Problem C: Vito's family 

Background

The world-known gangster Vito Deadstone is moving to New York. He hasa very big family there, all of them living in Lamafia Avenue. Sincehe will visit all his relatives very often, he is trying to find ahouse close to them.

Problem

Vito wants to minimize the total distance toall of them and has blackmailed you to write a program that solves his problem.

Input

The input consists of several test cases. The first line contains the number of test cases.

For each testcase you will be given the integer number of relatives r (0 < r < 500)and the street numbers (also integers) wherethey live (0 < si < 30000 ). Note that several relatives could live inthe same street number.

Output

For each test case your program must write the minimal sum ofdistances from the optimal Vito's house to each one of hisrelatives. The distance between two street numbers s i and s j is d ij= | s i- s j|.

Sample Input

2
2 2 4
3 2 4 6

Sample Output

2
4

题意:

给主人公找个安家的位置, 使得与所有邻居距离的和最小~

思路:

找中位数, 然后所有的距离减去中位数即可

水题~

AC代码:

#include<stdio.h>
#include<algorithm> using namespace std; int R[555]; int main() {
int T;
scanf("%d", &T);
while(T--) {
int r;
scanf("%d", &r);
for(int i = 0; i < r; i++)
scanf("%d", &R[i]);
sort(R, R+r);
int mid = R[r/2];
int sum = 0;
for(int i = 0; i < r; i++) {
if(R[i] > mid)
sum += R[i] - mid;
else
sum += mid - R[i];
}
printf("%d\n", sum);
}
return 0;
}

UVA 10041 (13.08.25)的更多相关文章

  1. UVA 10340 (13.08.25)

    Problem E All in All Input: standard input Output: standard output Time Limit: 2 seconds Memory Limi ...

  2. UVA 639 (13.08.25)

     Don't Get Rooked  In chess, the rook is a piece that can move any number of squaresvertically or ho ...

  3. UVA 10194 (13.08.05)

    :W Problem A: Football (aka Soccer)  The Problem Football the most popular sport in the world (ameri ...

  4. UVA 10499 (13.08.06)

    Problem H The Land of Justice Input: standard input Output: standard output Time Limit: 4 seconds In ...

  5. UVA 253 (13.08.06)

     Cube painting  We have a machine for painting cubes. It is supplied withthree different colors: blu ...

  6. UVA 156 (13.08.04)

     Ananagrams  Most crossword puzzle fans are used to anagrams--groupsof words with the same letters i ...

  7. UVA 573 (13.08.06)

     The Snail  A snail is at the bottom of a 6-foot well and wants to climb to the top.The snail can cl ...

  8. UVA 10025 (13.08.06)

     The ? 1 ? 2 ? ... ? n = k problem  Theproblem Given the following formula, one can set operators '+ ...

  9. UVA 465 (13.08.02)

     Overflow  Write a program that reads an expression consisting of twonon-negative integer and an ope ...

随机推荐

  1. IPTABLES 映射问题

    今天要做一个新的映射:将内网的一个8090口映射到外网的8087口. 在 /ETC/RC.LOCAL中最后插入: iptables -t nat -A PREROUTING -d outIP -p t ...

  2. poj 2533 Longest Ordered Subsequence(线性dp)

    题目链接:http://poj.org/problem?id=2533 思路分析:该问题为经典的最长递增子序列问题,使用动态规划就可以解决: 1)状态定义:假设序列为A[0, 1, .., n],则定 ...

  3. [置顶] Asp.Net底层原理(二、写自己的Asp.Net框架)

    我们介绍过了浏览器和服务器之间的交互过程,接下来介绍Asp.net处理动态请求. 写自己的Asp.Net框架,我们不会引用System.Web这个程序集,我们只需要创建要给自己的类库,所以在接下来的程 ...

  4. 获取图片中的文本--MODI

    http://www.aspsnippets.com/Articles/Read-Extract-Text-from-Image-OCR-in-ASPNet-using-C-and-VBNet.asp ...

  5. runtime的概念,message send如果寻找不到相应的对象,如何进行后续处理

    运行时刻是指一个程序在运行(或者在被执行)的状态.也就是说,当你打开一个程序使它在电脑上运行的时候,那个程序就是处于运行时刻.在一些编程语言中,把某些可以重用的程序或者实例打包或者重建成为“运行库”. ...

  6. BZOJ 1823: [JSOI2010]满汉全席( 2-sat )

    2-sat...假如一个评委喜好的2样中..其中一样没做, 那另一样就一定要做, 这样去建图..然后跑tarjan. 时间复杂度O((n+m)*K) ------------------------- ...

  7. BZOJ 100题留念

  8. oracle 集合变量以及自定义异常的用法

    oracle 集合变量以及自定义异常的用法, 在过程 record_practice 有record变量和自定义异常的用法实例.具体在3284行. CREATE OR REPLACE Package ...

  9. Sed简介 (转)

    Sed简介 sed 是一种在线编辑器,它一次处理一行内容.处理时,把当前处理的行存储在临时缓冲区中,称为“模式空间”(pattern space),接着用sed命令处理缓冲区中的内容,处理完成后,把缓 ...

  10. Visual Studio shortcut keys

    VS2010 快捷键   Ctrl+E,D ----格式化全部代码  Ctrl+E,F ----格式化选中的代码  CTRL + SHIFT + B生成解决方案  CTRL + F7 生成编译  CT ...