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poj2051&amp;&amp;UVALive 3135 水題

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poj2051&&UVALive 3135 水題

Argus Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 9716 Accepted: 4543

Description

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

Source

Beijing 2004

水題拉，就是學習下優先隊列結構體咋寫，注意重載時候大於其實是小的優先，因為stl默認是大根堆。。

/**
 * @author neko01
 */
//#pragma comment(linker, /STACK:102400000,102400000)
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include
using namespace std;
typedef long long LL;
#define min3(a,b,c) min(a,min(b,c))
#define max3(a,b,c) max(a,max(b,c))
#define pb push_back
#define mp(a,b) make_pair(a,b)
#define clr(a) memset(a,0,sizeof a)
#define clr1(a) memset(a,-1,sizeof a)
#define dbg(a) printf(%d
,a)
typedef pair pp;
const double eps=1e-8;
const double pi=acos(-1.0);
const int INF=0x7fffffff;
const LL inf=(((LL)1)<<61)+5;
struct item{
    int qnum,period,time;
    bool operator < (const item& a) const{
        return time>a.time||(time==a.time&&qnum>a.qnum); //這裡注意大於號實際是time小的優先
    }
};
char s[23];
int main()
{
    priority_queueq;
    while(~scanf(%s,s)&&strcmp(s,#)!=0)
    {
        item cur;
        scanf(%d%d,&cur.qnum,&cur.period);
        cur.time=cur.period;
        q.push(cur);
    }
    int k;
    scanf(%d,&k);
    while(k--)
    {
        item cur=q.top();
        q.pop();
        printf(%d
,cur.qnum);
        cur.time+=cur.period;
        q.push(cur);
    }
    return 0;
}