Problem:
Marathon Match

N people run a marathon.
There are M resting places on the way.
For each resting place, the i-th runner takes a break with probability P_i percent.
When the i-th runner takes a break, he gets rest for T_i time.

The i-th runner runs at constant speed V_i, and
the distance of the marathon is L.

You are requested to compute the probability for each runner to win the first place.
If a runner arrives at the goal with another person at the same time, they are not considered to win the first place.

Input

A dataset is given in the following format:

N M L

P_1 T_1 V_1

P_2 T_2 V_2

...

P_N T_N V_N

The first line of a dataset contains three integers N (1 \leq N \leq 100), M (0 \leq M \leq 50) and L (1 \leq L \leq 100,000).
N is the number of runners.
M is the number of resting places.
L is the distance of the marathon.

Each of the following N lines contains three integers P_i (0 \leq P_i \leq 100), T_i (0 \leq T_i \leq 100) and V_i (0 \leq V_i \leq 100) describing the i-th runner.
P_i is the probability to take a break.
T_i is the time of resting.
V_i is the speed.

Output

For each runner, you should answer the probability of winning.
The i-th line in the output should be the probability that the i-th runner wins the marathon.
Each number in the output should not contain an error greater than 10^{-5}.

Sample Input 1

2 2 50
30 50 1
30 50 2

Output for the Sample Input 1

0.28770000
0.71230000

Sample Input 2

2 1 100
100 100 10
0 100 1

Output for the Sample Input 2

0.00000000
1.00000000

Sample Input 3

3 1 100
50 1 1
50 1 1
50 1 1

Output for the Sample Input 3

0.12500000
0.12500000
0.12500000

Sample Input 4

2 2 50
30 0 1
30 50 2

Output for the Sample Input 4

0.51000000
0.49000000

Solution in JAVA (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	static Scanner sc = new Scanner(System.in);

	public static void main(String[] arg) {
		int N = sc.nextInt();
		int M = sc.nextInt();
		int L = sc.nextInt();
		double[] P = new double[N];
		int[] T = new int[N];
		int[] V = new int[N];
		for (int i = 0; i < N; ++i) {
			P[i] = sc.nextInt() * 0.01;
			T[i] = sc.nextInt();
			V[i] = sc.nextInt();
		}
		double[][] comb = new double[M + 1][M + 1];
		for (int i = 0; i <= M; ++i) {
			comb[i][0] = comb[i][i] = 1;
		}
		for (int i = 0; i <= M; ++i) {
			for (int j = 1; j < i; ++j) {
				comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
			}
		}
		double[][] time = new double[N][M + 1];
		double[][] prob = new double[N][M + 1];
		for (int i = 0; i < N; ++i) {
			for (int j = 0; j <= M; ++j) {
				time[i][j] = 1. * L / V[i] + j * T[i];
				prob[i][j] = comb[M][j] * Math.pow(P[i], j) * Math.pow(1 - P[i], M - j);
			}
		}
		double EPS = 1e-9;
		for (int i = 0; i < N; ++i) {
			double sum = 0;
			for (int j = 0; j <= M; ++j) {
				double t = time[i][j];
				double p = prob[i][j];
				for (int k = 0; k < N; ++k) {
					if (k == i) continue;
					double win = 1;
					for (int l = 0; l <= M; ++l) {
						if (time[k][l] - EPS > t) break;
						win -= prob[k][l];
					}
					p *= win;
				}
				sum += p;
			}
			System.out.println(sum);
		}
	}

}