Bus Fair

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 600    Accepted Submission(s): 293

Problem Description

You are now in Foolish Land. Once moving in Foolish Land you found that there is a strange Bus fair system. The fair of moving one kilometer by bus in that country is one coin. If you want to go to X km and your friend wants to go to Y km then you can buy a ticket of X+Y coins (you are also allowed to buy two or more tickets for you two).    Now as a programmer, you want to show your creativity in buying tickets! Suppose, your friend wants to go 1 km and you want to go 2 km. Then it’s enough for you to buy a 2coin ticket! Because both of you are valid passengers before crossing the first km. and when your bus cross the first km your friend gets down from the bus. So you have the ticket of 2km! And you can safely reach to your destination, 2km using that ticket.    Now, you have a large group of friends and they want to reach to different distance. You think that you are smart enough that you can buy tickets that should manage all to reach their destination spending the minimum amount of coins. Then tell us how much we should at least pay to reach our destination.

Input

There are multiple test cases. Each case start with a integer n, the total number of people in that group. 0<=n<=1000. Then comes n integers, each of them stands for a distance one of the men of the group wants to go to. You can assume that the distance a man wants to go is always less than 10000.

Output

Your program should print a single integer for a single case, the minimum amount of coins the group should spend to reach to the destination of all the members of that group.

Sample Input

2
1
2
2
2
3

Sample Output

2
4 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3420

 #include<stdio.h>
#include<stdlib.h>
int cmp(const void* a,const void* b)
{
return *(int *)a - *(int *)b;
}
int main()
{
int n,tmp,sum,i;
int num[];
while(~scanf("%d",&n))
{
for(i=;i<n;i++)
scanf("%d",&num[i]);
sum=;
qsort(num,n,sizeof(num[]),cmp);
for(i=;i<n;i++)
{
tmp=num[i]*(n-i);
if(tmp>sum)
sum=tmp;
}
printf("%d\n",sum);
}
return ;
}

HDU 3420 -- Bus Fair ACM的更多相关文章

  1. HDU 3420 Bus Fair [补]

    今天玩魔灵玩多了,耽误了时间,回去宿舍又没电. /*********************************************/ Bus Fair Time Limit: 2000/10 ...

  2. HDU 4046 Panda (ACM ICPC 2011北京赛区网络赛)

    HDU 4046 Panda (ACM ICPC 2011北京赛区网络赛) Panda Time Limit: 10000/4000 MS (Java/Others)    Memory Limit: ...

  3. hdu 5552 Bus Routes

    hdu 5552 Bus Routes 考虑有环的图不方便,可以考虑无环连通图的数量,然后用连通图的数量减去就好了. 无环连通图的个数就是树的个数,又 prufer 序我们知道是 $ n^{n-2} ...

  4. (并查集)Travel -- hdu -- 5441(2015 ACM/ICPC Asia Regional Changchun Online )

    http://acm.hdu.edu.cn/showproblem.php?pid=5441 Travel Time Limit: 1500/1000 MS (Java/Others)    Memo ...

  5. hdu 1690 Bus System(Dijkstra最短路)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1690 Bus System Time Limit: 2000/1000 MS (Java/Others ...

  6. 【线段树】HDU 5493 Queue (2015 ACM/ICPC Asia Regional Hefei Online)

    题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5493 题目大意: N个人,每个人有一个唯一的高度h,还有一个排名r,表示它前面或后面比它高的人的个数 ...

  7. hdu 4747 Mex (2013 ACM/ICPC Asia Regional Hangzhou Online)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4747 思路: 比赛打得太菜了,不想写....线段树莽一下 实现代码: #include<iost ...

  8. (二叉树)Elven Postman -- HDU -- 54444(2015 ACM/ICPC Asia Regional Changchun Online)

    http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Time Limit: 1500/1000 MS (Java/Others)  ...

  9. (字符串处理)Fang Fang -- hdu -- 5455 (2015 ACM/ICPC Asia Regional Shenyang Online)

    链接: http://acm.hdu.edu.cn/showproblem.php?pid=5455 Fang Fang Time Limit: 1500/1000 MS (Java/Others)  ...

随机推荐

  1. (转)linux内存源码分析 - 内存回收(整体流程)

    http://www.cnblogs.com/tolimit/p/5435068.html------------linux内存源码分析 - 内存回收(整体流程) 概述 当linux系统内存压力就大时 ...

  2. JAVA学习1:Maven3环境搭建

    好长时间不用Java,今天看了下,Maven集成成主流了,在技术水平与日俱进的同时,感叹下IT行业必须有活到老学到老的精神. 先说下环境: Maven:Maven 3.0.5 解压后路径:F:\Mav ...

  3. centos 7编译安装mysql 5.7.17

    1.进入MySQL官网下载MySQL源代码 依次点击Downloads -> Community -> MySQL Community Server 源代码1.Select Operati ...

  4. eclipse中springsource-tool-suite(sts)插件安装教程

    插件的下载参照:http://www.cnblogs.com/jepson6669/p/8540157.html 用过的eclipse不能安装成功,需要重新解压新的才能安装成功,不知道为什么? 解压上 ...

  5. Android控件之ListView的使用

    ListView是Android当中一个非常常用的数据显示控件. 第一种可以使用List<HashMap<String , Object>>,作为适配器的数据源来显示要显示的数 ...

  6. window server 搭建git服务器

    Git服务器Gogs简易安装-Windows环境   1.下载git for windows 1 https://github.com/git-for-windows/git/releases/dow ...

  7. Parsing Failure in config.xml: java.lang.IllegalArgumentException: In production mode, it's not allowed to set a clear text value to the property

    Step1). in your "setDomainEnv.sh" script set the "PRODUCTION_MODE=false" or use ...

  8. jquery 文本框回车与change事件

    文本框的改变用change事件 要用bind,两个是有区别的,change只是在失去焦点的时候出发,很多时候不能满足需要.  代码如下   $('#flowfromid').bind("pr ...

  9. 文档类型DTD,DOCTYPE和浏览器模式

    出处:http://blog.csdn.net/freshlover/article/details/11616563 浏览器从服务端获取网页后会根据文档的DOCTYPE定义显示网页,如果文档正确定义 ...

  10. MemcacheHelper.cs

    using Memcached.ClientLibrary; using System; using System.Collections.Generic; using System.Linq; us ...