[Usaco2005 Jan]Moo Volume 牛的呼声

[Usaco2005 Jan]Moo Volume 牛的呼声

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.

    约翰的邻居鲍勃控告约翰家的牛们太会叫．
    约翰的N(1≤N≤10000)只牛在一维的草场上的不同地点吃着草．她们都是些爱说闲话的奶牛，每一只同时与其他N-1只牛聊着天．一个对话的进行，需要两只牛都按照和她们间距离等大的音量吼叫，因此草场上存在着N（N-1）/2个声音．  请计算这些音量的和

Input

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

    第1行输入N，接下来输入N个整数，表示一只牛所在的位置．

Output

* Line 1: A single integer, the total volume of all the MOOs.

    一个整数，表示总音量．

Sample Input

5
1
5
3
2
4

INPUT DETAILS:

There are five cows at locations 1, 5, 3, 2, and 4.
Sample Output
40

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.

 

题目给出N个数，求每两个数之差的绝对值。

由于N(1≤N≤10000)，所以本题只能用O(N)的时间复杂度。

abs（num[a]-num[b]）=(num[a+1]-num[a])+（num[a+2]-num[a+1]）...+（num[b]-num[b-1]）。（a<b）

升序排列。

我们可以把abs（num[a]-num[b]）看做一小段一小段编号相邻的两个数距离（num[i+1]-num[i]）之和，那么abs（num[a]-num[b]）就是由这几段组成。我们只要得出每一段被加了几次，就可以得出答案了。

(num[i+1]-num[i])的左端点可以为第1,2...,(i-1)，k个数,共k个数;右端点可以为第(i+1)，（i+2）...,(n-1)，n个数,共(n-i)个数。所以跟据乘法原理，共有i（n-i）段，故ans=ans+i*(n-i)*(num[i+1]-num[i]）。

由于是双向的，所以abs（num[i+1]-num[i]）与abs(num[i]-num[i+1]）被看做是不同的，所以最后ans*2。

注意：因为N(1≤N≤10000)，且num[i]（0..1,000,000,000),所以ans需要开long long；因为i是int，所以要强行转long long。

 

#include<bits/stdc++.h>
using namespace std;
int n,num[10010];
long long ans=0;//注意ans要开long long 
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&num[i]);
	sort(num+1,num+1+n);//升序排列 
	for(int i=1;i<n;i++)
		ans=ans+(long long)(i)*(n-i)*(num[i+1]-num[i]);//强制转long long
	printf("%lld",ans*2);
	return 0;	
}