Codeforces Round #143 (Div. 2) D. Magic Box

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D. Magic Box

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

One day Vasya was going home when he saw a box lying on the road. The box can be represented as a rectangular parallelepiped. Vasya needed no time to realize that the box is special, as all its edges are parallel to the coordinate axes, one of its vertices is at point (0, 0, 0), and the opposite one is at point (x1, y1, z1). The six faces of the box contain some numbers a1, a2, ..., a6, exactly one number right in the center of each face.

The numbers are located on the box like that:

• number a1 is written on the face that lies on the ZOX plane;
• a2 is written on the face, parallel to the plane from the previous point;
• a3 is written on the face that lies on the XOY plane;
• a4 is written on the face, parallel to the plane from the previous point;
• a5 is written on the face that lies on the YOZ plane;
• a6 is written on the face, parallel to the plane from the previous point.

At the moment Vasya is looking at the box from point (x, y, z). Find the sum of numbers that Vasya sees. Note that all faces of the box are not transparent and Vasya can't see the numbers through the box. The picture contains transparent faces just to make it easier to perceive. You can consider that if Vasya is looking from point, lying on the plane of some face, than he can not see the number that is written on this face. It is enough to see the center of a face to see the corresponding number for Vasya. Also note that Vasya always reads correctly the ai numbers that he sees, independently of their rotation, angle and other factors (that is, for example, if Vasya sees some ai = 6, then he can't mistake this number for 9 and so on).

Input

The fist input line contains three space-separated integers x, y and z (|x|, |y|, |z| ≤ 106) — the coordinates of Vasya's position in space. The second line contains three space-separated integers x1, y1, z1 (1 ≤ x1, y1, z1 ≤ 106) — the coordinates of the box's vertex that is opposite to the vertex at point (0, 0, 0). The third line contains six space-separated integers a1, a2, ..., a6 (1 ≤ ai ≤ 106) — the numbers that are written on the box faces.

It is guaranteed that point (x, y, z) is located strictly outside the box.

Output

Print a single integer — the sum of all numbers on the box faces that Vasya sees.

Sample test(s)

input

2 2 2
1 1 1
1 2 3 4 5 6

output

12

input

0 0 10
3 2 3
1 2 3 4 5 6

output

4

Note

The first sample corresponds to perspective, depicted on the picture. Vasya sees numbers a2 (on the top face that is the darkest), a6(on the right face that is the lightest) and a4 (on the left visible face).

In the second sample Vasya can only see number a4.

题意：有一个人站在（x,y,z）点上，而他的前方有一个（0,0,0）点的正方体，（0,0,0）的对立点的坐标为（a1,y1,z1）.正方体的每一个面的正中央有一个值，如上图所示：zox面是：a1,其对立面是：a2; xoy面是：a3,其对立面是：a4;  yoz面是：a5其对立面是：a6;求你能看的的面的值的和。如果你的x坐标小于0的话，就能看到yoz面的a5，以此类推。

最近在学几何，本来打算大显身手的，结果发现是一个赤裸裸的水题。。。

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
  int x,y,z,x1,y1,z1;
  int a1,a2,a3,a4,a5,a6;
 scanf("%d%d%d%d%d%d",&x,&y,&z,&x1,&y1,&z1);
 cin>>a1>>a2>>a3>>a4>>a5>>a6;
int ans=0;
if(y<0)  ans+=a1;
if(y>y1) ans+=a2;
if(z<0)  ans+=a3;
if(z>z1) ans+=a4;
if(x<0)  ans+=a5;
if(x>x1) ans+=a6;
cout<<ans<<endl;
return 0;
}