POJ-2231 &AOJ-231 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.
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.
 

/************************************************************************************************************
题意：直线上有N头牛，求N头牛的距离之和 例如 1 3 5 则ans = (3-1)+(5-1)+|1-3|+(5-3)+|1-5|+|3-5|
思路：此题直接暴力在POJ上1000MS低空飘过 在AOJ上会TLE
暴力做法不多说 直接循环相加
比较叼的方法是求出每段出现的次数加起来 这我不太好说明 可以在草稿纸上写写
比如 1 3 5 7 9
先算吼声向右传的 首先 1吼一声 然后 1 3段记为d(1,3)出现了一次 d(3,5)出现了一次 d(5,7)出现了一次，以此类推
然后2吼一声 d(3,5)又出现了一次 此时为2次 以此类推
然后3吼一声 继续 推 能发现ans += (a[i] - a[i-1]) * i * (n-i)的规律

下面先附上暴力渣代码  最后是比较好的代码
************************************************************************************************************/
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
int a[10000+10];
int main()
{
int n;
scanf("%d", &n);
for(int i = 0 ; i < n ; i ++)
scanf("%d", &a[i]);
sort(a , a + n);
__int64 ans = 0;
for(int i = 0 ; i < n ; i ++)
{
for(int j = 0 ; j < n ; j ++)
ans += abs(a[j] - a[i]);
}
printf("%I64d\n", ans);
return 0;
}

#include <cstdio>
#include <algorithm>
using namespace std;
int a[10000+10];
int main()
{
int n;
scanf("%d", &n);
for(int i = 0 ; i < n ; i ++)
scanf("%d", &a[i]);
sort(a, a + n);
__int64 ans = 0;
for(int i = 1 ; i < n ; i ++)
ans += (__int64)(a[i] - a[i-1]) * i * (n - i);
printf("%I64d\n", ans * 2);
}