HDU 4006 The kth great number (基本算法-水题)
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The kth great number
Problem Description
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.
Input
There are several test cases. For each test case, 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.
Output
The output consists of one integer representing the largest number of islands that all lie on one line.
Sample Input
8 3I 1I 2I 3QI 5QI 4Q
Sample Output
123HintXiao Ming won't ask Xiao Bao the kth great number when the number of the written number is smaller than k. (1=<k<=n<=1000000).
Source
The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest
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题目大意:
解题思路:有m组操作,求第K大数。I为insert操作,即插入1个数,Q为询问,输出此时的第K大数。
解题代码:一道水题让我智商捉鸡了,居然笨到想去用线段树去做。后来发现1个set搞定,就存K个大数,超过的把小的踢掉,输出第一个即可。
#include <iostream>#include <cstdio>#include <set>#include <algorithm>using namespace std;int n,k;multiset <long long> mys;void solve(){ mys.clear(); char ch; int x; for(int i=0;i<n;i++){ scanf("\n"); scanf("%c",&ch); if(ch=='I'){ scanf("%lld",&x); mys.insert(x); if(mys.size()>k) mys.erase(mys.begin()); }else{ cout<<*mys.begin()<<endl; } }}int main(){ while(scanf("%d%d",&n,&k)!=EOF){ solve(); } return 0;}
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