# POJ2231总结

## Moo Volume

### 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

* 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.

```5
1
5
3
2
4```

`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.

### 我的代码

``````#include<stdio.h>
#include<algorithm>

using namespace std;

int locations;
long long sum;
long long ans;
int main()
{
int n;
scanf("%d",&n);
for (int i=0;i<n;i++){
scanf("%d",&locations[i]);
}
sort(locations,locations+n);
sum=locations;
ans=0;
//要开long long
for (long long int i=1;i<n;i++){
ans+=i*locations[i]-sum;
sum+=locations[i];
}
printf("%lld\n",ans*2);
return 0;
}``````