diff --git a/day12/homework/h1.c b/day12/homework/h1.c
new file mode 100644
index 0000000..6d89330
--- /dev/null
+++ b/day12/homework/h1.c
@@ -0,0 +1,14 @@
+// 给定一张图片的规格信息，请提取出图片的格式、宽和高。 char *s = "base64;image/png:320,480";
+// 输出结果： image / png 320 480
+#include <stdio.h>
+#include <string.h>
+
+int main()
+{
+    char *s = "base64;image/png:320,480";
+    char *format;
+    int width, height;
+    sscanf(s, "%*[^;];%[^:]:%d,%d", format, &width, &height);
+    printf("%s %d %d\n", format, width, height);
+    return 0;
+}
\ No newline at end of file
diff --git a/day12/homework/h2.c b/day12/homework/h2.c
new file mode 100644
index 0000000..f5cb4e6
--- /dev/null
+++ b/day12/homework/h2.c
@@ -0,0 +1,27 @@
+/*
+    找出以下程序的bug, 并修正 int main()
+    {
+        const int *a = malloc(20);
+        for (int i = 0; i < 20; i++)
+        {
+            *(a + i) = i;
+        }
+        printf("%d %d\n", a[0], a + 2);
+    }
+*/
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+int main()
+{
+    int *a = malloc(20); // 修改1: 去掉 const 修饰
+    for (int i = 0; i < 20; i++)
+    {
+        *(a + i) = i;
+    }
+    printf("%d %d\n", a[0], *(a + 2)); // 修改2: a+2 加上解引用 *, 以取得元素值
+    free(a);                           // 修改3: 释放手动分配的内存空间
+
+    return 0; // 修改4: 加上 return 0
+}
\ No newline at end of file
diff --git a/day12/homework/h3.c b/day12/homework/h3.c
new file mode 100644
index 0000000..0c1dc05
--- /dev/null
+++ b/day12/homework/h3.c
@@ -0,0 +1,21 @@
+/*
+    请找出程序的错误行数（   ）
+    int main()
+    {
+            int a='21';
+            const char *p = (char *) &a;
+            *p = 48;
+            printf("%s\n", p);
+    }
+*/
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+int main()
+{
+    int a = '21'; // 错误1: '21' 是字符常量，不是整型常量，应该写成 21
+    const char *p = (char *)&a;
+    *p = 48; // 错误2: *p 在上一行被 const 修饰为了只读变量，无法修改
+    printf("%s\n", p);
+}
\ No newline at end of file
diff --git a/day12/homework/h4.c b/day12/homework/h4.c
new file mode 100644
index 0000000..bb368a1
--- /dev/null
+++ b/day12/homework/h4.c
@@ -0,0 +1,23 @@
+// 编写一个程序，定义一个结构体表示学生的信息，包括姓名、年龄和分数。通过键盘输入学生的信息，然后输出这些信息。
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+typedef struct stu_s
+{
+    char *name;
+    int age;
+    float score;
+} STU;
+
+int main()
+{
+    STU stu1;
+    stu1.name = malloc(20); // 因为 name 是指针，所以需要手动分配内存空间
+    printf("请输入学生的姓名 年龄 分数:");
+    scanf("%s %d %f", stu1.name, &stu1.age, &stu1.score);
+    printf("姓名: %s, 年龄: %d, 分数: %.2f", stu1.name, stu1.age, stu1.score);
+    free(stu1.name); // 释放手动分配的内存空间
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day12/homework/h5.c b/day12/homework/h5.c
new file mode 100644
index 0000000..e74f5c4
--- /dev/null
+++ b/day12/homework/h5.c
@@ -0,0 +1,78 @@
+// 1. 2. 编写一个程序，定义一个结构体表示日期，包括年、月和日。编写一个函数，接受两个日期作为参数，并计算并返回它们之间的天数差。
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+typedef struct date_s
+{
+    int year;
+    int month;
+    int day;
+} DATE;
+
+int sumDays(int year, int month, int day)
+{
+    int sum = day;
+    switch (month)
+    {
+    case 12:
+        sum += 30;
+    case 11:
+        sum += 31;
+    case 10:
+        sum += 30;
+    case 9:
+        sum += 31;
+    case 8:
+        sum += 31;
+    case 7:
+        sum += 30;
+    case 6:
+        sum += 31;
+    case 5:
+        sum += 30;
+    case 4:
+        sum += 31;
+    case 3:
+        sum += 28;
