Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 11450 | Accepted: 5545 |
Description
Query-1: "Every five minutes, retrieve the maximum temperature over the past five minutes."
Query-2: "Return the average temperature measured on each floor over the past 10 minutes."
We have developed a Data Stream Management System called Argus, which processes the queries over the data streams. Users can register queries to the Argus. Argus will keep the queries running over the changing data and return the results to the corresponding user with the desired frequency.
For the Argus, we use the following instruction to register a query:
Register Q_num Period
Q_num (0 < Q_num <= 3000) is query ID-number, and Period (0 < Period <= 3000) is the interval between two consecutive returns of the result. After Period seconds of register, the result will be returned for the first time, and after that, the result will be returned every Period seconds.
Here we have several different queries registered in Argus at once. It is confirmed that all the queries have different Q_num. Your task is to tell the first K queries to return the results. If two or more queries are to return the results at the same time, they will return the results one by one in the ascending order of Q_num.
Input
The second part is your task. This part contains only one line, which is one positive integer K (<= 10000).
Output
Sample Input
Register 2004 200 Register 2005 300 # 5
Sample Output
2004 2005 2004 2004 2005
Source
Regionals 2004 >> Asia
- Beijing
问题链接:UVALive3135 UVA1203 POJ2051 ZOJ2212 Argus
问题简述:每个用户注册一个ID号,并且有一个间隔时间。对于每个用户,每隔一个间隔时间打印一次。注册的用户到“#”为止。之后给出k,是需要打印的用户ID数量。
问题分析:用优先队列来解决是最合适的。
程序说明:(略)
题记:(略)
参考链接:(略)
AC的C++语言程序如下:
/* UVALive3135 UVA1203 POJ2051 ZOJ2212 Argus */ #include <iostream> #include <queue> using namespace std; struct _node { int num, period, cnt; bool operator <(_node a)const { if(cnt != a.cnt) return cnt > a.cnt; else return num > a.num; } }; int main() { priority_queue<_node> q; _node t; string s; int k; while(cin >> s) { if(s == "#") break; else { cin >> t.num >> t.period; t.cnt = t.period; q.push(t); } } cin >> k; while(k--) { t = q.top(); q.pop(); cout << t.num << endl; t.cnt += t.period; q.push(t); } return 0; }