坑点1: 如果该用户的消费不满足有效条件 那么他的总消费也不必输出

坑点2：最后一组样例输出总消费时 应该考虑最后一组样例是否满足有效条件

有效条件: 第i个记录与第i+1条记录的名字相同并且第i个记录的状态是 "on-line" 第i+1个记录的状态是"off-line"；

kswl

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <iostream>
 4 #include <algorithm>
 5 using namespace std;
 6 const int maxn = 0x3f3f3f3f;
 7 typedef long long ll;
 8 int bill[25];
 9 struct bl{
10     string name;
11     int mon,day,hour,minu;
12     string st;// state on-line or off-line
13 }b[1005];
14 string a = "on-line";
15 string c = "off-line";
16 bool cmp(bl p,bl q) {//排序部分
17     if(p.name != q.name) return  p.name < q.name;
18     else if(p.mon != q.mon) return p.mon < q.mon;
19     else if(p.day != q.day) return p.day < q.day;
20     else if(p.hour != q.hour) return p.hour < q.hour;
21     return p.minu < q.minu;
22 }
23 int main()
24 {
25     for(int i = 0; i < 24; i++) {
26         cin >> bill[i];
27     }
28     int n;
29     cin >> n;
30     for(int i = 0; i < n ; i++) {
31         cin >> b[i].name;
32         scanf("%d:%d:%d:%d",&b[i].mon,&b[i].day,&b[i].hour,&b[i].minu);
33         cin >> b[i].st;
34     }
35     sort(b,b+n,cmp);
36     int t = 0;
37     double total = 0;
38     bool flag = false;
39     for(int i = 0; i < n-1; i++) {
40            double money = 0;
41             if(b[i].name == b[i+1].name && b[i].st == a && b[i+1].st == c) {
42                int sum = b[i+1].day*60*24 + b[i+1].hour*60 + b[i+1].minu - (b[i].day*60*24 + b[i].hour*60 + b[i].minu);
43                for(int p = b[i].day; p <= b[i+1].day; p++) {//一分一分的算消费
44                 int j = 0,jj = 23;
45                 if(p == b[i].day) j = b[i].hour;
46                 if(p == b[i+1].day) jj = b[i+1].hour;
47                for(; j <= jj; j++) {
48                    int k = 0,kk = 60;
49                    if(j == b[i].hour&&p == b[i].day) k = b[i].minu;
50                    if(j == b[i+1].hour&&p == b[i+1].day) kk = b[i+1].minu;
51                    for(;k < kk; k++) {
52                     money += bill[j];
53                    }
54                }
55                }
56                money = money*1.0/100.0;
57                total += money;
58                if(t == 0) {
59                 cout << b[i].name << " ";
60                 if(b[i].mon < 10) cout << 0;
61                 cout << b[i].mon << endl;
62                 t = 1;
63                }
64                printf("%02d:%02d:%02d %02d:%02d:%02d ",b[i].day,b[i].hour,b[i].minu,b[i+1].day,b[i+1].hour,b[i+1].minu);
65                printf("%d $%.2lf\n",sum,money);
66             } else if(b[i].name != b[i+1].name) {
67                t = 0;
68                if(total != 0){//坑点1:当total不等于0时同时也说明该用户的消费满足有效条件
69                printf("Total amount: $%.2lf\n",total);
70                }
71                total = 0;
72             }
73     }
74     if(t == 1)//坑点2：最后一组也要判断是否满足有效条件
75     printf("Total amount: $%.2lf\n",total);
76     return 0;
77 }