Codeforces Round #143 D: Magic Box

Hi All,

Problem Statement:
Given a parallelepiped with coordinates of diagonal as (0,0,0) and (x,y,z), another point (x,y,z) having an observer you need to find sum of all the values observer can see. Every face has a value written over it. 

For more details: http://codeforces.com/problemset/problem/231/D

My Approach: Observer can look at any face's value only and only if he is on opposite side of parallelepiped with respect to that face. For example: Observer can see value written on Z=0 plane only and only if coordinates of his value are (Z<0), otherwise he will always be on other side of the face and as parallelepiped is not transparent, he can't look at the number. 

Implementation:

	#include <iostream>
	#include <cstring>
	using namespace std;
	#define ll long long
	int main(){
	ll faceValues[6];
	ll observerCoordinate[3];
	ll endPoints[3];
	for(int i=0; i<3; i++) cin>>observerCoordinate[i];
	for(int i=0; i<3; i++) cin>>endPoints[i];
	for(int i=0; i<6; i++) cin>>faceValues[i];
	ll totalValue=0;
	if(observerCoordinate[1]<0) totalValue+=faceValues[0];
	if(observerCoordinate[1]>endPoints[1]) totalValue+=faceValues[1];
	if(observerCoordinate[2]<0) totalValue+=faceValues[2];
	if(observerCoordinate[2]>endPoints[2]) totalValue+=faceValues[3];
	if(observerCoordinate[0]<0) totalValue+=faceValues[4];
	if(observerCoordinate[0]>endPoints[0]) totalValue+=faceValues[5];
	cout << totalValue << endl;
	return 0;
	}

view raw Div2_143D.cpp hosted with ❤ by GitHub

Link: http://codeforces.com/contest/231/submission/41682545