Leha decided to move to a quiet town Vi?kopolis, 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 Vi?kopolis.

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

Though, Leha is too tired. Consequently he is not able to solve this task. Help the hacker to attend a date.

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 allxi are distinct.

Print a single integer — the required sum modulo 109 + 7.

2
4 7

3

3
4 3 1

9

There are three non-empty subsets in the first sample test:, and . The first and the second subset increase the sum by 0and 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: , , , . In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.

題目大意：

　　　　　輸入一集合S，定義F（a）函數（a為S的非空子集），F（a）=a集合中最大值與最小值的差

　　　　　計算所有F(a)的和。

解題思路：

　　　　　因為要求最大值與最小值的差，所以首先想到的是把S集合排序 （a[j]-a[i]）*2^(j-i)

　　　　　對於只有一個元素的子集可以不用考慮

　　　　　對於上面的式子

　　　　　最小值一共要被減去2^n-2^0次 也可以理解為加上2^0-2^n次

　　　　　最大值一共被加上2^n-2^0次 第i個元素被加了2^i-2^(n-i-1)

AC代碼：

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <math.h>
 4 #include <stdlib.h>
 5 #include <iostream>
 6 #include <algorithm>
 7 const int mod = 1e9+7;
 8 int a[300010];
 9 __int64 b[300010];
10 
11 using namespace std;
12 
13 int main ()
14 {
15     int n,i,j;
16     while (~scanf("%d",&n))
17     {
18         b[1] = 1;
19         for (i = 1; i <= n; i ++)
20         {
21             cin>>a[i];
22             b[i+1] = b[i]*2%mod;   // 2^i  打表
23         }
24         sort(a+1,a+n+1);
25         
26         __int64 sum = 0;
27         for (i = 1; i <= n; i ++)
28         {
29             sum += a[i]*(b[i]-b[n-i+1])%mod; // 直接代入公式
30             sum %= mod;    
31         }
32         cout<<sum<<endl;
33     }
34     return 0;
35 }
36             scanf("%d%d%d",&l,&r,&x);
37             for(i = l; i <= r; i ++)
38                 if (p[i] < p[x])
39                     sum ++;
40             if (x-l == sum)
41                 printf("Yes\n");
42             else
43                 printf("No\n");
44         }
45     }
46     return 0;
47 }

Codeforces Round #415 (Div. 2) C. Do you want a date?