C. Do you want a date?
time limit per test
memory limit per test
input
output
n
1 to n. So the i-th hacked computer is located at the point xi. Moreover the coordinates of all computers are distinct.
Leha is determined to have a little rest after a hard week. Therefore he is going to invite his friend Noora to a restaurant. However the girl agrees to go on a date with the only one condition: Leha have to solve a simple task.
F(a) for all a, where a is a non-empty subset of the set, that consists of all hacked computers. Formally, let's denote A the set of all integers from 1 to n. Noora asks the hacker to find value of the expression
. Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally,
. Since the required sum can be quite large Noora asks to find it modulo 109.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
n (1 ≤ n ≤ 3·105)
n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi
Output
109.
Examples
input
24 7
output
3
input
34 3 1
output
9
将数组从小到大排序后，若集合中最小的数位d[i], 最大的数位d[j], 则这样的集合有2^(j-i-1)个
#include <bits/stdc++.h>
#define maxn 300005
#define MOD 1000000007
using namespace std;
typedef long long ll;
ll d[maxn];
int main() {
//freopen("in.txt", "r", stdin);
ll mm = (MOD + 1) / 2;
int n;
scanf("%d", &n);
for(int i = 0; i < n; i++)
scanf("%I64d", d+i);
sort(d, d+n);
ll k = 1, p = 0, f = 0;
for(int i = 0; i <= n - 2; i++) {
(p += k) %= MOD;
(f += d[i+1] * k % MOD) %= MOD;
(k <<= 1) %= MOD;
}
ll ans = 0;
for(int i = 0; i <= n-2; i++) {
ans += ((f - p * d[i]) % MOD + MOD) % MOD;
ans %= MOD;
p--;
(p += MOD) %= MOD;
(p *= mm) %= MOD;
f -= d[i+1];
f += MOD;
f %= MOD;
(f *= mm) %= MOD;
}
cout << ans << endl;
return 0;
}