Лимит времени 2000/4000/4000/4000 мс. Лимит памяти 65000/65000/65000/65000 Кб.  bfs. 
  
 
Plane Segmentation
Question:  
The plane equation in three dimensional space can be defined as: 
ax + by +cz + d=0 
Where a, b and c are plane parameters and d is distance of plane from the origin. 
A point lies on plane if  ax + by +cz + d=0 
Write a program which counts how many of this points lie on the plane. 
Input Specifications: In the first line you will be given 4 coefficients of the plane (a, b, c, d) than in the following three lines x, y, z coordinates of four points. where -20000<=a,b,c<=20000, -18000000<=d<=18000000, and -100<=x,y,z<=100 . 
Output Specifications: Show how many points lie on the plane. 
  
Sample Input I  
 -4 2 3 -3 2 2 1 2 0 3 3 2 5   | 
Sample Input II 
 2 2 2 -18 1 2 6 5 1 5 3 2 4   | 
 
Sample Output I 
 0  | 
Sample Output II 
 2  | 
 
 
  
Sample Input II Explanation: 
Based on plane equation we check if the points lie on plane or not. 
Check Point 1 
equation= ax + by +cz + d=1*2 + 2*2 + 2*6 + (-18)=2+4+12-18=0 
Check Point 2 
equation= ax + by +cz + d=2*5 + 2*1 + 2*5 + (-18)=10+2+5-18=4 
Check Point 2 
equation= ax + by +cz + d=2*3 + 2*2 + 2*4 + (-18)=6+4+8-18=0 
Total number of points lying on plane is 2 
 
  
  Для отправки решений необходимо выполнить вход.
  
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