帮我写一个查看学生申请并完成什么活动的C语言代码

好的，以下是一个简单的查看学生申请并完成活动的C语言代码：

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

#define MAX_STUDENTS 100
#define MAX_ACTIVITIES 10

typedef struct {
    char name[50];
    int age;
    char gender[10];
    char activity[MAX_ACTIVITIES][50];
    int num_activities;
} Student;

int main() {
    Student students[MAX_STUDENTS];
    int num_students = 0;
    int i, j;

    // 添加学生
    printf("请输入学生信息：\n");
    while (num_students < MAX_STUDENTS) {
        Student s;
        printf("姓名：");
        scanf("%s", s.name);
        printf("年龄：");
        scanf("%d", &s.age);
        printf("性别：");
        scanf("%s", s.gender);
        printf("申请的活动（最多%d个，以空格分隔）：", MAX_ACTIVITIES);
        s.num_activities = 0;
        while (s.num_activities < MAX_ACTIVITIES) {
            scanf("%s", s.activity[s.num_activities]);
            if (strcmp(s.activity[s.num_activities], "") == 0) {
                break;
            }
            s.num_activities++;
        }
        students[num_students] = s;
        num_students++;
        printf("是否继续添加学生？（y/n）");
        char c;
        scanf(" %c", &c);
        if (c == 'n' || c == 'N') {
            break;
        }
    }

    // 查看学生信息
    printf("所有学生信息如下：\n");
    for (i = 0; i < num_students; i++) {
        printf("姓名：%s\n", students[i].name);
        printf("年龄：%d\n", students[i].age);
        printf("性别：%s\n", students[i].gender);
        printf("申请的活动：");
        for (j = 0; j < students[i].num_activities; j++) {
            printf("%s ", students[i].activity[j]);
        }
        printf("\n");
    }

    // 查看某个活动的申请情况
    printf("请输入要查看的活动名称：");
    char activity_name[50];
    scanf("%s", activity_name);
    printf("申请了 %s 活动的学生有：\n", activity_name);
    for (i = 0; i < num_students; i++) {
        for (j = 0; j < students[i].num_activities; j++) {
            if (strcmp(students[i].activity[j], activity_name) == 0) {
                printf("%s\n", students[i].name);
                break;
            }
        }
    }

    return 0;
}

希望这个代码能够帮到你！