Moo Volume

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other)

Total Submission(s) : 6   Accepted Submission(s) : 2

Problem Description

Farmer John has received a noise complaint from his neighbor, Farmer Bob, stating that his cows are making too much noise. 

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

* Line 1: N <br> <br>* Lines 2..N+1: The location of each cow (in the range 0..1,000,000,000).

 

Output

There are five cows at locations 1, 5, 3, 2, and 4.

 

Sample Input

5 1 5 3 2 4

 

Sample Output

40

 

Source

PKU

 题目大意：

数轴上共有N头牛，每头牛有一个坐标。每头牛都要向其他所有牛发出一个叫声，因此一共有 N*(N-1) 个叫声。

每个叫声的大小等于两头牛之间坐标差的绝对值。输出所有叫声大小的和。

思路：

只计算单向距离，最后结果乘以2即可

如：1 2 3 4 5   （这里代表的是点的序号）

以 2-3 这一段为例， 1 （ 2 - 3 ） 4 5

计算2-3这一段在总的距离中出现的次数，因为只有跨越2-3这一段的长度才包含这一段，从1到3、4、5，2到3、4、5都包含这一段，可以得出这一段出现的次数为该段左侧的点数乘右侧的点数，以上计算的只是从左侧端点到右侧端点的次数，由于是双向的，反过来的结果是等效的，所以最后还要乘2

设当前一段的端点为a[i]和a[i+1]，则这一段出现的次数表达式为：i * ( n - i ) * 2

题目中要求的是距离的总和，把每一段出现的次数乘该段的长度再求和即可

sum = sum+（ ( a[i+1] - a[i] ) * i * ( n - i )）   （ i = 1, 2, ..., n )

#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
long long  a[10002];

int main(){
	int  n;
	while(scanf("%d",&n)!=EOF){
	
	memset(a,0,sizeof(a));

	
	for(int i=0;i<n;i++)
	{
		scanf("%lld",&a[i]);	
	}
	sort(a,a+n);
		
	long long sum=0;

	for(int i=1;i<n;i++)
	{		
		sum+=(a[i]-a[i-1])*i*(n-i);
	}
	
	printf("%lld\n",sum*2);
				
	}

}