PAT乙级1073-----多选题常见计分法 (20分)

Posted 2020-11-27 a982961222

1073 多选题常见计分法 (20分)

 

 

 

思路：

1.错选的和未选的都计入错误信息

2.错误信息的输出应该是  MAX某道题某选项的错误次数+题号+选项

 

首次通过代码:

 1 #include<stdio.h>
 2 #include<stdlib.h>
 3 #include<string.h>
 4 
 5 typedef struct answer{
 6      int s_sum;
 7      int r_s_sum;
 8      int score;
 9      int w_sum;
10      int right_answer[6];
11      int wrong_answer[6];
12 }answer,*a;
13 
14 int main(){
15 int student_sum;
16 int answer_sum;
17 scanf("%d %d",&student_sum,&answer_sum);
18 a answer_list[101]={NULL};
19 for(int i=0;i<answer_sum;i++){
20     a a1=(a)malloc(sizeof(answer));a1->w_sum=0;
21     for(int j=0;j<6;j++){
22         a1->right_answer[j]=0;
23         a1->wrong_answer[j]=0;
24     }
25     scanf("%d %d %d",&a1->score,&a1->s_sum,&a1->r_s_sum);
26      for(int j=0;j<a1->r_s_sum;j++){
27          char k[2];
28          scanf("%s",k);
29          a1->right_answer[k[0]-‘a‘]=1;
30      }
31      answer_list[i]=a1;
32 }
33 getchar();
34 int all_right=0;
35 for(int i=0;i<student_sum;i++){
36     char input[1000];double s_score=0;int flag=1;
37     gets(input);
38     int counter=0;int j=0;
39     while(1){
40         int k=input[j]-‘0‘;
41          if(k>=0&&k<=9) {
42             int c1=0,c2=0;int flag1=1;
43             while(1){
44                 int k=input[++j]-‘a‘;
45                 if(k>=0&&k<=9){
46                 if(answer_list[counter]->right_answer[k]==1) {
47                      c1++;
48                      answer_list[counter]->wrong_answer[k]++;
49                 }
50                 else {answer_list[counter]->wrong_answer[k]++;
51                      flag=0;
52                      if(flag1){
53                          flag1=0;
54                          answer_list[counter]->w_sum++;
55                      }
56                      c2++;}
57                 }
58                 else if(input[j]==‘)‘||input[j]==‘‘) break;
59             }
60             if(c1==answer_list[counter]->r_s_sum&&c2==0) s_score+=answer_list[counter]->score;
61             else if(c1<answer_list[counter]->r_s_sum&&c2==0) s_score+=0.5*(double)answer_list[counter]->score;
62             counter++;
63         }
64         if(counter==answer_sum||input[j]==‘‘)  break;
65         j++;
66     }
67     if(flag&&counter==answer_sum) all_right++;
68     printf("%.1lf
",s_score);
69 }
70 if(all_right==student_sum) {
71     printf("Too simple");
72     return 0;
73 }
74 int max=0;
75 for(int i=0;i<answer_sum;i++)
76     for(int j=0;j<6;j++){
77         if(answer_list[i]->right_answer[j]==1&&student_sum-answer_list[i]->wrong_answer[j]>max) 
78         max=student_sum-answer_list[i]->wrong_answer[j];
79         else if(answer_list[i]->right_answer[j]==0&&answer_list[i]->wrong_answer[j]>max)
80         max=answer_list[i]->wrong_answer[j];
81     }
82 for(int i=0;i<answer_sum;i++){
83     for(int j=0;j<6;j++){
84             if(answer_list[i]->right_answer[j]==1&&student_sum-answer_list[i]->wrong_answer[j]==max)
85             printf("%d %d-%c
",max,i+1,j+‘a‘);
86             else if(answer_list[i]->right_answer[j]==0&&answer_list[i]->wrong_answer[j]==max)
87             printf("%d %d-%c
",max,i+1,j+‘a‘);
88         
89     }
90 }
91 return 0;
92 }

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