1016 Phone Bills -----15分

其实自己写的时候是一点思路都没有的，看了柳婼的分析才勉强能写出样例15分。。。

分析：将给出的数据先按照姓名排序，再按照时间的先后顺序排列，这样遍历的时候，前后两个名字相同且前面的状态为on-line后面一个的状态为off-line的就是合格数据～

注意：【关于最后一个测试点】计算费用从00:00:00到dd:hh:mm计算可以避免跨天的问题，比如01:12:00到02:02:00

```cpp
#include<iostream>
#include<algorithm>
#include<iomanip>
using namespace std;

struct client
{
    string name;
    int month;
    int day;
    int hour;
    int minute;
    string statue;
}per[1000];

int fee[24];

bool cmp1(client a,client b)
{
    if(a.name!=b.name)
        return a.name<b.name;
    else if(a.month!=b.month)
        return a.month<b.month;
    else if(a.day!=b.day)
        return a.day<b.day;
    else if(a.hour!=b.hour)
        return a.hour<b.hour;
    else if(a.minute!=b.minute)
        return a.minute<b.minute;
}

int main()
{
    cout<<fixed<<setprecision(2);
    for(int i=0;i<24;i++)
        cin>>fee[i];
    int n;
    cin>>n;

    string name,statue;
    int month,day,hour,minute;
    for(int i=0;i<n;i++)
    {
        cin>>name;
        scanf("%d:%d:%d:%d",&month,&day,&hour,&minute);
        cin>>statue;
        per[i].name=name;
        per[i].month=month;
        per[i].day=day;
        per[i].hour=hour;
        per[i].minute=minute;
        per[i].statue=statue;
    }

    sort(per,per+n,cmp1);
    float singlesum=0,totalsum=0;
    int totalminute=0;
    for(int i=0;i<n;i++)
    {
        singlesum=0;
        totalsum=0;
        cout<<per[i].name<<' ';
        printf("%02d\n",per[i].month);
        while(per[i].name==per[i+1].name)
        {
            totalminute=0;
            singlesum=0;
            if(per[i].statue=="on-line"&&per[i+1].statue=="off-line")
            {
                printf("%02d:%02d:%02d ",per[i].day,per[i].hour,per[i].minute);
                printf("%02d:%02d:%02d ",per[i+1].day,per[i+1].hour,per[i+1].minute);
                for(int j=per[i].hour;j<=per[i+1].hour;j++)
                {
                    if(j==per[i].hour)
                    {
                        totalminute+=60-per[i].minute;
                        singlesum+=(60-per[i].minute)*fee[j];
                    }

                    else if(j==per[i+1].hour)
                    {
                        totalminute+=per[i+1].minute;
                        singlesum+=(per[i+1].minute)*fee[j];
                    }

                    else
                    {
                        totalminute+=60;
                        singlesum+=60*fee[j];
                    }

                }
                if(per[i].day!=per[i+1].day)
                {
                    float onedaymoney=0;
                    for(int k=0;k<24;k++)
                        onedaymoney+=60*fee[k];
                    onedaymoney=onedaymoney;
                    singlesum+=(per[i+1].day-per[i].day)*onedaymoney;
                    totalminute+=24*60*(per[i+1].day-per[i].day);
                }
                totalsum+=singlesum*0.01;
                cout<<totalminute<<" $"<<singlesum*0.01<<endl;
                i++;
            }
            else
                i++;
        }
        cout<<"Total amount: $"<<totalsum<<endl;
    }

    return 0;
}

说实话，能写出样例就很费劲了。。。测试点太难了

柳婼的：
这代码太巧妙了，学不来。

#include <iostream>
#include <map>
#include <vector>
#include <algorithm>
using namespace std;
struct node {
    string name;
    int status, month, time, day, hour, minute;
};
bool cmp(node a, node b) {
    return a.name != b.name ? a.name < b.name : a.time < b.time;
}
double billFromZero(node call, int *rate) {
    double total = rate[call.hour] * call.minute + rate[24] * 60 * call.day;
    for (int i = 0; i < call.hour; i++)
        total += rate[i] * 60;
    return total / 100.0;
}
int main() {
    int rate[25] = {0}, n;
    for (int i = 0; i < 24; i++) {
        scanf("%d", &rate[i]);
        rate[24] += rate[i];
    }
    scanf("%d", &n);
    vector<node> data(n);
    for (int i = 0; i < n; i++) {
        cin >> data[i].name;
        scanf("%d:%d:%d:%d", &data[i].month, &data[i].day, &data[i].hour, &data[i].minute);
        string temp;
        cin >> temp;
        data[i].status = (temp == "on-line") ? 1 : 0;
        data[i].time = data[i].day * 24 * 60 + data[i].hour * 60 + data[i].minute;
    }
    sort(data.begin(), data.end(), cmp);
    map<string, vector<node> > custom;
    for (int i = 1; i < n; i++) {
        if (data[i].name == data[i - 1].name && data[i - 1].status == 1 && data[i].status == 0) {
            custom[data[i - 1].name].push_back(data[i - 1]);
            custom[data[i].name].push_back(data[i]);
        }
    }
    for (auto it : custom) {
        vector<node> temp = it.second;
        cout << it.first;
        printf(" %02d\n", temp[0].month);
        double total = 0.0;
        for (int i = 1; i < temp.size(); i += 2) {
            double t = billFromZero(temp[i], rate) - billFromZero(temp[i - 1], rate);
            printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n", temp[i - 1].day, temp[i - 1].hour, temp[i - 1].minute, temp[i].day, temp[i].hour, temp[i].minute, temp[i].time - temp[i - 1].time, t);
            total += t;
        }
        printf("Total amount: $%.2f\n", total);
    }
    return 0;
}