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 sequencea1, 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.

The first line contains integer n (2 ≤ n ≤ 105). Next line contains n distinct integers a1, a2, ..., an (1 ≤ ai ≤ 107).

Output two integers — the numerator and denominator of a fraction which is equal to the wanted average number. The fraction must be irreducible.

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  =  = .

题意：求出所有走的方案的平均花费。。

思路：先找出规律。我们发现，num中每两个组合都会出现(n - 1)!次，而分母是n!次，所以抵消，分母为n，分子为求出每个数字和其他数字（以及0）的差的绝对值之和即可，直接暴力枚举为O(n^2)会超时。所以这样：先把num从小到大排序，然后算出总和，然后用一个now来记录第i个前i个和，然后总和为num[i] * i - now (i的前半部分) + sum - now -num[i] * (n - i) - num[i](后半部分)，这样一来时间复杂度只有O(n)。

要注意的是要用longlong。由于当时编译器只有渣渣vc6.0所以就用__int64了。

代码：

#include <stdio.h>

#include <string.h>

#include <algorithm>

using namespace std;

int main() {

	__int64 sum = 0, now = 0, ans = 0, num[100005], n, i;

	num[0] = 0;

	scanf("%I64d", &n);

	for (i = 1; i <= n; i ++) {

		scanf("%I64d", &num[i]);

		sum += num[i];

	}

	sort(num, num + 1 + n);

	for (i = 1; i <= n; i ++) {

		ans += 2 * num[i] * i - 2 * now + sum - num[i] * (n + 1);

		now += num[i];

	}

	__int64 a, b;

	a = ans; b = n;

	if (a < b) {

		__int64 t = b;

		b = a;

		a = t;

	}

	while (b) {

		__int64 sb = b;

		b = a % b;

		a = sb;

	}

	printf("%I64d %d\n", ans / a, n / a);

	return 0;

}