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 Ruby (judge verdict: Accepted)

def upper_bound(arr, v)
  l = -1
  r = arr.size
  while l + 1 < r
    m = (l + r) / 2
    if arr[m] <= v then l = m
    else r = m
    end
  end
  r
end

EPS = 1e-9
n, m, l = gets.split.map(&:to_i)
p, t, v = n.times.map{ gets.split.map(&:to_i) }.transpose
@combination = (0..m).map{ [0] * (m + 1) }
(0..m).each{ |i| @combination[i][0] = @combination[i][i] = 1 }
(2..m).each do |i|
  (1..m).each do |j|
    @combination[i][j] = @combination[i-1][j] + @combination[i-1][j-1]
  end
end

time_set = n.times.map{ [0] * (m + 1) }
prob_set = n.times.map{ [0] * (m + 1) }
n.times do |i|
  s = v[i] == 0 ? Float::INFINITY : l / v[i].to_f
  (m + 1).times do |j|
    time_set[i][j] = s + t[i] * j
    prob_set[i][j] = (p[i] / 100.0) ** j * (1 - p[i] / 100.0) ** (m - j) * @combination[m][j]
  end
end

prob_sum = prob_set.map{ |ps| ps.reverse.each_with_object([0]){ |p, arr| arr << arr[-1] + p }[1..-1].reverse }

n.times do |i|
  ret = 0
  (m + 1).times do |j|
    prob = prob_set[i][j]
    n.times do |k|
      next if i == k
      idx = upper_bound(time_set[k], time_set[i][j] + EPS)
      if idx == m + 1
        prob = 0
        break
      end
      prob *= prob_sum[k][idx]
    end
    ret += prob
  end
  puts '%.8f' % ret
end