浙大PAT乙级1095（C语言）测试点3、4过不了

Question

编程小菜鸡一枚，请教如何修改能通过测试点3、4。

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
char g1;//对cmp1排序所输入的全局字符
char g3[7];//对cmp3排序所输入的全局字符串
struct student{
    char grade;//第 1 位是级别，即 T 代表顶级；A 代表甲级；B 代表乙级；
    char room[4];//第 2~4 位是考场编号，范围从 101 到 999；
    char date[7];//第 5~10 位是考试日期，格式为年、月、日顺次各占 2 位；
    char id[4];//最后 11~13 位是考生编号，范围从 000 到 999
    int marks;//分数
};

struct room{
    char id[4];
    int students;//考场人数
};

int cmp1(const void *a, const void *b){
    struct student c = *(struct student *)a;
    struct student d = *(struct student *)b;
    if (c.grade!=d.grade)
    {
        if (c.grade==g1)
            return -1;
        else if (d.grade==g1)
            return 1;
        else
            return c.grade-d.grade;
    }
    else//对于分数并列的考生，按其准考证号的字典序递增输出
    {
        if (c.marks!=d.marks)
            return d.marks-c.marks;
        if (strcmp(c.room,d.room)!=0)
            return strcmp(c.room,d.room);
        else
        {
            if (strcmp(c.date,d.date)!=0)
                return strcmp(c.date,d.date);
            else
                return strcmp(c.id,d.id);
        }
    }
}

int cmp2(const void *a, const void *b){
    struct room c = *(struct room *)a;
    struct room d = *(struct room *)b;
    if (c.students!=d.students)//输出按人数非递增顺序
        return d.students-c.students;
    else //若人数并列则按考场编号递增顺序输出
        return strcmp(c.id,d.id);
}

int cmp3(const void *a, const void *b){
    struct student c = *(struct student *)a;
    struct student d = *(struct student *)b;
    if (strcmp(c.date,d.date)!=0)
    {
        if (strcmp(c.date,g3)==0)
            return -1;
        else if (strcmp(d.date,g3)==0)
            return 1;
        else
            return strcmp(c.date,d.date);
    }
    else
        return strcmp(c.room,d.room);//考场编号递增顺序
}

int main(){
    char temp[20];
    int i,j,M,N,temp1;//考生人数M和统计要求的个数N
    scanf("%d %d",&M,&N);
    getchar();//接下来要读入字符串的一律用getchar
    struct student S[M];
    struct room R[M];
    for (i=0;i<M;i++)//读入学生的信息
    {
        R[i].students = 0;//初始化
        scanf("%s %d",temp,&S[i].marks);
        getchar();
        S[i].grade = temp[0];
        for (j=0;j<3;j++)
            S[i].room[j] = temp[1+j];
        S[i].room[3] = '\0';
        for (j=0;j<6;j++)
            S[i].date[j] = temp[4+j];
        S[i].date[6] = '\0';
        for (j=0;j<3;j++)
            S[i].id[j] = temp[10+j];
        S[i].id[3] = '\0';
    }
    for (i=0;i<N;i++)
    {
        scanf("%d %s",&temp1,temp);
        printf("Case %d: %d %s\n",i+1,temp1,temp);
        switch (temp1){
            case 1:
                {
                    int isfind = 0;
                    g1 = temp[0];
                    qsort(S,M,sizeof(S[0]),cmp1);
                    for (j=0;j<M;j++)
                    {
                        if (S[j].grade!=g1)
                            break;
                        else
                        {
                            printf("%c%s%s%s %d\n",S[j].grade,S[j].room,S[j].date,S[j].id,S[j].marks);
                            isfind = 1;
                        }
                    }
                    if (isfind==0)
                        printf("NA\n");
                    break;
                }
            case 2:
                {
                    int num = 0;
                    int sum = 0;
                    for (j=0;j<M;j++)
                    {
                        if (strcmp(S[j].room,temp)==0)
                        {
                            num++;
                            sum+=S[j].marks;
                        }
                    }
                    if (num==0)
                        printf("NA\n");
                    else
                        printf("%d %d\n",num,sum);
                    break;
                }
            case 3:
                {
                    int k = -1;//统计符合日期的不同的考场个数
                    strcpy(g3,temp);
                    qsort(S,M,sizeof(S[0]),cmp3);
                    for (j=0;j<M;j++)
                    {
                        if (strcmp(S[j].date,temp)!=0)//日期不相等说明不符号要求
                            break;
                        if (j==0)
                        {
                            k++;
                            strcpy(R[k].id,S[j].room);
                            R[k].students++;
                        }
                        else
                        {
                            if (strcmp(S[j].room,R[k].id)==0)
                                R[k].students++;
                            else
                            {
                                k++;
                                strcpy(R[k].id,S[j].room);
                                R[k].students++;
                            }
                        }
                    }
                    if (k==-1)
                        printf("NA\n");
                    else
                    {
                        k++;//下标转换为数组的长度
                        qsort(R,k,sizeof(R[0]),cmp2);
                        for (j=0;j<k;j++)
                            printf("%s %d\n",R[j].id,R[j].students);
                        break;
                    }
                    break;
                }
            default:break;
        }
    }
    return 0;
}