PAT 甲级 1014 Waiting in Line
发布日期:2021-07-01 03:08:50 浏览次数:2 分类:技术文章

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1014 Waiting in Line (30 point(s))

Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:

  • The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line behind the yellow line.
  • Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
  • Customer​i​​ will take T​i​​ minutes to have his/her transaction processed.
  • The first N customers are assumed to be served at 8:00am.

Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.

For example, suppose that a bank has 2 windows and each window may have 2 custmers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, customer​1​​ is served at window​1​​ while customer​2​​ is served at window​2​​. Customer​3​​ will wait in front of window​1​​ and customer​4​​ will wait in front of window​2​​. Customer​5​​ will wait behind the yellow line.

At 08:01, customer​1​​ is done and customer​5​​ enters the line in front of window​1​​ since that line seems shorter now. Customer​2​​ will leave at 08:02, customer​4​​ at 08:06, customer​3​​ at 08:07, and finally customer​5​​ at 08:10.

Input Specification:

Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (≤20, number of windows), M (≤10, the maximum capacity of each line inside the yellow line), K (≤1000, number of customers), and Q (≤1000, number of customer queries).

The next line contains K positive integers, which are the processing time of the K customers.

The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.

Output Specification:

For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output Sorry instead.

Sample Input:

2 2 7 51 2 6 4 3 534 23 4 5 6 7

Sample Output:

08:0708:0608:1017:00Sorry

Experiential Summing-up

This question is a simple simulation of daily line up. Please care about the people who after the yellow line, what method he use to line up. Of course that is select the queue with shortest length. The method of we realize is by comparing windows' first people's end time. The smallest end time means this window will be the shortest one earliestly. There is nothing else to say~

(The purpose of using English to portray my solution is that to exercise the ability of my expression of English and accommodate PAT advanced level's style.We can make progress together by reading and comprehending it. Please forgive my basic grammar's and word's error. Of course, I would appreciated it if you can point out my grammar's and word's error in comment section.( •̀∀•́ ) Furthermore, Big Lao please don't laugh at me because I just a English beginner settle for CET6    _(:з」∠)_  )

Accepted Code

#include 
#include
using namespace std;const int maxn=1010;const int INF=0x3fffffff;struct customer{ int t,start;}C[maxn];int search(queue
q[],int n){ int min=INF,u=-1; for(int i=1;i<=n;++i) { if(q[i].front()
W[30]; scanf("%d %d %d %d",&n,&m,&k,&q); for(int i=1;i<=k;++i) { scanf("%d",&C[i].t); } int num=1; for(int i=0;i
k) printf("Sorry\n"); else { int h=C[temp].start/60; int m=C[temp].start%60; h+=8; if(h>=17) { printf("Sorry\n"); } else { printf("%02d:%02d\n",(C[temp].start+C[temp].t)/60+8,(C[temp].start+C[temp].t)%60); } } } return 0;}

 

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哈哈,博客排版真的漂亮呢~
[***.90.31.176]2024年04月25日 08时28分20秒