F - Moo Volume

最新推荐文章于 2024-04-05 14:43:08 发布

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最新推荐文章于 2024-04-05 14:43:08 发布

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解决农场中牛群因位置不同而产生的哞叫噪音问题，通过计算每头牛与其他牛之间的距离总和来评估整体噪音水平。

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

Sample Output

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.

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

Sample Output

Hint

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

Sample Output

Hint

#include<iostream>
#include<cstring>
#include<string>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<iomanip>
using namespace std;
int main()
{
    /*int n,a[10001],i,j;
    long long sum;
    while(scanf("%d",&n)!=EOF)
    {
       for(i=1;i<=n;i++)
        //cin>>a[i];
        scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        sum=0;
        if(n%2==0)
        {
            for(i=1;i<=n/2;i++)
            {
                for(j=1;j<=n;j++)
                    sum+=fabs(a[j]-a[i]);
            }
            //cout<<sum*2<<endl;
            printf("%lld\n",sum*2);
        }
        else
        {
            for(i=1;i<=n/2;i++)
            {
                for(j=1;j<=n;j++)
                        sum+=fabs(a[j]-a[i]);
            }
            sum=sum*2;
            for(i=1;i<=n;i++)
            sum+=fabs(a[i]-a[n/2+1]);
            //cout<<sum<<endl;
            printf("%lld\n",sum);
        }
    }
}*/
    int n,a[10001],i,j;
    long long sum;
    while(cin>>n)
    {
        for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
        sum=0;
        sort(a+1,a+n+1);
        for(i=n;i>=1;i--)
        {
            for(j=i-1;j>=1;j--)
                sum+=a[i]-a[j];
        }
        cout<<sum*2<<endl;
    }
}