Iahub is a big fan of tourists. He wants to become a tourist himself, so he planned a trip. There are n destinations on a straight road that Iahub wants to visit. Iahub starts the excursion from kilometer 0. The n destinations are described by a non-negative integers sequence a1, a2, ..., an. The number ak represents that the kth destination is at distance ak kilometers from the starting point. No two destinations are located in the same place.
Iahub wants to visit each destination only once. Note that, crossing through a destination is not considered visiting, unless Iahub explicitly wants to visit it at that point. Also, after Iahub visits his last destination, he doesn't come back to kilometer 0, as he stops his trip at the last destination.
The distance between destination located at kilometer x and next destination, located at kilometer y, is |x - y| kilometers. We call a "route" an order of visiting the destinations. Iahub can visit destinations in any order he wants, as long as he visits all n destinations and he doesn't visit a destination more than once.
Iahub starts writing out on a paper all possible routes and for each of them, he notes the total distance he would walk. He's interested in the average number of kilometers he would walk by choosing a route. As he got bored of writing out all the routes, he asks you to help him.
Input
The first line contains integer n (2 ≤ n ≤ 105). Next line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ 107).
Output
Output two integers — the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.
Examples
Input
3
2 3 5
Output
22 3
Note
Consider 6 possible routes:
- [2, 3, 5]: total distance traveled: |2 – 0| + |3 – 2| + |5 – 3| = 5;
- [2, 5, 3]: |2 – 0| + |5 – 2| + |3 – 5| = 7;
- [3, 2, 5]: |3 – 0| + |2 – 3| + |5 – 2| = 7;
- [3, 5, 2]: |3 – 0| + |5 – 3| + |2 – 5| = 8;
- [5, 2, 3]: |5 – 0| + |2 – 5| + |3 – 2| = 9;
- [5, 3, 2]: |5 – 0| + |3 – 5| + |2 – 3| = 8.
The average travel distance is 
思路：
zxz大佬一眼就看出来了，是要求点的贡献，我想想好像真的是这样，可是不会啊。只好找了大佬的博客。
题目思路解法来自：大佬的博客 文章内容也是在大佬博客的基础上，加了一点自己的想法。
假设
所以对于任意一个排列
然后就是考虑每一个
由上面的公式我们可以看出某个元素的贡献度和他周围相邻点的贡献度是有关系的，比如
首先考虑的是第一个点对结果的贡献度，因此假设
接下来是考虑最后一个位置的贡献度（思路同第一步），假设
最后是考虑在中间n-2中任何一个位置的贡献，假设
求得了
即
总共的路径条数n!种（就是n个数随机排列形成一个序列的种类数），要求期望，所以要除以n!。
进而能够求得期望的公式为：
ac代码：
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<map>
#include<string.h>
#define ll long long
#define ull unsigned long long
#define mod 123456789
using namespace std;
ll a[101000];
int main()
{
ll n;
cin>>n;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
sort(a+1,a+n+1);//排序
ll ans=0;
for(int i=1;i<=n;i++)
{
ans+=(4*i-2*n-1)*a[i];
}
ll g=__gcd(ans,n);
cout<<ans/g<<" "<<n/g<<endl;
return 0;
}
以下是别的大佬用别的方法做的：
http://www.cnblogs.com/sineatos/p/3522288.html
https://blog.csdn.net/jeremy1149/article/details/56281920