#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
const int N=1e6+,M=4e6+,inf=1e9+,mod=1e9+;
const ll INF=1e18+;
struct is
{
int x;
int pos;
}a[N];
int cmp1(is a,is b)
{
if(a.x!=b.x)
return a.x<b.x;
return a.pos<b.pos;
}
int cmp2(is a,is b)
{
if(a.x!=b.x)
return a.x>b.x;
return a.pos>b.pos;
}
int cmp3(is a,is b)
{
return a.pos>b.pos;
}
int cmp4(is a,is b)
{
return a.pos<b.pos;
}
int main()
{
int n,k1,k2,cas=;
while(~scanf("%d%d%d",&n,&k1,&k2))
{
if(n==&&k1==&&k2==)
break;
for(int i=;i<=n;i++)
scanf("%d",&a[i].x),a[i].pos=i;
sort(a+,a+n+,cmp1);
sort(a+,a+k1+,cmp4);
printf("Case %d\n",cas++);
for(int i=;i<=k1;i++)
printf("%d%c",a[i].pos,(i!=k1)?' ':'\n');
sort(a+,a+n+,cmp2);
sort(a+,a+k2+,cmp3);
for(int i=;i<=k2;i++)
printf("%d%c",a[i].pos,(i!=k2)?' ':'\n');
}
return ;
}

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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
 
Source

hdu 4163 Stock Prices 水的更多相关文章

  1. hdu 4163 Stock Prices 花式排序

    Stock Prices Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Tot ...

  2. hdu 4940 数据太水...

    http://acm.hdu.edu.cn/showproblem.php?pid=4940 给出一个有向强连通图,每条边有两个值分别是破坏该边的代价和把该边建成无向边的代价(建立无向边的前提是删除该 ...

  3. hdu 1106:排序(水题,字符串处理 + 排序)

    排序 Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submissi ...

  4. HDU 4950 Monster (水题)

    Monster 题目链接: http://acm.hust.edu.cn/vjudge/contest/123554#problem/I Description Teacher Mai has a k ...

  5. HDU 4813 Hard Code 水题

    Hard Code Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.act ...

  6. HDU 4593 H - Robot 水题

    H - RobotTime Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hust.edu.cn/vjudge/contest/view.act ...

  7. HDOJ/HDU 2560 Buildings(嗯~水题)

    Problem Description We divide the HZNU Campus into N*M grids. As you can see from the picture below, ...

  8. HDOJ(HDU) 1859 最小长方形(水题、、)

    Problem Description 给定一系列2维平面点的坐标(x, y),其中x和y均为整数,要求用一个最小的长方形框将所有点框在内.长方形框的边分别平行于x和y坐标轴,点落在边上也算是被框在内 ...

  9. hdu 5753 Permutation Bo 水题

    Permutation Bo 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5753 Description There are two sequen ...

随机推荐

  1. NEON简介【转】

    转自:http://blog.csdn.net/fengbingchun/article/details/38020265 版权声明:本文为博主原创文章,未经博主允许不得转载. “ARM Advanc ...

  2. 为什么anylase和scenaio中的平均响应时间差别会这么大?

    场景里的响应时间截图如下所示: 结果里的响应时间截图如下所示:

  3. 自定义tableViewCell

    http://my.oschina.net/joanfen/blog/137601 效果如下图:可触发按钮事件 1.创建一个Empty Application 2.新建一个TableViewContr ...

  4. 23、jQuery九类选择器/jQuery常用Method-API/jQuery常用Event-API

      1)掌握jQuery九类选择器及应用 2)掌握jQuery常用Method-API 3)掌握jQuery常用Event-API 一)jQuery九类选择器[参见jQueryAPI.chm手册] 目 ...

  5. 19、文件上传与下载/JavaMail邮件开发

    回顾: 一. 监听器 生命周期监听器 ServletRequestListener HttpSessionListener ServletContextListener 属性监听器 ServletRe ...

  6. linux命令介绍:df使用介绍

    linux中df命令参数功能:检查文件系统的磁盘空间占用情况.可以利用该命令来获取硬盘被占用了多少空间,目前还剩下多少空间等信息. 语法:df [选项] 说明:linux中df命令可显示所有文件系统对 ...

  7. java 基本类型之间的转换

    基本数据类型从低级到高级是:byte  short int long float double ,char 类型比int 类型之后的都要低 下面通过一个例子说明: import javax.swing ...

  8. Android处理图片OOM的若干方法小结 (推荐)

    众所周知,每个Android应用程序在运行时都有一定的内存限制,限制大小一般为16MB或24MB(视平台而定).因此在开发应用时需要特别关注自身的内存使用量,而一般最耗内存量的资源,一般是图片.音频文 ...

  9. python 数据加密以及生成token和token验证

    代码如下: # -*- coding: utf-8 -*- from passlib.apps import custom_app_context as pwd_context import conf ...

  10. linux epoll 学习

    一.epoll介绍 epoll是linux内核为处理大批量句柄而作的改进的poll,是linux下IO多路复用select.poll的增强版,它能显著减少程序在大量并发连接中只有少量活跃的情况下的系统 ...