贪心算法 A

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.
做题思路：这道题主要是计算N头奶牛分别对N-1头奶牛的距离差之和，这题感觉和贪心没有多大关系，用递推关系就可以解决，先用sort将坐标从小到大排序，再根据f[i]=f[i-1]-(n-i)*(a[i]-a[i-1])+(n-2)*(a[i]-a[i-1]),只需先求出f[1],就可以递推求出我们所需的各个奶牛距离之和，最终相加即为我们所求的答案。

#include
#include
#include
using namespace std;
long long a[10010],f[10010];   //定义是一定要定义long long
int n;
long long ans=0;

int main(){
	cin>>n;
	for (int i=1;i<=n;i++)	scanf("%d",&a[i]);

	sort(a+1,a+n+1);//数组a从1开始输入也要从一开始排序

	for (int i=2;i<=n;i++) f[1]+=a[i]-a[1];    
	ans=f[1];//将f[1]罗列出来进行下面的递推

	for (int i=2;i<=n;i++){
		f[i]= f[i-1] - ((n-i)*(a[i]-a[i-1])) + (i-2)*(a[i]-a[i-1]);
		ans+=f[i];
	}
	cout<