HDU ACMSTEPS 1.3.4

最先看到这题，犹豫了下，发现这题的输入需要做比较复杂的字符串处理。

随后还是编了代码出来。

 

版本1：AC。

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

#define STU_MAX 1000
#define NAME_LEN 11
#define TIME_LEN 11

#ifdef DEBUG
#include <assert.h>
#endif

typedef struct Student {
    
    char name[NAME_LEN];
    int acnum;
    int time;

};

int cmp(const void* a, const void* b) 
{
    Student* stu1 = (Student*)a;
    Student* stu2 = (Student*)b;

    if (stu1->acnum != stu2->acnum)
        return (stu2->acnum - stu1->acnum);
    if (stu1->acnum == stu2->acnum && stu1->time != stu2->time)
        return (stu1->time - stu2->time);
    return strcmp(stu1->name,stu2->name); 
}

int main(int argc, char* argv[])
{

    //topic number & punish score
    int topic_num,score_minus;

    scanf("%d %d",&topic_num,&score_minus);

    //malloc space for storage
    Student *stud_list = (Student*)malloc(sizeof(Student)*STU_MAX);
    memset(stud_list,0,sizeof(Student)*STU_MAX);

    //asume that only six student
    int stu_cnt = 0;

    char name[NAME_LEN] = {0};
    
    for (; scanf("%s",name) != EOF; stu_cnt++) {
        
        memcpy(stud_list[stu_cnt].name,name,strlen(name));

        //malloc two division of array
        //for topic cost time storage
        char** topics = (char**)malloc(sizeof(char*)*topic_num);
        memset(topics,0,sizeof(char*)*topic_num);
        
        for(int j=0; j<topic_num; j++) {

            //storage the input cost time for
            //every topic
            topics[j] = (char*)malloc(sizeof(char)*TIME_LEN);
            memset(topics[j],0,sizeof(char)*TIME_LEN);
            
            scanf("%s",topics[j]);
        }

        //cnt ac number
        //cnt cost time
        int ac_cnt = 0;
        int ac_cost = 0;

        for (int j=0; j<topic_num; j++) {

            //find braket
            char* start_braket = NULL;
            char* end_braket = NULL;

            start_braket = strstr(topics[j],"(");
            if (start_braket != topics[j]) {
                end_braket = strstr(topics[j],")");
            }

            //deal ac time with braket
            if (start_braket != NULL && end_braket != NULL) {

                //deal punish time
                char* sub_tmp = (char*)malloc(sizeof(char)*TIME_LEN);
                memset(sub_tmp,0,sizeof(char)*TIME_LEN);
                memcpy(sub_tmp,start_braket+1,strlen(start_braket)-strlen(end_braket)-1);
                int minus_time = atoi(sub_tmp)*score_minus;

                //deal ac time
                memset(sub_tmp,0,sizeof(char)*TIME_LEN);
                memcpy(sub_tmp,topics[j],strlen(topics[j])-strlen(start_braket));
                int ac_time = atoi(sub_tmp);

                //free memory
                free(sub_tmp);

                ac_cnt++;
                ac_cost += ac_time+minus_time;

            } else {
                int ac_time = atoi(topics[j]);
                if(ac_time > 0) {
                    ac_cnt++;
                    ac_cost += ac_time;
                }
            }
        }
        
        stud_list[stu_cnt].acnum = ac_cnt;
        stud_list[stu_cnt].time = ac_cost;

        //free memory
        free(topics);

    }

    //sort the student
    qsort(stud_list,stu_cnt,sizeof(Student),cmp);
    
    //print result
    for (int i = 0; i < stu_cnt; ++i) {
        printf("%-10s %2d %4d\n", stud_list[i].name,stud_list[i].acnum,stud_list[i].time);
    }

    //free memory
    free(stud_list);

    return 0;
}

 

这个版本的执行耗时为 0MS，执行内存为 900K，略大。 代码大小为 2943 字节。

 

后来发现，在输入的处理上，可以用sscanf来代替。遂，重写一个版本。

版本2：AC。

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

#define STU_MAX 1000
#define NAME_LEN 11

typedef struct Student{
    char name[NAME_LEN];
    int num;
    int cost;
};

int cmp(const void *a, const void *b)
{
    Student* stu1 = (Student*)a;
    Student* stu2 = (Student*)b;
    if (stu1->num != stu2->num) 
        return (stu2->num - stu1->num);
    if (stu1->cost != stu2->cost)
        return (stu1->cost - stu2->cost);
    return strcmp(stu1->name,stu2->name);
}

int main(int argc, char** argv)
{

    int n,m;
    scanf("%d %d",&n,&m);

    char name[NAME_LEN] = {0};

    int stuindex = 0;

    Student* stus = (Student*)malloc(sizeof(Student)*STU_MAX);
    memset(stus,0,sizeof(Student)*STU_MAX);

    while(scanf("%s",name) != EOF) {

        memcpy(stus[stuindex].name,name,strlen(name));

        for(int i=0; i<n; i++) {
            
            int t,d;
            
            char *time = (char*)malloc(sizeof(char)*20);
            scanf("%s",time);

            int res = sscanf(time,"%d(%d)",&t,&d);
            if (res == 1 && t > 0) {
                stus[stuindex].num++;
                stus[stuindex].cost += t;
            } else if (res == 2) {
                stus[stuindex].num++;
                stus[stuindex].cost += t+d*m;
            }
        }

        stuindex++;

    }

    //score stus!
    qsort(stus,stuindex,sizeof(Student),cmp);

    //print
    for (int i = 0; i < stuindex; ++i) {
        printf("%-10s %2d %4d\n", stus[i].name,stus[i].num,stus[i].cost);
    }

    return 0;
}

 

代码确实简洁不少，而且不用处理复杂的字符串格式。

这个版本的代码，执行耗时为 15MS，执行内存为290K，正常。 代码大小为 1372 字节。

 

2个版本，各有优劣吧。 从编码复杂度来说，版本2更为合适。