Stock Prices

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 987    Accepted Submission(s): 397

Problem Description
Buy
low, sell high. That is what one should do to make profit in the stock
market (we will ignore short selling here). Of course, no one can tell
the price of a stock in the future, so it is difficult to know exactly
when to buy and sell and how much profit one can make by repeatedly
buying and selling a stock.

But if you do have the history of
price of a stock for the last n days, it is certainly possible to
determine the maximum profit that could have been made. Instead, we are
interested in finding the k1 lowest prices and k2 highest prices in the
history.

 
Input
The
input consists of a number of cases. The first line of each case starts
with positive integers n, k1, and k2 on a line (n <= 1,000,000, k1 +
k2 <= n, k1, k2 <= 100). The next line contains integers giving
the prices of a stock in the last n days: the i-th integer (1 <= i
<= n) gives the stock price on day i. The stock prices are
non-negative. The input is terminated by n = k1 = k2 = 0, and that case
should not be processed.
 
Output
For
each case, produce three lines of output. The first line contains the
case number (starting from 1) on one line. The second line specifies the
days on which the k1 lowest stock prices occur. The days are sorted in
ascending order. The third line specifies the days on which the k2
highest stock prices occur, and the days sorted in descending order. The
entries in each list should be separated by a single space. If there
are multiple correct lists for the lowest prices, choose the
lexicographically smallest list. If there are multiple correct lists for
the highest prices, choose the lexicographically largest list.
 
Sample Input
10 3 2
1 2 3 4 5 6 7 8 9 10
10 3 2
10 9 8 7 6 5 4 3 2 1
0 0 0
 
Sample Output
Case 1
1 2 3
10 9
Case 2
8 9 10
2 1
 
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1000001
const int inf=0x7fffffff; //无限大
struct node
{
int x;
int y;
};
node a[maxn];
node b[maxn];
bool cmp(node c,node d)
{
if(c.x==d.x)
return c.y<d.y;
return c.x<d.x;
}
bool cmp1(node c,node d)
{
return c.y>d.y;
}
bool cmp2(node c,node d)
{
return c.y<d.y;
}
int main()
{
int n,k1,k2;
int cas=;
while(scanf("%d%d%d",&n,&k1,&k2)!=EOF)
{
if(n==)
break;
for(int i=;i<n;i++)
{
scanf("%d",&a[i].x);
a[i].y=i+;
}
sort(a,a+n,cmp);
sort(a,a+k1,cmp2);
sort(a+n-k2,a+n,cmp1);
printf("Case %d\n",cas++);
int first=;
for(int i=;i<k1;i++)
{
if(first)
{
printf("%d",a[i].y);
first=;
}
else
printf(" %d",a[i].y);
}
printf("\n");
first=;
for(int i=n-k2;i<n;i++)
{
if(first)
{
printf("%d",a[i].y);
first=;
}
else
printf(" %d",a[i].y);
}
printf("\n");
}
return ;
}

hdu 4163 Stock Prices 花式排序的更多相关文章

  1. hdu 4163 Stock Prices 水

    #include<bits/stdc++.h> using namespace std; #define ll long long #define pi (4*atan(1.0)) #de ...

  2. 题解报告:hdu 2647 Reward(拓扑排序)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2647 Problem Description Dandelion's uncle is a boss ...

  3. hdu 5098 双队列拓扑排序

    http://acm.hdu.edu.cn/showproblem.php?pid=5098 软件在安装之后需要重启才能发挥作用,现在给你一堆软件(有的需要重启有的不需要)以及安装这个软件之前需要哪些 ...

  4. HDU 5811 Colosseo(拓扑排序+单调DP)

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5811 [题目大意] 给出 一张单向图,现在将其划分成了两个部分,问划分之后的点是否分别满足按照一定 ...

  5. HDU 5738 Eureka(极角排序)

    [题目链接] http://acm.hdu.edu.cn/showproblem.php?pid=5738 [题目大意] 给出平面中一些点,在同一直线的点可以划分为一个集合,问可以组成多少包含元素不少 ...

  6. (hdu) 4857 逃生 (拓扑排序+优先队列)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4857 Problem Description 糟糕的事情发生啦,现在大家都忙着逃命.但是逃命的通道很窄 ...

  7. HDU 2647 Reward(拓扑排序+判断环+分层)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2647 题目大意:要给n个人发工资,告诉你m个关系,给出m行每行a b,表示b的工资小于a的工资,最低工 ...

  8. hdu 5188 zhx and contest [ 排序 + 背包 ]

    传送门 zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Othe ...

  9. HDU 5695 Gym Class 拓扑排序

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5695 题解: 求出字典序最大的拓扑序.然后把求好的数列翻转过来就是满足条件的数列,然后模拟求一下va ...

随机推荐

  1. linux音频alsa-uda134x驱动文档阅读之一转自http://blog.csdn.net/wantianpei/article/details/7817293

    前言 目前,linux系统常用的音频驱动有两种形式:alsa oss alsa:现在是linux下音频驱动的主要形式,与简单的oss兼容.oss:过去的形式而我们板子上的uda1341用的就是alsa ...

  2. 【转载】ajaxFileUpload 报这错jQuery.handleError is not a function

    今天刚打个一个技术群,里面有人问标题上的问题,嘿,我恰好遇过,现在大家至少也在用jquery1.9以上的版本,ajaxfileupload的版本早就不更新了,大家可以下载看:地址这里,它例子里使用的J ...

  3. 【前端】h5音乐播放demo 可关闭可播放

    复制如下代码,直接可预览(记得把超链接换成自己本地路径) <!DOCTYPE html> <html> <head> <meta charset=" ...

  4. java 二叉树遍历

    package com.lever; import java.util.LinkedList;import java.util.Queue; /** * 二叉树遍历 * @author lckxxy ...

  5. Linux入门(一)root密码设置和用户切换

    从这学期开始,本人将会亲自开一个Linux 专题学习包括Linux 常用命令,常见问题的一些解决方法,以及Linux系统下C和C++一些学习经验 下面这张图片是首次安装Ubuntu后第一次设置root ...

  6. 洛谷P1342请柬

    传送门啦 核心思想:两遍最短路. 1号点去各地的时间直接套最短路模板,各地到1号点时间用逆向思维,视为求1号点沿反边到各地的时间即可. #include <iostream> #inclu ...

  7. linux用户操作

    1.用户种类 Linux具有三种用户: 超级管理员root:具有最高权限,UID=0  GID=0伪用户(System Account):(UID=1~499)普通用户(login-Account): ...

  8. 一个带bash,带glibc,中国时区,非root用户可运行crond命令的基于alpine镜像的Dockerfile

    这个镜像现在说起来简单, 带bash(增加执行脚本的兼容性,带GLIBC,中国时区,非root用户可运行crond命令-安全) 但让我开始陷入时,真的让我有段时间有点爆了. 比如,将filebeat文 ...

  9. HTML--1

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  10. day6 shelve模块

        shelve模块 shelve模块是一个简单的k,v将内存数据通过文件持久化的模块,可以持久化任何pickle可支持的python数据格式,shelve模块是对pickle模块的补充.我们知道 ...