Codeforces 313C Ilya and Matrix【思维】

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本文介绍了一个关于矩阵美的计算问题，探讨如何通过递归算法排列数值以获得矩阵的最大美观度，并提供了具体的实现思路与代码。

Ilya is a very good-natured lion. He likes maths. Of all mathematical objects, his favourite one is matrices. Now he's faced a complicated matrix problem he needs to solve.

He's got a square 2ⁿ × 2ⁿ-sized matrix and 4ⁿ integers. You need to arrange all these numbers in the matrix (put each number in a single individual cell) so that the beauty of the resulting matrix with numbers is maximum.

The beauty of a 2ⁿ × 2ⁿ-sized matrix is an integer, obtained by the following algorithm:

Find the maximum element in the matrix. Let's denote it as m.
If n = 0, then the beauty of the matrix equals m. Otherwise, a matrix can be split into 4 non-intersecting 2^n - 1 × 2^n - 1-sized submatrices, then the beauty of the matrix equals the sum of number m and other four beauties of the described submatrices.

As you can see, the algorithm is recursive.

Help Ilya, solve the problem and print the resulting maximum beauty of the matrix.

Input

The first line contains integer 4ⁿ (1 ≤ 4ⁿ ≤ 2·10⁶). The next line contains 4ⁿ integers a_i (1 ≤ a_i ≤ 10⁹) — the numbers you need to arrange in the 2ⁿ × 2ⁿ-sized matrix.

Output

On a single line print the maximum value of the beauty of the described matrix.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

1
13

Output

Input

4
1 2 3 4

Output

Note

Consider the second sample. You need to arrange the numbers in the matrix as follows:

1 2
3 4

Then the beauty of the matrix will equal: 4 + 1 + 2 + 3 + 4 = 14.

题目大意：

一个4^n一个矩形，让你将这些数都放在每一个格子中。

定义一个记矩阵的值是这个矩阵中最大价值+一分为四的四个小矩阵的价值。

很显然这是一个递归的过程。

问最大价值。

思路：

我们只要规划出每个数字将会作为矩阵中最大值的次数即可。

很显然，最大值会出现n+1次，然后接下来次大值会出现n次，次次大值也会出现n次，次次次大值也会出现n次，次次次次大值会出现n-1次..........................

依次类推我们可以得到一个规律，统计和即可。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<math.h>
using namespace std;
int a[3000600];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        reverse(a+1,a+1+n);
        __int64 output=0;
        int now=0;
        while(1)
        {
            int tmp=pow(4,now);
            now++;
            if(tmp>n)break;
            else
            {
                for(int i=1;i<=tmp;i++)
                {
                    output+=a[i];
                }
            }
        }
        printf("%I64d\n",output);
    }
}