POJ 2231 Moo Volume
Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu
Description
FJ's N cows (1 <= N <= 10,000) all graze at various locations on a long one-dimensional pasture. The cows are very chatty animals. Every pair of cows simultaneously carries on a conversation (so every cow is simultaneously MOOing at all of the N-1 other cows). When cow i MOOs at cow j, the volume of this MOO must be equal to the distance between i and j, in order for j to be able to hear the MOO at all. Please help FJ compute the total volume of sound being generated by all N*(N-1) simultaneous MOOing sessions.
Input
* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).
Output
Sample Input
5
1
5
3
2
4
Sample Output
40
Hint
There are five cows at locations 1, 5, 3, 2, and 4.
OUTPUT DETAILS:
Cow at 1 contributes 1+2+3+4=10, cow at 5 contributes 4+3+2+1=10, cow at 3 contributes 2+1+1+2=6, cow at 2 contributes 1+1+2+3=7, and cow at 4 contributes 3+2+1+1=7. The total volume is (10+10+6+7+7) = 40.
#include<cstdio>
#include<iostream>
#include<cstring>
#include<stdlib.h>
#include<algorithm>
#include<cmath>
using namespace std;
#define N 10005
int c[N];
int main()
{
int n,i,j;
long long ans;
while(scanf("%d",&n)!=EOF)
{
ans=;
for(i=;i<n;i++)
{
scanf("%d",&c[i]);
}
sort(c,c+n);
for(i=;i<n-;i++)
{
for(j=i+;j<n;j++)
{
ans+=*abs(c[i]-c[j]);
}
}
printf("%lld\n",ans);}
return ;
}
POJ 2231 Moo Volume的更多相关文章
- Poj 2232 Moo Volume(排序)
题目链接:http://poj.org/problem?id=2231 思路分析:先排序,再推导计算公式. 代码如下: #include <iostream> #include <a ...
- BZOJ1679: [Usaco2005 Jan]Moo Volume 牛的呼声
1679: [Usaco2005 Jan]Moo Volume 牛的呼声 Time Limit: 1 Sec Memory Limit: 64 MBSubmit: 723 Solved: 346[ ...
- POJ 2010 Moo University - Financial Aid( 优先队列+二分查找)
POJ 2010 Moo University - Financial Aid 题目大意,从C头申请读书的牛中选出N头,这N头牛的需要的额外学费之和不能超过F,并且要使得这N头牛的中位数最大.若不存在 ...
- BZOJ 1679: [Usaco2005 Jan]Moo Volume 牛的呼声( )
一开始直接 O( n² ) 暴力..结果就 A 了... USACO 数据是有多弱 = = 先sort , 然后自己再YY一下就能想出来...具体看code --------------------- ...
- Poj2231 Moo Volume 2017-03-11 22:58 30人阅读 评论(0) 收藏
Moo Volume Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22104 Accepted: 6692 Descr ...
- Moo Volume POJ - 2231
Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are ...
- poj -2010 Moo University - Financial Aid (优先队列)
http://poj.org/problem?id=2010 "Moo U"大学有一种非常严格的入学考试(CSAT) ,每头小牛都会有一个得分.然而,"Moo U&quo ...
- B - Moo Volume
Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are ...
- 【BZOJ】1679: [Usaco2005 Jan]Moo Volume 牛的呼声(数学)
http://www.lydsy.com/JudgeOnline/problem.php?id=1679 水题没啥好说的..自己用笔画画就懂了 将点排序,然后每一次的点到后边点的声音距离和==(n-i ...
随机推荐
- java.lang.IllegalAccessError: tried to access method com.google.common.base.Stopwatch.<init>()V from 解决
在用spark的yarn-cluster模式跑fpgrowth进行频繁项集挖掘的时候,报如下错误: ERROR yarn.ApplicationMaster: User class threw exc ...
- SQL注入(三)
邮给我一个密码 我们意识到虽然不能添加一条新的记录在members表中,但我们可以通过修改一个存在的记录, 这也获得了我们的证明是可行的. 从先前的步骤中,我们知道bob@example.com在系统 ...
- 【CSS学习笔记】CSS选择器
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xht ...
- iOS Quartz2D画图
对于刚接触Quartz2D的同学来说,先了解 上下文 的概念,再从最基础的画线来具体体验Quartz2D的画图步骤 介绍Quart2D :是苹果官方的二维(平面)绘图引擎,同时支持iOS和macOS系 ...
- mysql5.7.17安装问题
在根目录新建data文件夹和my.ini,把ini复制到bin目录下才可以
- 2.开启TFTP,NFS,SAMBA,SSH服务
一.TFTP (1)dpkg -s tftp-hpa查看服务器端是否安装 (2)如果没安装 sudo apt-get install tftpd-hpa sudo apt-get install tf ...
- XTU 1249 Rolling Variance
$2016$长城信息杯中国大学生程序设计竞赛中南邀请赛$G$题 前缀和. 把公式化开来,会发现只要求一段区间的和以及一段区间的平方和就可以了. #pragma comment(linker, &quo ...
- 第一百二十二节,JavaScript表单处理
JavaScript表单处理 学习要点: 1.表单介绍 2.文本框脚本 3.选择框脚本 为了分担服务器处理表单的压力,JavaScript提供了一些解决方案,从而大大打破了处处依赖服务器的局面. 一. ...
- Python学习笔记——进阶篇【第八周】———进程、线程、协程篇(Socket编程进阶&多线程、多进程)
本节内容: 异常处理 Socket语法及相关 SocketServer实现多并发 进程.线程介绍 threading实例 线程锁.GIL.Event.信号量 生产者消费者模型 红绿灯.吃包子实例 mu ...
- Java多线程的信号量
Java的信号量主要的作用是控制线程对资源的访问例如我一个线程池里面有100个线程等待执行,但是我允许最多同时运行5个线程,这5个线程只有其中一个线程执行完毕后,在线程池中等待的线程才能进入开始执行, ...