Codeforces Round #138 (Div. 2) A. Parallelepiped

A. Parallelepiped

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You've got a rectangular parallelepiped with integer edge lengths. You know the areas of its three faces that have a common vertex. Your task is to find the sum of lengths of all 12 edges of this parallelepiped.

Input

The first and the single line contains three space-separated integers — the areas of the parallelepiped's faces. The area's values are positive ( > 0) and do not exceed 10⁴. It is guaranteed that there exists at least one parallelepiped that satisfies the problem statement.

Output

Print a single number — the sum of all edges of the parallelepiped.

Sample test(s)

Input

1 1 1

Output

12

Input

4 6 6

Output

28

Note

In the first sample the parallelepiped has sizes 1 × 1 × 1, in the second one — 2 × 2 × 3.

     题目和算法分析：本题给出 一个长方体的 有“共同顶点”的3个面的面积 ，请你计算出 该长方体的12条棱的长度之和。

                           假设：3个面的面积分别是：x, y, z;  长宽高分别是：a, b, c;

                                   --->(1) a*b=x;   (2) a*c=y;  (3) b*c=z; 

                                   --->先让(1)与(2) 式 相除， --->  b/c = x/y ;  将此式与(3)式 联立 ：---> c^2=(z*y)/x;

                                    同理可得： a^2=(x*y)/z;     b^2=(x*z)/y;

                                    然后用数学函数sqrt() 开方一下，计算出a, b, c, 再将它们加起来的和乘以4 输出就可以了！

     

#include <iostream>
#include <string>
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>

using namespace std;


int main()
{
    int x, y, z;
    int a, b, c;

    while(scanf("%d %d %d", &x, &y, &z)!=EOF)
    {
        a=sqrt((y*x)/z);
        b=sqrt( (x*z)/y );
        c=sqrt( (z*y)/x);
        int n;
        n=(a+b+c)*4;
        printf("%d
", n);
    }
    return 0;
}