PAT(甲) 1016 Phone Bills (25)（详解）

1016 Phone Bills (25)

题目描述：

A long-distance telephone company charges its customers by the following rules:
Making a long-distance call costs a certain amount per minute, depending on the time of day when the call is made. When a customer starts connecting a long-distance call, the time will be recorded, and so will be the time when the customer hangs up the phone. Every calendar month, a bill is sent to the customer for each minute called (at a rate determined by the time of day). Your job is to prepare the bills for each month, given a set of phone call records.

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• 输入格式：
  Each input file contains one test case. Each case has two parts: the rate structure, and the phone call records.
  The rate structure consists of a line with 24 non-negative integers denoting the toll (cents/minute) from 00:00 - 01:00, the toll from 01:00-02:00, and so on for each hour in the day.
  The next line contains a positive number N (<= 1000), followed by N lines of records. Each phone call record consists of the name of the customer (string of up to 20 characters without space), the time and date (mm:dd:hh:mm), and the word “on-line” or “off-line”.
  For each test case, all dates will be within a single month. Each “on-line” record is paired with the chronologically next record for the same customer provided it is an “off-line” record. Any “on-line” records that are not paired with an “off-line” record are ignored, as are “off-line” records not paired with an “on-line” record. It is guaranteed that at least one call is well paired in the input. You may assume that no two records for the same customer have the same time. Times are recorded using a 24-hour clock.

• 输出格式：
  For each test case, you must print a phone bill for each customer.
  Bills must be printed in alphabetical order of customers’ names. For each customer, first print in a line the name of the customer and the month of the bill in the format shown by the sample. Then for each time period of a call, print in one line the beginning and ending time and date (dd:hh:mm), the lasting time (in minute) and the charge of the call. The calls must be listed in chronological order. Finally, print the total charge for the month in the format shown by the sample.

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题目大意：
这道题一开始我这么都读不懂。。后来才发现CYLL和CYJJ是两个人。。一开始还以为是一个人。
题目的要求就是按照名字字典序升序，时间升序排列，然后找到其中一前一后正好是同一个人，并且前面是on-line后面是off-line。

解题方法：
这边在排序方面可以自己写个函数，然后sort一下就好。在求花费的时候，要充分考虑到跨天、跨小时的情况。代码有详细注释（超详细的那种），直接看吧~

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程序：

#include <stdio.h>
#include <algorithm>
#include <string.h>
#include <vector>
using namespace std;

struct call
{
    char name[21], status[10];
    int month, D, H, M, time, flag;
};

bool cmp(call c1, call c2)
{   /* 按名字升序，如果相等再按时间升序 */
    return strcmp(c1.name, c2.name) == 0 ? c1.time < c2.time : strcmp(c1.name, c2.name) < 0;  
}

int main(int argc, char const *argv[])
{
    int prize[25], N, countnameflag = 0;
    prize[24] = 0;
    vector <call> v;    /* 存放配对的call */
    for (int i = 0; i < 24; i++)
    {
        scanf("%d", &prize[i]);
        prize[24] += prize[i];  /* 用于存放一天总和 */
    }
    scanf("%d", &N);
    call B[N];   /* 电话账单数组，保存每一条记录 */
    for (int i = 0; i < N; i++)
    {
        scanf("%s %d:%d:%d:%d %s", B[i].name, &B[i].month, &B[i].D, &B[i].H, &B[i].M, B[i].status);
        B[i].time = B[i].D * 24 * 60 + B[i].H * 60 + B[i].M;
        if (strcmp(B[i].status, "on-line") == 0)
            B[i].flag = 1;   /* on-line */
        else
            B[i].flag = 0;   /* off-line */
    }
    sort(B, B+N, cmp);    /* 按题意排序 */
    for (int i = 1; i < N; i++)
        if (strcmp(B[i].name, B[i-1].name) == 0)    /* 如果是同一个用户 */
            if (B[i-1].flag && !B[i].flag)    /*如果前后正好配对 */ 
            {   /* 将结果存入向量中 */
                v.push_back(B[i-1]);
                v.push_back(B[i]);
                i++;    /* 下一次比较 跳过 i+1与i */
            }
    char name[21];
    strcpy(name, v[0].name);    
    printf("%s %02d\n", v[0].name, v[0].month);
    double TotalSum = 0;
    for (int i = 0; i < v.size(); i = i+2)
    {
        if (strcmp(v[i].name, name) == 0)
            countnameflag = 1;  /* 将flag置为1 防止重复打印相同的名字 */
        else    /* 如果当前客户的名字不同于前面客户的名字 */
        {
            countnameflag = 0;
            strcpy(name, v[i].name);  
        }        
        if (countnameflag == 0) /* 如果是新用户，打印名字信息并且打印上个用户的total */
        {   
            printf("Total amount: $%.2lf\n", TotalSum);
            TotalSum = 0;
            printf("%s %02d\n", v[i].name, v[i].month);
        }
        int p = 0;
        double sum = 0;
        if (v[i].H < v[i+1].H || v[i].D < v[i+1].D)  /* 如果通话小时跨度超过2或跨越一天 */
        {
            /* 先把零头计算一下 */
            sum += prize[v[i].H]*(60-v[i].M) + prize[v[i+1].H]*v[i+1].M;
            if (v[i].D < v[i+1].D)
            {   /* 如果电话的D值之差超过1，也即不是在一天内打完（比如昨晚23点到今天1点） */
                for (int j = v[i].H+1; j < 24; j++)
                    p += prize[j];
                for (int j = 0; j < v[i+1].H; j++)
                    p += prize[j];
                p += (v[i+1].D - v[i].D - 1) * prize[24]; 
            }
            else
                for (int j = v[i].H + 1; j < v[i+1].H; j++)
                    p += prize[j];
            sum += p * 60; 
        }   
        else
            sum += prize[v[i].H] * (v[i+1].M - v[i].M);
        sum /= 100;     /* 将cent->dollar */
        printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2lf\n", v[i].D, v[i].H, v[i].M, v[i+1].D, v[i+1].H, v[i+1].M, v[i+1].time-v[i].time, sum);
        TotalSum += sum;
    }
    printf("Total amount: $%.2lf\n", TotalSum);
    return 0;
}

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