PAT乙级1095-----解码PAT准考证 (25分)

1095 解码PAT准考证 (25分)

输入样例：

8 4
B123180908127 99
B102180908003 86
A112180318002 98
T107150310127 62
A107180908108 100
T123180908010 78
B112160918035 88
A107180908021 98
1 A
2 107
3 180908
2 999

 

输出样例：

Case 1: 1 A
A107180908108 100
A107180908021 98
A112180318002 98
Case 2: 2 107
3 260
Case 3: 3 180908
107 2
123 2
102 1
Case 4: 2 999
NA

思路：
1.对于类型1，使用结构体数组直接排序即可
2.对于类型2，用数组下标作为考场号记录人数和成绩
3.对于类型3，利用年月日考场号作为数组的四个下标进行输入，输出时将年月日对应的考场信息进行排序即可
4.大数组要定义为全局变量
5.类型2若人数为0要输出NA

首次通过代码:

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
   int a3[100][13][32][1000]={0};
struct student{
    char id[15];
    int score;
};

struct exam{
    int num;
    int p_sum;
};

void style1(struct student a1[],char x,int a1_sum[]){
    switch(x){
        case 'A':if(a1_sum[0]==0) {
                 printf("NA");
                 break;
               }
               else {
                  for(int j=0;j<a1_sum[0];j++) {
                      printf("%s %d",a1[j].id,a1[j].score);
                      if(j!=a1_sum[0]-1) printf("
");
                  }
               }
                  break;
        case 'B':if(a1_sum[1]==0) {
                 printf("NA");
                 break;
               }
               else {
                  for(int j=a1_sum[0];j<a1_sum[0]+a1_sum[1];j++) {
                      printf("%s %d",a1[j].id,a1[j].score);
                      if(j!=a1_sum[0]+a1_sum[1]-1) printf("
");
                  }
               }
                  break;
        case 'T':if(a1_sum[2]==0) {
                 printf("NA");
                 break;
               }
               else {
                  for(int j=a1_sum[0]+a1_sum[1];j<a1_sum[0]+a1_sum[1]+a1_sum[2];j++) {
                      printf("%s %d",a1[j].id,a1[j].score);
                      if(j!=a1_sum[0]+a1_sum[1]+a1_sum[2]-1) printf("
");
                  }
                  }
                  break;
            default:printf("NA");break;
       }
    }

void style2(int a2_score[],int a2_sum[],char x[]){
    int y=0;
    for(int i=0;i<strlen(x);i++){
        if(x[i]>='0'&&x[i]<='9') y=y*10+x[i]-'0';
        else {
            printf("NA");return;
        }
    }
    if(a2_sum[y]==0) printf("NA");
    else printf("%d %d",a2_sum[y],a2_score[y]);
    return ;
}



int cmp1(const void *a,const void *b){
   struct student a1=*(struct student *)a;
   struct student b1=*(struct student *)b;
   if(a1.id[0]<b1.id[0]) return -1;
   else if(a1.id[0]==b1.id[0]){
       if(a1.score>b1.score) return -1;
       else if(a1.score==b1.score){
           return strcmp(a1.id,b1.id);
       }
       else return 1;
   }
   else return 1;
}

int cmp2(const void *a,const void *b){
   struct exam a1=*(struct exam *)a;
   struct exam b1=*(struct exam *)b;
   if(a1.p_sum>b1.p_sum) return -1;
   else if(a1.p_sum==b1.p_sum){
       if(a1.num<b1.num) return -1;
       else return 1;
   }
   else return 1;
}

void style3(int a3[]){
    int num1=0;
    struct exam s[1000];
   for(int i=0;i<1000;i++){
       if(a3[i]>0) {
        s[num1].num=i;
        s[num1].p_sum=a3[i];
        num1++;
       }
   }
   if(num1==0) {
    printf("NA");
    return ;
   }  
    else {
        qsort(s,num1,sizeof(struct exam),cmp2);
    }
  for(int i=0;i<num1;i++) {
      printf("%d %d",s[i].num,s[i].p_sum);
      if(i!=num1-1) printf("
");
  }
}

int main(){
   struct student a1[10005];
   int a1_sum[3]={0};
   int a2_sum[1000]={0};
   int a2_score[1000]={0};

   int sum1,sum2;
   int num=0;
   scanf("%d %d",&sum1,&sum2);
   for(int i=0;i<sum1;i++){
       struct student s1;int x=0;
       scanf("%s %d",s1.id,&s1.score);
       a1[num++]=s1;
       for(int j=1;j<4;j++){
           x=x*10+s1.id[j]-'0';
       }
       a2_sum[x]++;
       a2_score[x]+=s1.score;
       int year=(s1.id[4]-'0')*10+(s1.id[5]-'0');
       int month=(s1.id[6]-'0')*10+(s1.id[7]-'0');
       int day=(s1.id[8]-'0')*10+(s1.id[9]-'0');
       a3[year][month][day][x]++;
   }
   qsort(a1,num,sizeof(struct student),cmp1);
   for(int i=0;i<num;i++) {
       if(a1[i].id[0]=='A') a1_sum[0]++;
       else if(a1[i].id[0]=='B') a1_sum[1]++;
       else if(a1[i].id[0]=='T') a1_sum[2]++;
   }
   for(int i=0;i<sum2;i++){
       int ins1;char ins2[10];
       scanf("%d %s",&ins1,ins2);
       printf("Case %d: %d %s
",i+1,ins1,ins2);
       switch(ins1){
           case 1:if(strlen(ins2)==1) style1(a1,ins2[0],a1_sum);
                   else  printf("NA");
                   break;
           case 2:style2(a2_score,a2_sum,ins2);
                   break;
           case 3:if(strlen(ins2)!=6)
                 printf("NA");
                 else{
                     int flag=1;
                     for(int j=0;j<6;j++) {
                       if(ins2[j]>='0'&&ins2[j]<='9') ;
                       else {
                           flag=0;break;
                       }
                     }
                     if(flag) {
                         int x[1000];
                         int year=(ins2[0]-'0')*10+(ins2[1]-'0');
                         int month=(ins2[2]-'0')*10+(ins2[3]-'0');
                         int day=(ins2[4]-'0')*10+(ins2[5]-'0');
                         for(int j=0;j<1000;j++) x[j]=a3[year][month][day][j];
                         style3(x);
                     }
                 } 
                    break;
           default:break;
       }
       if(i!=sum2) printf("
");
   }
}

 参考：

FROM：https://blog.csdn.net/qq_41325698/article/details/97903937#2.%E5%A4%A7%E6%95%B0%E7%BB%84%E4%B8%80%E5%AE%9A%E8%A6%81%E5%BC%80%E5%85%A8%E5%B1%80%EF%BC%8C%E8%80%8C%E4%B8%8D%E6%98%AF%E5%86%99%E5%9C%A8main%E5%87%BD%E6%95%B0%E9%87%8C%E9%9D%A2%E3%80%82