+    case 2:
+        sum += 31;
+    }
+
+    if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
+        sum += 1;
+
+    return sum;
+}
+
+int yearOFdays(int year1, int year2)
+{
+    int sum = (abs(year1 - year2) - 1) * 365;
+    int runYearCount = 0;                          // 闰年计数器
+    int minYear = (year1 < year2) ? year1 : year2; // 找出较小的年份
+    int maxYear = (year1 > year2) ? year1 : year2; // 找出较大的年份
+    for (; minYear <= maxYear; minYear++)
+        if ((minYear % 4 == 0 && minYear % 100 != 0) || minYear % 400 == 0)
+            runYearCount++;
+
+    sum += runYearCount; // 每多一个闰年多一天
+    return sum;
+}
+int main()
+{
+    DATE day1, day2;
+    printf("请输入第一个日期，输入格式为 年,月,日 ");
+    scanf("%d,%d,%d", &day1.year, &day1.month, &day1.day);
+    printf("请输入第二个日期，输入格式为 年,月,日 ");
+    scanf("%d,%d,%d", &day2.year, &day2.month, &day2.day);
+
+    int sumdays1 = sumDays(day1.year, day1.month, day1.day);
+    int sumdays2 = sumDays(day2.year, day2.month, day2.day);
+
+    int yearMoreDays = yearOFdays(day1.year, day2.year);
+
+    int days = abs(sumdays1 - sumdays2) + yearMoreDays;
+    printf("两个日期的天数差为: %d", days);
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day12/homework/h5_2.c b/day12/homework/h5_2.c
new file mode 100644
index 0000000..2938dd1
--- /dev/null
+++ b/day12/homework/h5_2.c
@@ -0,0 +1,73 @@
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+typedef struct date_s
+{
+    int year;
+    int month;
+    int day;
+} DATE;
+
+int daysOfMonth(int year, int month)
+{
+    switch (month)
+    {
+    case 2:
+        if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0)
+            return 29;
+        else
+            return 28;
+    case 4:
+    case 6:
+    case 9:
+    case 11:
+        return 30;
+    default:
+        return 31;
+    }
+}
+
+int sumDays(int year, int month, int day)
+{
+    int sum = day;
+    for (int i = 1; i < month; i++)
+    {
+        sum += daysOfMonth(year, i);
+    }
+    return sum;
+}
+
+int yearOFdays(int year1, int year2)
+{
+    int sum = 0;
+    int minYear = (year1 < year2) ? year1 : year2;
+    int maxYear = (year1 > year2) ? year1 : year2;
+    for (int i = minYear; i < maxYear; i++)
+    {
+        if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0)
+            sum += 366;
+        else
+            sum += 365;
+    }
+    return sum;
+}
+
+int main()
+{
+    DATE day1, day2;
+    printf("请输入第一个日期，输入格式为 年,月,日 ");
+    scanf("%d,%d,%d", &day1.year, &day1.month, &day1.day);
+    printf("请输入第二个日期，输入格式为 年,月,日 ");
+    scanf("%d,%d,%d", &day2.year, &day2.month, &day2.day);
+
+    int sumdays1 = sumDays(day1.year, day1.month, day1.day);
+    int sumdays2 = sumDays(day2.year, day2.month, day2.day);
+
+    int yearMoreDays = yearOFdays(day1.year, day2.year);
+
+    int days = abs(sumdays1 - sumdays2) + yearMoreDays;
+    printf("两个日期的天数差为: %d", days);
+
+    return 0;
+}
diff --git a/day12/homework/h6.c b/day12/homework/h6.c
new file mode 100644
index 0000000..87f6b8e
--- /dev/null
+++ b/day12/homework/h6.c
@@ -0,0 +1,37 @@
+// 编写一个程序，定义一个结构体表示员工的信息，包括姓名、工号和工资。编写一个函数，接受一个员工结构体数组和数组长度作为参数，并计算并返回所有员工的平均工资。
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+typedef struct employee_s
+{
+    char name[32];
+    int eid;
+    double salary;
+} EMPLOYEE;
+
+double avgSalary(EMPLOYEE *emps, int n)
+{
+    double avgS;
+    for (int i = 0; i < n; i++)
