C. Tourist Problem time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output 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. 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. Sample test(s) 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 = = . 題意：求出所有走的方案的平均花費。。 思路：先找出規律。我們發現，num中每兩個組合都會出現(n - 1)!次，而分母是n!次，所以抵消，分母為n，分子為求出每個數字和其他數字（以及0）的差的絕對值之和即可，直接暴力枚舉為O(n^2)會超時。所以這樣：先把num從小到大排序，然後算出總和，然後用一個now來記錄第i個前i個和，然後總和為num[i] * i - now (i的前半部分) + sum - now -num[i] * (n - i) - num[i](後半部分)，這樣一來時間復雜度只有O(n)。 要注意的是要用longlong。由於當時編譯器只有渣渣vc6.0所以就用__int64了。 代碼：
#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;
}
