1016 Phone Bills (25)
題目描述：
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.
輸入格式：
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.輸出格式：
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.
題目大意：
這道題一開始我這麼都讀不懂。。後來才發現CYLL和CYJJ是兩個人。。一開始還以爲是一個人。
題目的要求就是按照名字字典序升序，時間升序排列，然後找到其中一前一後正好是同一個人，並且前面是on-line後面是off-line。
解題方法：
這邊在排序方面可以自己寫個函數，然後sort一下就好。在求花費的時候，要充分考慮到跨天、跨小時的情況。代碼有詳細註釋（超詳細的那種），直接看吧~
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，如果相等再按時間升序 */
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; /* 存放配對的call */
for (int i = 0; i < 24; i++)
{
scanf("%d", &prize[i]);
prize[24] += prize[i]; /* 用於存放一天總和 */
}
scanf("%d", &N);
call B[N]; /* 電話賬單數組，保存每一條記錄 */
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); /* 按題意排序 */
for (int i = 1; i < N; i++)
if (strcmp(B[i].name, B[i-1].name) == 0) /* 如果是同一個用戶 */
if (B[i-1].flag && !B[i].flag) /*如果前後正好配對 */
{ /* 將結果存入向量中 */
v.push_back(B[i-1]);
v.push_back(B[i]);
i++; /* 下一次比較 跳過 i+1與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; /* 將flag置爲1 防止重複打印相同的名字 */
else /* 如果當前客戶的名字不同於前面客戶的名字 */
{
countnameflag = 0;
strcpy(name, v[i].name);
}
if (countnameflag == 0) /* 如果是新用戶，打印名字信息並且打印上個用戶的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) /* 如果通話小時跨度超過2或跨越一天 */
{
/* 先把零頭計算一下 */
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)
{ /* 如果電話的D值之差超過1，也即不是在一天內打完（比如昨晚23點到今天1點） */
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; /* 將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;
}如果對您有幫助，幫忙點個小拇指唄~