day3优先队列c++(priority_queue)

The kth great number

题目描述

Xiao Ming and Xiao Bao are playing a simple Numbers game. In a round Xiao Ming can choose to write down a number, or ask Xiao Bao what the kth great number is. Because the number written by Xiao Ming is too much, Xiao Bao is feeling giddy. Now, try to help Xiao Bao.

翻译:给你一堆数你告诉我第K大的数就行了

输入格式

The first line of input contains two positive integer n, k. Then n lines follow. If Xiao Ming choose to write down a number, there will be an " I" followed by a number that Xiao Ming will write down. If Xiao Ming choose to ask Xiao Bao, there will be a "Q", then you need to output the kth great number. Xiao Ming won't ask Xiao Bao the kth great number when the number of the written number is smaller than k.

翻译:第一行两个数n,k(1<=k<=n<=1000000).,接下来n行每行一个操作,如果是'I'表示插入一个数,如果是'Q'表示询问第k大的数。

保证数据合法就是当前数量比k小时不会询问。

输出格式

The output consists of one integer representing the largest number of islands that all lie on one line.

翻译:每次询问输出第K大的数,每次输出占一行

样例

输入数据 1

8 3
I 1
I 2
I 3
Q
I 5
Q
I 4
Q
 

输出数据 1

1
2
3
 

#include<bits/stdc++.h>
 
using namespace std;
 
priority_queue<int, vector<int>, greater<int> >a;//升序排列 小顶堆
 
int main () {
  char ch;
  int n,k;
  while (~scanf("%d%d",&n,&k)) {
    while (!a.empty())
        a.pop();
    while (n--) {
      int num;
      getchar();//读走换行
      scanf("%c", &ch);
      if (ch == 'I') {
        scanf("%d",&num);
        a.push(num);
        if(a.size()>k)
        a.pop();//弹出队头元素
      }else {
        num = a.top();//访问队头元素即最小元素
        printf("%d\n",num);
      }
    }
  }
  return 0;
}

评论 1
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值