CodeForces 810C - Do you want a date?(数学+排序)

最新推荐文章于 2017-10-09 18:09:34 发布

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本文介绍了一个计算机科学领域的算法问题，黑客Leha需要计算其控制的所有计算机之间的距离的最大值之和，以完成朋友Noora的挑战。文章详细解释了如何通过排序和数学公式优化计算过程。

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 $\text{[math]}$ . Here F(a) is calculated as the maximum among the distances between all pairs of computers from the set a. Formally, $\text{[math]}$ . 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

output
3

input
3
4 3 1

output
9

Note

There are three non-empty subsets in the first sample test: $\text{[math]}$ , $\text{[math]}$ and $\text{[math]}$ . 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: $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ , $\text{[math]}$ . In total the sum is (4 - 3) + (4 - 1) + (3 - 1) + (4 - 1) = 9.

对于已经从小到大排完序的aj-ai一共出现多少次，就是用集合的子集合的数量就可以算出来是2^(j-i+1)个。

但是这样还需要再求组合数来列出所有aj-ai的情况无疑是复杂的。

那么还有一种思路降低时间复杂度，就是求每一个ai作为最大值和最小值的次数。

分析：当ai作为最大值时,会出现2^(i-1)次，作为最小值时，会出现2^(n-i)次，只要将这些都加起来取模就行了。注意要用long long来存不然会wa。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define MAX_N 300500
#define MOD 1000000007
using namespace std;
long long mul[MAX_N],a[MAX_N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        long long sum=0;
        mul[0]=1;
        for(int i=1;i<=n;i++)
            cin>>a[i];
        sort(a+1,a+1+n);
        for(int i=1;i<=n;i++)
            mul[i]=mul[i-1]*2%MOD;
        for(int i=1;i<=n;i++)
            sum=(sum+(mul[i-1]-mul[n-i])*a[i]%MOD)%MOD;
        cout<<sum<<endl;
    }
    return 0;
}