+    {
+        avgS += emps[i].salary / n;
+    }
+
+    return avgS;
+}
+
+int main()
+{
+    int n; // 数组长度
+    printf("请输入数组长度: ");
+    scanf("%d", &n);
+    EMPLOYEE *employees = malloc(n * sizeof(EMPLOYEE));
+    employees[0].salary = 100;
+    employees[1].salary = 59;
+
+    printf("平均工资为: %.2lf", avgSalary(employees, n));
+    free(employees);
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day12/homework/h7.c b/day12/homework/h7.c
new file mode 100644
index 0000000..b10eb5e
--- /dev/null
+++ b/day12/homework/h7.c
@@ -0,0 +1,59 @@
+// 编写一个程序，定义一个枚举（enum）表示一周的每一天，包括周一至周日。编写一个函数，接受一个枚举值作为参数，并根据枚举值打印出相应的星期几的名称。例如，传递枚举值为2时，函数应该打印出 "Tuesday"。
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+enum
+{
+    Monday,
+    Tuesday,
+    Wednesday,
+    Thursday,
+    Friday,
+    Saturday,
+    Sunday
+};
+
+void print_weekday(int day)
+{
+    switch (day)
+    {
+    case Monday:
+        printf("Monday\n");
+        break;
+    case Tuesday:
+        printf("Tuesday\n");
+        break;
+    case Wednesday:
+        printf("Wednesday\n");
+        break;
+    case Thursday:
+        printf("Thursday\n");
+        break;
+    case Friday:
+        printf("Friday\n");
+        break;
+    case Saturday:
+        printf("Saturday\n");
+        break;
+    case Sunday:
+        printf("Sunday\n");
+        break;
+    default:
+        printf("Invalid day\n");
+        break;
+    }
+}
+
+int main()
+{
+    print_weekday(Monday);
+    print_weekday(Tuesday);
+    print_weekday(Wednesday);
+    print_weekday(Thursday);
+    print_weekday(Friday);
+    print_weekday(Saturday);
+    print_weekday(Sunday);
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day12/homework/h8.c b/day12/homework/h8.c
new file mode 100644
index 0000000..3d36516
--- /dev/null
+++ b/day12/homework/h8.c
@@ -0,0 +1,56 @@
+// 练习结构体数组排序　　 从键盘输入5个学生的信息（姓名、学号、成绩）,
+//     存入一个结构体数组中，计算平均分，并按成绩高低排序并输出
+#include <stdio.h>
+#include <stdlib.h>
+#include <string.h>
+
+typedef struct stu_s
+{
+    char name[32];
+    int sid;
+    float score;
+} STU;
+
+void arrAndOutput(STU *stus) // 排序并输出
+{
+    float avg = 0.0f;
+    for (int i = 0; i < 5; i++)
+    {
+        avg += stus[i].score / 5;
+    }
+
+    printf("平均分为: %.2f\n", avg);
+
+    // 冒泡排序
+    for (int i = 0; i < 5; i++)
+    {
+        for (int j = 0; j < 5 - i - 1; j++)
+        {
+            if (stus[j].score < stus[j + 1].score)
+            {
+                STU tmp = stus[j];
+                stus[j] = stus[j + 1];
+                stus[j + 1] = tmp;
+            }
+        }
+    }
+
+    // 输出
+    for (int i = 0; i < 5; i++)
+    {
+        printf("姓名: %s, 学号: %d, 成绩: %.2f\n", stus[i].name, stus[i].sid, stus[i].score);
+    }
+}
+
+int main()
+{
+    STU s[5];
+    for (int i = 0; i < 5; i++)
+    {
+        printf("请输入第 %d 个学生的,格式为:(姓名 学号 成绩)", i + 1);
+        scanf("%s %d %f", s[i].name, &s[i].sid, &s[i].score);
+    }
+    arrAndOutput(s);
+
+    return 0;
+}
\ No newline at end of file
diff --git a/day12/homework/h9.c b/day12/homework/h9.c
new file mode 100644
index 0000000..cac3778
--- /dev/null
+++ b/day12/homework/h9.c
@@ -0,0 +1,118 @@
+/*
+了解GPS相关知识
+$GPRMC协议格式
+　　例：$GPRMC, 024813.640, A, 3158.4608, N, 11848.3737, E, 10.05, 324.27, 150706, , , A * 50
+　　字段0：$GPRMC，语句ID，表明该语句为Recommended Minimum Specific GPS / TRANSIT Data（RMC）推荐最小定位信息
