1073. 多选题常见计分法(20)

1.不太想做这种题目
2.关键是输出：漏选的，多选的，都属于易错选项，分两次循环查找，同时保存好max值，开个数组，保存每个选项的错误次数即可，输出的时候根据mmax对应输出

#include <stdio.h>
#include <string.h>
struct stem{
    int score;
    int total;
    int right;
    char str[60];
    int a[5];
    int max;
}puzzle[101];
int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    for(int i=0;i<=m-1;i++){
        scanf("%d %d %d",&puzzle[i].score,&puzzle[i].total,&puzzle[i].right);
        gets(puzzle[i].str);
    }
    int wrong=0;
    int mmax=-1;
    for(int i=0;i<=n-1;i++){
        double count=0;
        for(int j=0;j<=m-1;j++){
            int ge;
            char sss[100];
            scanf("(%d%[^)]",&ge,sss);
            scanf(")");
            getchar();
            int flag=1;//right or no
            if(strcmp(puzzle[j].str,sss)==0){
                count+=puzzle[j].score;
            }
            else{
                wrong++;
                flag=0;
            }
            if(!flag){
                int judge=0;
                for(int s=1;s<=2*ge-1;s+=2){
                    int f=0;
                    for(int q=1;q<=2*puzzle[j].right-1;q+=2){
                        if(sss[s]==puzzle[j].str[q]){
                            judge++;
                            f=1;
                            break;
                        }
                    }
                    if(f==0){
                        puzzle[j].a[sss[s]-'a']++;
                        if(puzzle[j].a[sss[s]-'a']>mmax)mmax=puzzle[j].a[sss[s]-'a'];
                    }       
                }
                for(int qq=1;qq<=2*puzzle[j].right-1;qq+=2){
                    int ffff=0;
                    for(int la=1;la<=2*ge-1;la+=2){
                        if(puzzle[j].str[qq]==sss[la]){
                            ffff=1;
                            break;
                        }
                    }
                    if(ffff==0){
                        puzzle[j].a[puzzle[j].str[qq]-'a']++;
                        if(puzzle[j].a[puzzle[j].str[qq]-'a']>mmax)mmax=puzzle[j].a[puzzle[j].str[qq]-'a'];
                    }
                }
                if(judge==ge&&ge<puzzle[j].right){
                    count+=1.0*puzzle[j].score/2;
                }
            }
        }
        printf("%.1lf\n",count);
    }
    if(wrong==0)printf("Too simple\n");
    else{
        for(int i=0;i<=m-1;i++){
            for(int j=0;j<=4;j++){
                if(mmax==puzzle[i].a[j])printf("%d %d-%c\n",mmax,i+1,'a'+j);
            }
        }
    }
    return 0;
}