From 2011 facebook is organizing a Facebook hacker cup every year around feb, visit Facebook hacker cup homepage for more details
This question is the third question(Carried 45 marks out of 100) for the Round 1 of 2013 hacker cup. only top 500 contestants will be moved to Round 2.
Problem Statement
John’s friend Peter purchases a new high resolution monitor with dimension W * H where W is the number of pixels in each row (i.e. width) and H is the number of pixels in each column (i.e. height).
However, there are N dead pixels on the monitor. The i-th dead pixel is located at (x[i], y[i]). (0, 0) is the top-left pixel and (W – 1, H – 1) is the bottom-right pixel. The locations of the dead pixels could be generated by 6 given integers X, Y, a, b, c and d by the following rules. If 2 pixels are at the same location, they are considered the same. It is possible that there are less than N distinct dead pixels.
- x[0] = X
- y[0] = Y
- x[i] = (x[i – 1] * a + y[i – 1] * b + 1) % W (for 0 < i < N)
- y[i] = (x[i – 1] * c + y[i – 1] * d + 1) % H (for 0 < i < N)
Peter connects his monitor to his computer and opens an image with dimension P (width) * Q (height). How many unique positions can the image be placed so that it can be displayed perfectly (i.e. all pixels of the picture are shown on the monitor)? The image cannot be rotated.
Input
The first line contains an integer T, which is the number of test cases. Then T test cases follow. Each test case contains 11 integers W, H, P, Q, N, X, Y,a, b, c, d.
Output
For each of the test cases numbered in order from 1 to T, output “Case #”, followed by the case number (with 1 being the first test case), followed by “: “, followed by an integer which is the number of different possible positions for the poster.
Constraints
- 1 ≤ T ≤ 20
- 1 ≤ W, H ≤ 40 000
- 1 ≤ P ≤ W
- 1 ≤ Q ≤ H
- 1 ≤ N ≤ min(1 000 000, W * H)
- 1 ≤ a, b, c, d ≤ 100
- 0 ≤ X < W
- 0 ≤ Y < H
Example Input
5 4 4 2 2 1 0 2 1 2 3 4 4 4 1 1 3 1 1 2 2 2 2 6 10 3 2 2 0 0 5 4 3 2 16 18 5 1 5 10 8 21 27 29 87 14 15 12 4 4 3 5 84 74 53 68
Example Output
Case #1: 7 Case #2: 15 Case #3: 32 Case #4: 197 Case #5: 16
Solution
Basically we have to see how many total rectangles of size PxQ size are possible such that there is no dead pixels inside those rectangles. We can first count the total number of rectangles and subtract the rectangles with dead pixels.
Total number of PxQ rectangles possible in WxH LED screen = (W-P+1)*(H-Q+1);
For any one dead pixel(x, y) all rectangles with top left position in the range for X =x-P+1 to x and Y = y-Q+1 to y are damaged rectangles.
This is the Java implementation of above explanation:-
import java.util.HashSet;
import java.util.Set;
public class DeadPixels {
public int getUniqueWays(int W, int H, int P, int Q, int N, int X, int Y, int a, int b, int c, int d){
int currX = X;
int currY = Y;
Set<String> damagedRectangles = new HashSet<String>();
addDamagedRectangles(currX, currY, W, H, P, Q, damagedRectangles);
for(int i = 1; i < N; i++){
int tempX = getNextX(currX, currY, a, b, W);
int tempY = getNextY(currX, currY, c, d, H);
currX = tempX;
currY = tempY;
addDamagedRectangles(currX, currY, W, H, P, Q, damagedRectangles);
}
System.out.println("Total damaged Rectangles = " + damagedRectangles.size());
System.out.println("Total possible rectangles = " + ((W-P+1)*(H-Q+1) - damagedRectangles.size()));
return ((W-P+1)*(H-Q+1) - damagedRectangles.size());
}
public void addDamagedRectangles(int x, int y, int W, int H, int P, int Q, Set<String> damagedRectangles){
for(int i = x; i > x - P ; i--){
for(int j = y; j > y- Q; j--){
if(j >= 0 && i >=0){
damagedRectangles.add(i +"," + j);
}
}
}
}
public int getNextX(int x, int y, int a, int b, int W){
return ((x*a + y*b + 1)%W);
}
public int getNextY(int x, int y, int c, int d, int H){
return ((x*c + y*d + 1)%H);
}
}