+　　字段1：UTC时间，hhmmss.sss格式(UTC时间是世界统一时间，中国的时间是UTC + 8，比如：上面的时间是02 : 48 : 13而中国的时间是10 : 48 : 13)
+　　字段2：状态，A = 定位，V = 未定位
+　　字段3：纬度ddmm.mmmm，度分格式（前导位数不足则补0） 
+　　字段4：纬度N（北纬）或S（南纬）
+　　字段5：经度dddmm.mmmm，度分格式（前导位数不足则补0）
+　　字段6：经度E（东经）或W（西经）
+　　字段7：速度，节，Knots
+　　字段8：方位角，度
+　　字段9：UTC日期，DDMMYY格式
+　　字段10：磁偏角，（000 - 180）度（前导位数不足则补0）
+　　字段11：磁偏角方向，E = 东W = 西
+　　字段16：校验值
+
+【提示】利用strtok切割、sscanf或者strncpy、atoi、atof
+*/
+#include <stdio.h>
+#include <string.h>
+
+typedef struct
+{
+    char time[10];        // UTC时间，格式为hhmmss.sss
+    char status;          // 状态，A表示定位，V表示未定位
+    double latitude;      // 纬度，格式为ddmm.mmmm
+    char latitude_dir;    // 纬度半球，N表示北半球，S表示南半球
+    double longitude;     // 经度，格式为dddmm.mmmm
+    char longitude_dir;   // 经度半球，E表示东经，W表示西经
+    double speed;         // 速度，以节为单位
+    double course;        // 方位角，以度为单位
+    char date[7];         // UTC日期，格式为DDMMYY
+    double declination;   // 磁偏角，以度为单位
+    char declination_dir; // 磁偏角方向，E表示东偏，W表示西偏
+} gprmc_t;
+
+int parse_gprmc(const char *str, gprmc_t *gprmc)
+{
+    char *token;
+    int i = 0;
+    token = strtok((char *)str, ",");
+    while (token != NULL)
+    {
+        switch (i)
+        {
+        case 0:
+            if (strcmp(token, "$GPRMC") != 0)
+            {
+                return -1;
+            }
+            break;
+        case 1:
+            sscanf(token, "%2s", gprmc->time);
+            sscanf(token + 2, "%2lf%lf", &gprmc->time[2], &gprmc->time[4]);
+            sscanf(token + 4, "%lf", &gprmc->time[6]);
+            break;
+        case 2:
+            gprmc->status = *token;
+            break;
+        case 3:
+            sscanf(token, "%lf", &gprmc->latitude);
+            sscanf(token + 2, "%lf", &gprmc->latitude);
+            break;
+        case 4:
+            gprmc->latitude_dir = *token;
+            break;
+        case 5:
+            sscanf(token, "%lf", &gprmc->longitude);
+            sscanf(token + 3, "%lf", &gprmc->longitude);
+            break;
+        case 6:
+            gprmc->longitude_dir = *token;
+            break;
+        case 7:
+            sscanf(token, "%lf", &gprmc->speed);
+            break;
+        case 8:
+            sscanf(token, "%lf", &gprmc->course);
+            break;
+        case 9:
+            sscanf(token, "%2s", gprmc->date);
+            sscanf(token + 2, "%2lf%lf", &gprmc->date[2], &gprmc->date[4]);
+            sscanf(token + 4, "%lf", &gprmc->date[6]);
+            break;
+        case 10:
+            sscanf(token, "%lf", &gprmc->declination);
+            break;
+        case 11:
+            gprmc->declination_dir = *token;
+            break;
+        default:
+            break;
+        }
+        token = strtok(NULL, ",");
+        i++;
+    }
+    return 0;
+}
+
+int main()
+{
+    const char *str = "$GPRMC,024813.640,A,3158.4608,N,11848.3737,E,10.05,324.27,150706,,,A*50";
+    gprmc_t gprmc;
+    if (parse_gprmc(str, &gprmc) == 0)
+    {
+        printf("UTC时间：%s\n", gprmc.time);
+        printf("状态：%c\n", gprmc.status);
+        printf("纬度：%lf %c\n", gprmc.latitude, gprmc.latitude_dir);
+        printf("经度：%lf %c\n", gprmc.longitude, gprmc.longitude_dir);
+        printf("速度：%lf\n", gprmc.speed);
+        printf("方位角：%lf\n", gprmc.course);
+        printf("UTC日期：%s\n", gprmc.date);
+        printf("磁偏角：%lf %c\n", gprmc.declination, gprmc.declination_dir);
+    }
+    return 0;
+}