Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and ther…

1017. Queueing at Bank (25) 时间限制 400 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yue Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the custo…

1017 Queueing at Bank (25)(25 分)提问 Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is…

由PAT1017例题展开: Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be s…

Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and ther…

Suppose a bank has K windows open for service.  There is a yellow line in front of the windows which devides the waiting area into two parts.  All the customers have to wait in line behind the yellow line, until it is his/her turn to be served and th…

1017 Queueing at Bank (25 分)   Suppose a bank has K windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. All the customers have to wait in line behind the yellow line, until it is his…

PAT甲级1017. Queueing at Bank 题意: 假设一家银行有K台开放服务.窗前有一条黄线,将等候区分为两部分.所有的客户都必须在黄线后面排队,直到他/她轮到服务,并有一个可用的窗口. 假设一个客户不能占用1个小时以上的窗口. 现在考虑到每个客户的到达时间T和处理时间P,您应该告诉所有客户的平均等待时间. 输入: 每个输入文件包含一个测试用例.对于每种情况, 第一行包含2个数字:N(<= 10000) - 客户总数,K(<= 100) - 窗口数.然后N行跟随,每个包含2次:H…

有n个客户和k个窗口,给出n个客户的到达时间和需要的时长有空闲的窗口就去办理,没有的话就需要等待,求客户的平均时长.如果在8点前来的,就需要等到8点.如果17点以后来的,则不会被服务,无需考虑. 按客户的到达时间排序建立一个优先级队列,一开始放入k个窗口,初始结束时间为8*3600然后for循环客户,每次从优先级队列中取出最早结束时间的窗口如果客户比结束时间来的早,就需要等待如果客户比结束时间来的晚,就无需等待最后只要统计那些到达时间在17*3600之前的客户即可. #include <iost…

简单模拟. #include<iostream> #include<cstring> #include<cmath> #include<algorithm> #include<cstdio> #include<map> #include<queue> using namespace std; struct X { long long hh,mm,ss; long long len; long long st,ml; }s[…