题目如下: 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 an…

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…

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…

problem 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…

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 ther…

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[…

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…

题意: 输入两个正整数N,K(N<=10000,k<=100)分别表示用户的数量以及银行柜台的数量,接下来N行输入一个字符串(格式为HH:MM:SS)和一个正整数,分别表示一位用户到达银行的时间以及该用户占用银行资源的时间,求用户的平均等待时间.(当一位用户早于早晨八点到达银行不开门,他需要等待,当一位用户晚于下午五点到达银行关门,该用户不计入等待时间计算)[答案保留一位小数] AAAAAccepted code: #include<bits/stdc++.h> using nam…

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…

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) 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…

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…

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 th…

https://pintia.cn/problem-sets/994805342720868352/problems/994805491530579968 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…

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…

题意:银行有K个窗口用于服务,给出所有人的达到时间T和服务时间P,计算所有被服务的客户的平均等待时间.任何客户的服务时间不得超过60分钟.早于08:00到的,要等到08:00:在17:00:01及之后到的,不给予服务,不参与计算平均等待时间. 思路:首先,对所有顾客的到达时间进行排序:其次,令数组windows[i]表示窗口i当前顾客的结束服务的时间,初始化为08:00,这样的话,每次要安排下一个顾客前往某窗口接受服务时,就遍历所有窗口,找出当前最快结束的那个窗口(记为minidx),然后让这位…

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…

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 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 ther…

[思路] 1:将所有满足条件的(到来时间点在17点之前的)客户放入结构体中,结构体的长度就是需要服务的客户的个数.结构体按照到达时间排序. 2:wend数组表示某个窗口的结束时间,一开始所有窗口的值都为8点整.每一个客户到来的时候,选择最早结束时间的窗口.如果最早结束时间比客户到得还早,那么他一来就能被服务,更新wend的值:如果最早结束时间比他晚,他需要等待,累加等待的时间,然后更新wend的值. [坑] 测试点5:需要服务的客户数validn可能为0.此时它不能作为除数,所以要分开写. 测试…

1017. Queueing at Bank (25) 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/he…

博主欢迎转载,但请给出本文链接,我尊重你,你尊重我,谢谢~http://www.cnblogs.com/chenxiwenruo/p/6102219.html特别不喜欢那些随便转载别人的原创文章又不给出链接的所以不准偷偷复制博主的博客噢~~ 时隔两年,又开始刷题啦,这篇用于PAT甲级题解,会随着不断刷题持续更新中,至于更新速度呢,嘿嘿,无法估计,不知道什么时候刷完这100多道题. 带*的是我认为比较不错的题目,其它的难点也顶多是细节处理的问题~ 做着做着,发现有些题目真的是太水了,都不想写题解了…

早期部分代码用 Java 实现.由于 PAT 虽然支持各种语言,但只有 C/C++标程来限定时间,许多题目用 Java 读入数据就已经超时,后来转投 C/C++.浏览全部代码:请戳 本文谨代表个人思路,欢迎讨论;) 1011. World Cup Betting (20) 题意 给定一个 3*3 的矩阵,找到每行的最大值,格式化输出一个运算结果. 分析 非常简单的模拟题. 1012. The Best Rank (25) 题意 给定学生的学号和三个科目的分数,查询输出对应学生单门科目排名和总分排…

准备每天刷两题PAT真题.(一句话题解) 1001 A+B Format  模拟输出,注意格式 #include <cstdio> #include <cstring> #include <iostream> #include <algorithm> using namespace std; string ans = ""; int main() { ; cin >> a >> b; c = a + b; ) {…

树(23) 备注 1004 Counting Leaves   1020 Tree Traversals   1043 Is It a Binary Search Tree 判断BST,BST的性质 1053 Path of Equal Weight   1064 Complete Binary Search Tree 完全二叉树的顺序存储,BST的性质 1066 Root of AVL Tree 构建AVL树,模板题,需理解记忆 1079 Total Sales of Supply Chain…

A1017 Queueing at Bank (25 分) 题目内容 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…

专题一  字符串处理 A1001 Format(20) #include<cstdio> int main () { ]; int a,b,sum; scanf ("%d %d",&a,&b); sum=a+b; ) { printf ("-"); sum=-sum; } ; ) { s[top++]=; } ) { s[top++]=sum%; sum/=; } ;i>=;i--) { printf ("%d"…