B

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

* 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

Hint

INPUT DETAILS:

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.

 

贴两种方法：

第一种：

就是找规律，这个特别难想。

先对a数组进行排序，然后求出相邻的差分

之后就根据n头牛，和第i头牛直接差分用的次数找到规律，直接计算。

 

第二种，比较简单

就是也要排序

然后第i头牛的音量就是第i-1头牛再加上（i-1-1)*d  (d是i和i-1的距离） 再减去 （n-i）*d

其实每头之间牛音量的不同就在于他们的距离，所以对他们的距离进行处理即可。

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;//要找规律！！！
ll a[11000];
int main()
{
    int n;
    cin>>n;
    for(int i=0;i<n;i++) scanf("%I64d",&a[i]);
    sort(a,a+n);
    ll sum=0;
    for(int i=1;i<n;i++)
    {
        sum+=(a[i]-a[i-1])*i*(n-i)*2;
    }
    cout<<sum<<endl;
    return 0;
}

　　

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
ll a[11000],b[11000];
int main()
{
    int n;
    cin>>n;
    for(int i=1;i<=n;i++) scanf("%I64d",&a[i]);
    sort(a+1,a+n+1);
    for(int i=2;i<=n;i++)
    {
        b[1]+=abs(a[i]-a[1]);
    }
    ll sum=b[1];
    for(int i=2;i<=n;i++)
    {
        ll d=a[i]-a[i-1];
        b[i]=b[i-1]+(i-1-1)*d-(n-i)*d;
        sum+=b[i];
    }
    cout<<sum<<endl;
    return 0;
}