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大佬一眼就看出來了，是要求點的貢獻，我想想好像真的是這樣，可是不會啊。只好找了大佬的博客。
題目思路解法來自：大佬的博客 文章內容也是在大佬博客的基礎上，加了一點自己的想法。
假設
所以對於任意一個排列
然後就是考慮每一個
由上面的公式我們可以看出某個元素的貢獻度和他周圍相鄰點的貢獻度是有關係的，比如
首先考慮的是第一個點對結果的貢獻度，因此假設
接下來是考慮最後一個位置的貢獻度（思路同第一步），假設
最後是考慮在中間n-2中任何一個位置的貢獻，假設
求得了
即
總共的路徑條數n!種（就是n個數隨機排列形成一個序列的種類數），要求期望，所以要除以n!。
進而能夠求得期望的公式爲：
ac代碼：
#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;
}
以下是別的大佬用別的方法做的：
http://www.cnblogs.com/sineatos/p/3522288.html
https://blog.csdn.net/jeremy1149/article/details/56281920