Moo Volume POJ - 2231

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.

 

 

求任意两个点之间的距离总和；

这题数据大肯定不能直接求，要找规律；

总共有(n-i-1)*(a[n-i-1]-a[i])算出了每两个点的距离 

这总共是一半的距离和 所以要乘以2

 

 

 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #include<cctype>
 using namespace std;
 long long a[];
 int main() {
     int n;
     while(scanf("%d",&n)!=EOF){
         for (int i= ;i<n ;i++)
             scanf("%lld",&a[i]);
         sort(a,a+n);
         long long sum=;
         for (int i= ;i<n ;i++ )
             sum+=(n--i)*(a[n--i]-a[i]);
         printf("%lld\n",sum*);
     }
     return ;
 }