LA 3135

Argus

Time limit: 3.000 seconds

A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, financial tickers, on-line auctions, and transaction logs such as Web usage logs and telephone call records. Likewise, queries over streams run continuously over a period of time and incrementally return new results as new data arrives. For example, a temperature detection system of a factory warehouse may run queries like the following.

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 first part of the input are the register instructions to Argus, one instruction per line. You can assume the number of the instructions will not exceed 1000, and all these instructions are executed at the same time. This part is ended with a line of #.

The second part is your task. This part contains only one line, which is one positive integer K (≤ 10000).

Output

You should output the Q_num of the first K queries to return the results, one number per line.

Sample Input

Register 2004 200
Register 2005 300
#
5

Sample Output

2004
2005
2004
2004
2005

 

数据结构中的优先队列。

/*
* @author  Panoss
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<vector>
#include<ctime>
#include<stack>
#include<map>
#include<cmath>
#include<queue>
#include<list>
using namespace std;
#define DBG 1
#define fori(i,a,b) for(int i = (a); i < (b); i++)
#define forie(i,a,b) for(int i = (a); i <= (b); i++)
#define ford(i,a,b) for(int i = (a); i > (b); i--)
#define forde(i,a,b) for(int i = (a); i >= (b); i--)
#define forls(i,a,b,n) for(int i = (a); i != (b); i = n[i])
#define mset(a,v) memset(a, v, sizeof(a))
#define mcpy(a,b) memcpy(a, b, sizeof(a))
#define dout  DBG && cerr << __LINE__ << " >>| "
#define checkv(x) dout << #x"=" << (x) << " | "<<endl
#define checka(array,a,b) if(DBG) { \
    dout << #array"[] | " << endl;     forie(i, a, b) cerr << "[" << i << "]=" << array[i] << " |" << ((i - (a)+1) % 5 ? " " : "\n"); if (((b)-(a)+1) % 5) cerr << endl; }
#define redata(T, x) T x; cin >> x
#define MIN_LD -2147483648
#define MAX_LD  2147483647
#define MIN_LLD -9223372036854775808
#define MAX_LLD  9223372036854775807
#define MAX_INF 18446744073709551615
inline int  reint() { int d; scanf("%d", &d); return d; }
inline long relong() { long l; scanf("%ld", &l); return l; }
inline char rechar() { scanf(" "); return getchar(); }
inline double redouble() { double d; scanf("%lf", &d); return d; }
inline string restring() { string s; cin >> s; return s; }

struct Event
{
    int Q_num, Period;
    int Time;   ///下一次事件的时间

    bool operator < (const Event & X) const
    {
        return ((Time == X.Time && Q_num > X.Q_num) || (Time > X.Time));
    }
};

int main()
{
    priority_queue<Event> Q;
    char s[20];
    while(scanf("%s",s)==1&&s[0]!=‘#‘)
    {
        Event E;
        scanf("%d%d",&E.Q_num,&E.Period);
        E.Time = E.Period;
        Q.push(E);
    }
    int k;
    scanf("%d",&k);
    while(k--)
    {
        Event E = Q.top();
        Q.pop();
        printf("%d\n",E.Q_num);
        E.Time += E.Period;
        Q.push(E);
    }
    return 0;
}

LA 3135

时间: 2024-11-05 15:13:19

LA 3135的相关文章

la 3135 Argus Data Structure

// la 3135 Argus // 学习一下优先队列的使用吧,题目还是比较简单的 // 刘老师的训练指南p188. // 继续练吧.... #include <algorithm> #include <bitset> #include <cassert> #include <cctype> #include <cfloat> #include <climits> #include <cmath> #include &l

(算法竞赛入门经典 优先队列)LA 3135(前K条指令)

A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, financial tickers, on-line auctions, and transaction logs such as Web usage logs and telephone call records. Likewise, queries o

(算法入门经典大赛 优先级队列)LA 3135(之前K说明)

A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensor data, Internet traffic, financial tickers, on-line auctions, and transaction logs such as Web usage logs and telephone call records. Likewise, queries o

LA 3135 (优先队列) Argus

将多个有序表合并成一个有序表就是多路归并问题,可用优先队列来解决. 1 #include <cstdio> 2 #include <queue> 3 using namespace std; 4 5 const int maxn = 1000 + 10; 6 7 struct Node 8 { 9 int time, period, num; 10 bool operator < (const Node& rhs) const 11 { 12 return time

LA 3135 Argus (优先队列)

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1136 #include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<queue> using namespace

LA 3135 Argus (优先队列的简单应用)

A data stream is a real-time, continuous, ordered sequence of items. Some examples include sensordata, Internet traffic, nancial tickers, on-line auctions, and transaction logs such as Web usage logsand telephone call records. Likewise, queries over

hdu 5745 la vie en rose

这道题的官方题解是dp,但是可以暴力出来.改天再研究怎么dp. 暴力的时候,如果计算sum的时候,调用strlen函数会超时,可见这个函数并不是十分的好.以后能不用尽量不用. La Vie en rose Time Limit: 14000/7000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 861    Accepted Submission(s): 461 Problem

让MAC OS也能使用LL LA L等LS的别名

linux下默认ll是ls -l的别名.OS X下默认不支持.习惯了linux下使用ll,我们同样也可以将习惯搬到os x下的shell中. 再当前用户家目录下新建.bash_profile文件.根据你的习惯,添加下面格式内容即可. 1 2 3 alias ll='ls -l' alias la='ls -a' alias l='ls -la' 然后执行:source .bash_profile你还可以添加你喜欢的其他别名.

LA 3942 Remember the Word (Trie)

Remember the Word 题目:链接 题意:给出一个有S个不同单词组成的字典和一个长字符串.把这个字符串分解成若干个单词的连接(单词可以重复使用),有多少种方法? 思路:令d[i]表示从字符i开始的字符串(后缀s[i..L])的分解数,这d[i] = sum{d(i+len(x)) | 单词x是其前缀}.然后将所有单词建成一个Trie树,就可以将搜索单词的复杂度降低. 代码: #include<map> #include<set> #include<queue>