PAT甲级 1016 Phone Bills (25分)（坑点）

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

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 0x3f3f3f3f;
typedef long long ll;
int bill[25];
struct bl{
    string name;
    int mon,day,hour,minu;
    string st;// state on-line or off-line
}b[1005];
string a = "on-line";
string c = "off-line";
bool cmp(bl p,bl q) {//排序部分
    if(p.name != q.name) return  p.name < q.name;
    else if(p.mon != q.mon) return p.mon < q.mon;
    else if(p.day != q.day) return p.day < q.day;
    else if(p.hour != q.hour) return p.hour < q.hour;
    return p.minu < q.minu;
}
int main()
{
    for(int i = 0; i < 24; i++) {
        cin >> bill[i];
    }
    int n;
    cin >> n;
    for(int i = 0; i < n ; i++) {
        cin >> b[i].name;
        scanf("%d:%d:%d:%d",&b[i].mon,&b[i].day,&b[i].hour,&b[i].minu);
        cin >> b[i].st;
    }
    sort(b,b+n,cmp);
    int t = 0;
    double total = 0;
    bool flag = false;
    for(int i = 0; i < n-1; i++) {
           double money = 0;
            if(b[i].name == b[i+1].name && b[i].st == a && b[i+1].st == c) {
               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);
               for(int p = b[i].day; p <= b[i+1].day; p++) {//一分一分的算消费
                int j = 0,jj = 23;
                if(p == b[i].day) j = b[i].hour;
                if(p == b[i+1].day) jj = b[i+1].hour;
               for(; j <= jj; j++) {
                   int k = 0,kk = 60;
                   if(j == b[i].hour&&p == b[i].day) k = b[i].minu;
                   if(j == b[i+1].hour&&p == b[i+1].day) kk = b[i+1].minu;
                   for(;k < kk; k++) {
                    money += bill[j];
                   }
               }
               }
               money = money*1.0/100.0;
               total += money;
               if(t == 0) {
                cout << b[i].name << " ";
                if(b[i].mon < 10) cout << 0;
                cout << b[i].mon << endl;
                t = 1;
               }
               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);
               printf("%d $%.2lf\n",sum,money);
            } else if(b[i].name != b[i+1].name) {
               t = 0;
               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
               printf("Total amount: $%.2lf\n",total);
               }
               total = 0;
            }
    }
    if(t == 1)//坑点2：最后一组也要判断是否满足有效条件
    printf("Total amount: $%.2lf\n",total);
    return 0;
}