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
- 汇农PC个人中心总结
1. 技巧总结 1. 使用 padding 编写灵活的 登录 | 注册, '|' 竖线, 参考: http://www.imooc.com/learn/710 font-size: 0; border ...
- 2014蓝桥杯问题 C: 神奇算式
没做完,先搞答题了 #include <stdio.h> #include<string.h> #include<stdlib.h> int comp(const ...
- Extjs控件属性
Ext.form.TimeField: 配置项: maxValue:列表中允许的最大时间 maxText:当时间大于最大值时的错误提示信息 ...
- CSS菜单横竖布局要点
菜单纵向:把ul 元素的边框属性去除,li元素用border-top 上边框显示分离,把a 元素用display:block text-decoration:none 去除默认下划线 菜单横向: ...
- Alyona and copybooks
题目连接 题意: 给 n,a,b,c四个数,n为已有的书的数目,问再买k本书所需花费最少是多少,(k+n)%4==0: 有三种套餐 第一种只有一本书,花费a 第二种有两本书,花费b, 第三种有三本书, ...
- 《Intel汇编第5版》 汇编调用子过程
一.Call和Ret指令 二.在子过程中需要自己保存可能会修改掉的寄存器值,这里可以使用USES伪指令来生成 三.一个数组求和的汇编例子 TITLE Call a Proc Demo INCLUDE ...
- Gentoo网络管理方法总结
OpenRC/netifrc Netifrc is a collection of modules created to configure and manage network interfaces ...
- wpf listview 行变色
<ListView x:Name="listView_Date" Grid.Row="3" BorderBrush="{x:Null}" ...
- S.O.L.I.D五大原则之单一职责SRP
转自 : 汤姆大叔的blog Bob大叔提出并发扬了S.O.L.I.D五大原则,用来更好地进行面向对象编程,五大原则分别是: The Single Responsibility Principle(单 ...
- keyboard添加down按钮
self.textView.inputAccessoryView = [self addToolbar]; - (UIToolbar *)addToolbar { UIToolbar *toolbar ...