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。
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 5 #define STU_MAX 1000 6 #define NAME_LEN 11 7 #define TIME_LEN 11 8 9 #ifdef DEBUG 10 #include <assert.h> 11 #endif 12 13 typedef struct Student { 14 15 char name[NAME_LEN]; 16 int acnum; 17 int time; 18 19 }; 20 21 int cmp(const void* a, const void* b) 22 { 23 Student* stu1 = (Student*)a; 24 Student* stu2 = (Student*)b; 25 26 if (stu1->acnum != stu2->acnum) 27 return (stu2->acnum - stu1->acnum); 28 if (stu1->acnum == stu2->acnum && stu1->time != stu2->time) 29 return (stu1->time - stu2->time); 30 return strcmp(stu1->name,stu2->name); 31 } 32 33 int main(int argc, char* argv[]) 34 { 35 36 //topic number & punish score 37 int topic_num,score_minus; 38 39 scanf("%d %d",&topic_num,&score_minus); 40 41 //malloc space for storage 42 Student *stud_list = (Student*)malloc(sizeof(Student)*STU_MAX); 43 memset(stud_list,0,sizeof(Student)*STU_MAX); 44 45 //asume that only six student 46 int stu_cnt = 0; 47 48 char name[NAME_LEN] = {0}; 49 50 for (; scanf("%s",name) != EOF; stu_cnt++) { 51 52 memcpy(stud_list[stu_cnt].name,name,strlen(name)); 53 54 //malloc two division of array 55 //for topic cost time storage 56 char** topics = (char**)malloc(sizeof(char*)*topic_num); 57 memset(topics,0,sizeof(char*)*topic_num); 58 59 for(int j=0; j<topic_num; j++) { 60 61 //storage the input cost time for 62 //every topic 63 topics[j] = (char*)malloc(sizeof(char)*TIME_LEN); 64 memset(topics[j],0,sizeof(char)*TIME_LEN); 65 66 scanf("%s",topics[j]); 67 } 68 69 //cnt ac number 70 //cnt cost time 71 int ac_cnt = 0; 72 int ac_cost = 0; 73 74 for (int j=0; j<topic_num; j++) { 75 76 //find braket 77 char* start_braket = NULL; 78 char* end_braket = NULL; 79 80 start_braket = strstr(topics[j],"("); 81 if (start_braket != topics[j]) { 82 end_braket = strstr(topics[j],")"); 83 } 84 85 //deal ac time with braket 86 if (start_braket != NULL && end_braket != NULL) { 87 88 //deal punish time 89 char* sub_tmp = (char*)malloc(sizeof(char)*TIME_LEN); 90 memset(sub_tmp,0,sizeof(char)*TIME_LEN); 91 memcpy(sub_tmp,start_braket+1,strlen(start_braket)-strlen(end_braket)-1); 92 int minus_time = atoi(sub_tmp)*score_minus; 93 94 //deal ac time 95 memset(sub_tmp,0,sizeof(char)*TIME_LEN); 96 memcpy(sub_tmp,topics[j],strlen(topics[j])-strlen(start_braket)); 97 int ac_time = atoi(sub_tmp); 98 99 //free memory 100 free(sub_tmp); 101 102 ac_cnt++; 103 ac_cost += ac_time+minus_time; 104 105 } else { 106 int ac_time = atoi(topics[j]); 107 if(ac_time > 0) { 108 ac_cnt++; 109 ac_cost += ac_time; 110 } 111 } 112 } 113 114 stud_list[stu_cnt].acnum = ac_cnt; 115 stud_list[stu_cnt].time = ac_cost; 116 117 //free memory 118 free(topics); 119 120 } 121 122 //sort the student 123 qsort(stud_list,stu_cnt,sizeof(Student),cmp); 124 125 //print result 126 for (int i = 0; i < stu_cnt; ++i) { 127 printf("%-10s %2d %4d\n", stud_list[i].name,stud_list[i].acnum,stud_list[i].time); 128 } 129 130 //free memory 131 free(stud_list); 132 133 return 0; 134 }
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。
1 #include <stdio.h> 2 #include <string.h> 3 #include <stdlib.h> 4 5 #define STU_MAX 1000 6 #define NAME_LEN 11 7 8 typedef struct Student{ 9 char name[NAME_LEN]; 10 int num; 11 int cost; 12 }; 13 14 int cmp(const void *a, const void *b) 15 { 16 Student* stu1 = (Student*)a; 17 Student* stu2 = (Student*)b; 18 if (stu1->num != stu2->num) 19 return (stu2->num - stu1->num); 20 if (stu1->cost != stu2->cost) 21 return (stu1->cost - stu2->cost); 22 return strcmp(stu1->name,stu2->name); 23 } 24 25 int main(int argc, char** argv) 26 { 27 28 int n,m; 29 scanf("%d %d",&n,&m); 30 31 char name[NAME_LEN] = {0}; 32 33 int stuindex = 0; 34 35 Student* stus = (Student*)malloc(sizeof(Student)*STU_MAX); 36 memset(stus,0,sizeof(Student)*STU_MAX); 37 38 while(scanf("%s",name) != EOF) { 39 40 memcpy(stus[stuindex].name,name,strlen(name)); 41 42 for(int i=0; i<n; i++) { 43 44 int t,d; 45 46 char *time = (char*)malloc(sizeof(char)*20); 47 scanf("%s",time); 48 49 int res = sscanf(time,"%d(%d)",&t,&d); 50 if (res == 1 && t > 0) { 51 stus[stuindex].num++; 52 stus[stuindex].cost += t; 53 } else if (res == 2) { 54 stus[stuindex].num++; 55 stus[stuindex].cost += t+d*m; 56 } 57 } 58 59 stuindex++; 60 61 } 62 63 //score stus! 64 qsort(stus,stuindex,sizeof(Student),cmp); 65 66 //print 67 for (int i = 0; i < stuindex; ++i) { 68 printf("%-10s %2d %4d\n", stus[i].name,stus[i].num,stus[i].cost); 69 } 70 71 return 0; 72 }
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。