Sign In and Sign Out
题目的意思就是有一些人, 来者有先后, 给出每个人的来的时间, 和去的时间,求解第一个来的人和最后一个走的人
我的思路是直接qsort 对时间进行两次排序
附代码:
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<stdlib.h>
#define Max_len 20
#define Max_num 1000
int cmp(const void *a, const void *b);
int cmp1(const void *a, const void *b);
typedef struct come{
char str[Max_len];
int hour, min, sec;
}come;
typedef struct back{
char str[Max_len];
int hour, min, sec;
}back;
int main(){
come co[Max_num];
back ba[Max_num];
char temp[Max_len];
int m, i, hourc, minc, secc, hourb, minb, secb;
scanf("%d", &m);
for(i = 0; i < m; i++){
scanf("%s %d:%d:%d%d:%d:%d", temp, &hourc, &minc, &secc, &hourb, &minb, &secb);
strcpy(co[i].str, temp); co[i].hour = hourc; co[i].min = minc; co[i].sec = secc;
strcpy(ba[i].str, temp); ba[i].hour = hourb; ba[i].min = minb; ba[i].sec = secb;
}
qsort(co, m, sizeof(co[0]), cmp);
qsort(ba, m, sizeof(ba[0]), cmp1);
printf("%s %s\n", co[0].str, ba[0].str);
return 0;
}
int cmp(const void *a, const void *b){
come *c = (come *)a;
come *d = (come *)b;
if(c -> hour != d -> hour){
return c -> hour - d -> hour;
}
else if(c -> min != d -> min){
return c -> min - d -> min;
}
else{
return c -> sec - d -> sec;
}
}
int cmp1(const void *a, const void *b){
back *c = (back *)a;
back *d = (back *)b;
if(c -> hour != d -> hour){
return d -> hour - c -> hour;
}
else if(c -> min != d -> min){
return d -> min - c -> min;
}
else{
return d -> sec - c -> sec;
}
}