C. Do you want a date?
time limit per test：2 seconds
memory limit per test：256 megabytes
input：standard input
output：standard output
Leha decided to move to a quiet town Vičkopolis, because he was tired by living in Bankopolis. Upon arrival he immediately began to expand his network of hacked computers. During the week Leha managed to get access to n computers throughout the town. Incidentally all the computers, which were hacked by Leha, lie on the same straight line, due to the reason that there is the only one straight street in Vičkopolis.
Let's denote the coordinate system on this street. Besides let's number all the hacked computers with integers from 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.
Leha should calculate a sum of 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 + 7.
Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.
Input
The first line contains one integer n (1 ≤ n ≤ 3·105) denoting the number of hacked computers.
The second line contains n integers x1, x2, ..., xn (1 ≤ xi ≤ 109) denoting the coordinates of hacked computers. It is guaranteed that all xi are distinct.
Output
Print a single integer — the required sum modulo 109 + 7.
Examples
Input
2
4 7 複制
Output
3 複制
Input
3
4 3 1 複制
Output
9 複制
Note
There are three non-empty subsets in the first sample test:
{4}，{7} and {4,7} . The first and the second subset increase the sum by 0 and the third subset increases the sum by 7 - 4 = 3. In total the answer is 0 + 0 + 3 = 3.
There are seven non-empty subsets in the second sample test. Among them only the following subsets increase the answer:
{4,3}，{4,1}，{3,1}，{4,3,1}
In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.
題目連結：http://codeforces.com/contest/810/problem/C
分析：還是因為我太菜了，想了半天都沒想出來怎麼去寫，上網查題解半天沒有搞懂解題思路，過了一個小時靈光一現，答案出來了！
題意：給定集合，讓我們求出所有子集中最大值最小值差的和
思路：很容易我們會想到直接排序，然後暴力枚舉a[i]和a[j],那麼他們之間的包含i位和j位的子集個數一定是2^(j-i-1)個，ans = 2^(j-i-1)*(a[j]-a[i])
但是一定會逾時的啊，複雜度為O(n^2),是以我們考慮
長度為1結果是ans1 = a[1]-a[0]+a[2]-a[1] ....a[n-1]-a[n-2] = (a[1]+a[2]+...+a[n-1] )- (a[0]+a[1]+....+a[n-2])
長度為2是結果是ans2 = a[2]-a[0]+a[3]-a[1]+.....+a[n-1]-a[n-1-2]
這樣我們可以清楚的發現ans = ans1+ans2+ans(n-1) = sum[n-1]-sum[i]-sum[n-1-i-1]，這裡sum[i]代表前i個的和
下面給出AC代碼：
1 #include<bits/stdc++.h>
2 using namespace std;
3 #define MAX 300000
4 #define mod 1000000007
5 long long a[MAX+5],b[MAX+5],ans=0;
6 int main()
7 {
8 int n,i;
9 scanf("%d",&n);
10 b[0]=1;
11 for(i=1;i<=n;i++)
12 {
13 scanf("%d",&a[i]);
14 b[i]=(2*b[i-1])%mod;
15 }
16 sort(a+1,a+n+1);
17 for(i=1;i<=n;i++)
18 {
19 ans+=(b[i-1]-b[n-i])*a[i]%mod;
20 ans%=mod;
21 }
22 cout<<ans<<endl;
23 } 複制