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: Wrong Answer)

class Array
	def accum
		inject([0]){|a, x| a + [a[-1] + x]}
	end

	def accum!
		(1..size-1).each{|i|
			self[i] += self[i-1]
		}
		self.unshift(0)
	end

	def accum_2d
		ac = map{|r| r.accum}
		(1..size-1).each{|i|
			(1..ac[0].size-1).each{|j|
				ac[i][j] += ac[i-1][j]
			}
		}
		ac.unshift([0] * ac[0].size)
		ac
	end

	def accum_2d!
		each{|r| r.accum!}
		(1..size-1).each{|i|
			(1..self[0].size-1).each{|j|
				self[i][j] += self[i-1][j]
			}
		}
		unshift([0] * self[0].size)
	end
		
	def bsearch_first_index(l = 0, r = size - 1, &pred)
		return nil if l > r
		c = (l + r) / 2 
		if pred[self[c]]
			bsearch_first_index(l, c-1, &pred) || c
		else
			bsearch_first_index(c+1, r, &pred)
		end
	end

	def bsearch_first_item(l = 0, r = size - 1, &pred)
		i = bsearch_first_index(l, r, &pred)
		i ? self[i] : nil
	end

	def bsearch_last_index(l = 0, r = size - 1, &pred)
		return nil if l > r
		c = (l + r) / 2 
		if pred[self[c]]
			bsearch_last_index(c+1, r, &pred) || c
		else
			bsearch_last_index(l, c-1, &pred)
		end
	end

	def bsearch_last_item(l = 0, r = size - 1, &pred)
		i = bsearch_last_index(l, r, &pred)
		i ? self[i] : nil
	end
end

n, m, l = gets.split.map &:to_i

def comb(n, k)
	c = 1	
	(1..k).each do |i|
		c = c * (n - i + 1) / i
	end
	c
end

a = []
(0..n-1).each do |i|
	a[i] = []
	p, r, v = gets.split.map &:to_i		
	p /= 100.0
	a[i] = (0..m).map do |k|
		prob = p ** k * (1 - p) ** (m - k) * comb(m, k)
		[(v == 0 ? Float::INFINITY : l.to_f / v + r * k), prob]
	end
	
	tmp = a[i].map(&:last).reverse.accum[1..-1].reverse
	(0..m).each {|k| a[i][k] << tmp[k] }
end

(0..n-1).each do |i|
	total_prob = 0
	(0..m).map do |k|
		time, prob = a[i][k]
		(0..n-1).each do |j|
			next if j == i
			dist = a[j]
			item = dist.bsearch_first_item {|t, _, _| t > time}	
			prob *= (item ? item[-1] : 0)
		end
		total_prob += prob
	end
	p total_prob
end