# Cut Tiles

This is a bin packing problems. As we can easily find, in this problem every larger tiles can perfectly contain an integer number of smaller tiles, that hint we can apply greedy strategies to put as many as possible larger tiles first and then put smaller tiles in areas remain. In this case we can divide the free area in two rectangle area follow one edge of current used tiles and there is no arrangement that requires we cut smaller tiles across the boundaries of the rectangles we split. We use recursive ways to greedily cut as many as tiles from the largest material tile given and add a new material tile until we get all tiles we want.

import java.io.*;
import java.util.*;

public class CutTiles {
private static boolean cutTiles(int[] count, int i, long W, long H) {
long h = Math.max(W, H);
long w = Math.min(W, H);
if (w < 1) return false;
for (int j = i; j >= 0; j--) {
if (count[j] == 0) continue;
long subw = ((long)1) << j;
if (w < subw) continue;
int cost = (int)Math.min(count[j], h / subw);
count[j] -= cost;
cutTiles(count, j, h - cost * subw, subw);
cutTiles(count, j, w - subw, h);
return true;
}
return false;
}
private static void solve(int[] count, long M) {
int num = 0;
while(true) {
if (!cutTiles(count, 32, M, M)) break;
num++;
}
System.out.println(num);
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for (int t = 0; t < T; t++) {
int N = in.nextInt();
long M = in.nextLong();
int[] count = new int;
for (int i = 0; i < N; i++) {
count[in.nextInt()]++;
}
System.out.printf("Case #%d: ", t + 1);
solve(count, M);
}
}
}