我的PAT系列文章更新重心已移至Github，欢迎来看PAT题解的小伙伴请到Github Pages浏览最新内容。此处文章目前已更新至与Github Pages同步。欢迎star我的repo。
题目
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.
Input Specification:
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 (
), followed by
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.
Output Specification:
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.
Sample Input:
10 10 10 10 10 10 20 20 20 15 15 15 15 15 15 15 20 30 20 15 15 10 10 10
10
CYLL 01:01:06:01 on-line
CYLL 01:28:16:05 off-line
CYJJ 01:01:07:00 off-line
CYLL 01:01:08:03 off-line
CYJJ 01:01:05:59 on-line
aaa 01:01:01:03 on-line
aaa 01:02:00:01 on-line
CYLL 01:28:15:41 on-line
aaa 01:05:02:24 on-line
aaa 01:04:23:59 off-line
Sample Output:
CYJJ 01
01:05:59 01:07:00 61 $12.10
Total amount: $12.10
CYLL 01
01:06:01 01:08:03 122 $24.40
28:15:41 28:16:05 24 $3.85
Total amount: $28.25
aaa 01
02:00:01 04:23:59 4318 $638.80
Total amount: $638.80
思路
大体思路:
- 题目给了不同人的不同通话记录, 要求输出每个人的账单. 那么最好我们在读取了数据之后, 立即对数据进行排序. 首先对名称排序, 其次对时间排序. 这样后面就好处理了.
- 具体处理数据, 对比每相邻的两个记录, 以确定是否是同一个人/是否是同一组通话记录
- 如果两组记录姓名不同, 则为一个新的账单, 此时需要输出总结, 将相关变量归零
- 如果不是上述情况, 并且两组记录分为为'on-line'和'off-line', 那么则找到一组通话记录. 此时需计算通话费用, 累计总费用以及输出此记录信息.
数据结构:
-
struct[N + 1]: 结构体包括名称, 月, 日, 时, 分, 时间(归算为分钟)和通话状态. 数组要比数据多1是因为要用最后一个作为空白对照, 以减少特殊情况的处理
注意的点:
- 计算费用
- (在其他博客看到的方法)计算从0到开始/结束时间的费用, 两者相减. 我觉得这种方法是最简单的, 实现容易.
- 我的方法是从开始到结束, 对每小时内的情况进行处理, 照顾到不足一小时的情况, 具体实现见
calccharge函数.
- 输出, 除了上面思路中谈到的何时输出什么, 还有一点应该注意
- 对于每个人, 如果没有需要输出的记录, 那么就什么都不要输出, 所以只能在明确有合理通话记录时, 再输出账单开头的姓名和月份的信息
代码
最新代码@github，欢迎交流
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
/* Structure to store phone records */
typedef struct {
char name[21];
int month, day, hour, min, time, state;
}Record, *pRecord;
/* Compare first by name, then by date and time */
int cmp(const void *record1, const void *record2)
{
pRecord r1 = *(pRecord*)record1, r2 = *(pRecord*)record2;
return strcmp(r1->name, r2->name) ?
strcmp(r1->name, r2->name) : r1->time - r2->time;
}
/* Calculate the charge of the call with start record p1 and end record p2 */
int calccharge(pRecord p1, pRecord p2, int toll[])
{
int charge = 0, start = p1->time, end = p2->time, h, time1, time2;
for(time1 = start; time1 < end; time1 = time2)
{ /* Add the charge hour by hour */
time2 = (time1 / 60 + 1) * 60; /* time2 will be the time of next hour */
h = time1 / 60 % 24; /* h will be the index of the hour */
charge += ((time2 > end ? end : time2) - time1) * toll[h];
}
return charge;
}
int main()
{
char state[9];
int N, toll[24], charge, charge_total = 0;
Record records[1001] = {0};
pRecord precords[1001] = {0}, *p = precords;
/* Read data */
for(int i = 0; i < 24; i++)
scanf("%d", toll + i);
scanf("%d", &N);
for(int i = 0; i < N; i++, p++)
{
*p = records + i;
scanf("%s %d:%d:%d:%d %s", (*p)->name,
&(*p)->month, &(*p)->day, &(*p)->hour, &(*p)->min, state);
(*p)->time = ((*p)->day * 24 + (*p)->hour) * 60 + (*p)->min;
(*p)->state = strcmp(state, "on-line") ? 0 : 1;
}
/* Sort first by name, then by date and time */
qsort(precords, N, sizeof(pRecord), cmp);
/* Print phone bill one by one */
for(p = precords + 1; *p; p++)
{
if(strcmp((*p)->name, (*(p - 1))->name))
{ /* A new customer, print last total amount if any */
if(charge_total)
printf("Total amount: $%.2f\n", charge_total * 1e-2);
charge_total = 0;
}
else if((*(p - 1))->state == 1 && (*p)->state == 0)
{ /* Still the same customer, finding on/off record pair */
if(charge_total == 0)
printf("%s %02d\n", (*p)->name, (*p)->month);
charge = calccharge(*(p - 1), *p, toll);
charge_total += charge;
/* Print info of this call */
printf("%02d:%02d:%02d %02d:%02d:%02d %d $%.2f\n",
(*(p - 1))->day, (*(p - 1))->hour, (*(p - 1))->min,
(*p)->day, (*p)->hour, (*p)->min,
(*p)->time - (*(p - 1))->time, charge * 1e-2);
}
}
if(charge_total)
printf("Total amount: $%.2f\n", charge_total * 1e-2);
return 0;
}
