_len = strlen(x[i].password);
for (j = 0; j < pass_len; j++) {
printf("*");
}
for (j = pass_len; j < 20; j++) {
printf(" ");
}
printf(" %s\n", role_names[x[i].type]);
}
}
task6截图：
task7
task7代码：
#include <stdio.h>
#include <string.h>
#include<stdlib.h>
typedef struct {
char name[20]; // 姓名
char phone[12]; // 手机号
int vip; // 是否为紧急联系人，是取1；否则取0
} Contact;
// 函数声明
void set_vip_contact(Contact x[], int n, char name[]); // 设置紧急联系人
void output(Contact x[], int n); // 输出x中联系人信息
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
#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);
system("pause");
return 0;
}
void set_vip_contact(Contact x[], int n, char name[])
{
int i;
for (i = 0; i < n; ++i) {
if (!strcmp(x[i].name, name))
x[i].vip = 1;
}
}
void display(Contact x[], int n)
{
int i, j, k;
Contact t;
for (i = 0; i < n; i++) {
for (j = 0; j < n - i - 1; j++) {
if (strcmp(x[j].name, x[j + 1].name) > 0) {
t = x[j];
x[j] = x[j + 1];
x[j + 1] = t;
}
}
}
for (i = n - 1; i >= 0; i--) {
for (k = 0; k < i; ++k) {
if (x[i].vip) {
for (j = i; j > 0; j--) {
t = x[j];
x[j] = x[j - 1];
x[j - 1] = t;
}
}
}
}
output(x, 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");
}
}
task7截图：
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