# New Years Eye

This is a interesting problem. Instead of calculate the the amount of wite a bottle holds at last, we calculate the total amount of wine that flows over a bottle and distribute the flow of current bottle of this level to next level. At last, if a bottle has an amount of flow that is greater or equal to 250.0ml, the amount it holds at last is 250.0ml. Or the amount left is the amount of flow itself.

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

public class NewYearsEye {
private static double solve(int B, int L, int N) {
double[][] dp = new double[L][L];
double[][] last = new double[L][L];
last[0][0] = 750.0 * B;
for (int l = 1; l < L; l++) {
for (int i = 0; i < l; i++) {
for (int j = 0; j <= i; j++) {
if (last[i][j] > 250.0) {
double share = (last[i][j] - 250.0) / 3.0;
dp[i][j] += share;
dp[i + 1][j] += share;
dp[i + 1][j + 1] += share;
}
}
}

double[][] tmp = last;
last = dp;
dp = tmp;

for (int i = 0; i <= l; i++) {
for (int j = 0; j <= i; j++) {
dp[i][j] = 0.0;
}
}
}

int id = 0;
for (int i = 0; i < L; i++) {
for (int j = 0; j <= i; j++) {
if ((++id) == N) {
return Math.min(250.0, last[i][j]);
}
}
}
return 0.0;
}
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int T = in.nextInt();
for (int t = 0; t < T; t++) {
int B = in.nextInt(), L = in.nextInt(), N = in.nextInt();
System.out.printf("Case #%d: %.7f\n", t + 1, solve(B, L, N));
}
}
}