Cattle off network PAT training grounds - census

Topic Address: https://www.nowcoder.com/pat/6/problem/4054

Solution: sequencing structure to

 1 /**
 2 * Copyright(c)
 3 * All rights reserved.
 4 * Author : YCute
 5 * Date : 2019-11-19-20.13.37
 6 * Description : 结构体排序
 7 */
 8 #include<iostream>
 9 #include<cstdio>
10 #include<cmath>
11 #include<cstring>
12 #include<algorithm>
13 using namespace std;
14 
15 struct bir{
16     char name[10];
17     int year;
18     int month;
19     int day;
20 };
21 
22 int cmp(struct bir a,struct bir b){
23     if(a.year<b.year) return 1;
24     else if(a.year==b.year){
25         if(a.month<b.month) return 1;
26         else if(a.month==b.month){
27             if(a.day<b.day) return 1;
28             else return 0;
29         }else return 0;
30     }else{
31         return 0;
32     }
33 }
34 
35 int main(){
36     int n;
37     struct bir b[100005];
38     scanf("%d",&n);
39     getchar();
40     for(int i=0;i<n;i++){
41         scanf("%s %4d/%2d/%2d",b[i].name,&b[i].year,&b[i].month,&b[i].day);
42     }
43     sort(b,b+n,cmp);
44     int i=0;
45     while(1){
46         if(b[i].year<1814){
47                 i++;
48                 continue;
49         }
50         else if(b[i].year==1814){
51             if(b[i].month<9){
52                     i++;
53                     continue;
54             }
55             else if(b[i].month==9){
56                 if(b[i].day<6){
57                         i++;
58                         continue;
59                 }
60                 else break;
61             }else{
62                 break;
63             }
64         }else{
65             break;
66         }
67     }
68     int j=n-1;
69     while(j){
70         if(b[j].year>2014){
71                 j--;
72                 continue;
73         }
74         else if(b[j].year==2014){
75             if(b[j].month>9){
76                 j--;
77                 continue;
78             }else if(b[j].month==9){
79                 if(b[j].day>6){
80                     j--;
81                     continue;
82                 }
83                 else break;
84             }else break;
85         }else {
86             break;
87         }
88 
89     }
90     printf("%d %s %s",j-i+1,b[i].name,b[j].name);
91     return 0;
92 }