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 C++14 (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using pii = pair<int, int>;

ll comb[51][51];
int N, M, L;
int P[110], T[110], V[110];
double pos[110][51], cpos[110][51];
double t[110][51];

void init() {
  for (int i = 0; i <= 50; i++) {
    comb[i][0] = comb[i][i] = 1;
    for (int j = 1; j < i; j++) {
      comb[i][j] = comb[i-1][j-1] + comb[i-1][j];
    }
  }
}

int main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(10);
  
  init();
  cin >> N >> M >> L;
  for (int i = 0; i < N; i++) {
    cin >> P[i] >> T[i] >> V[i];
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j <= M; j++) {
      double tmp = 1.0;
      for (int k = 0; k < j; k++) {
        tmp *= P[i] / 100.0;
      }
      for (int k = 0; k < M - j; k++) {
        tmp *= (100 - P[i]) / 100.0;
      }
      pos[i][j] = comb[M][j] * tmp; 
    }
    cpos[i][M] = pos[i][M];
    for (int j = M-1; j >= 0; j--) {
      cpos[i][j] = pos[i][j] + cpos[i][j+1];
    }
  }
  for (int i = 0; i < N; i++) {
    for (int j = 0; j <= M; j++) {
      t[i][j] = (double)L / V[i] + j * T[i];
    }
  }

  for (int i = 0; i < N; i++) {
    double sum = 0;
    for (int j = 0; j <= M; j++) {
      double tmp = pos[i][j];
      for (int k = 0; k < N; k++) {
        if (k == i) continue;
        bool cant = true;
        for (int l = 0; l <= M; l++) {
          if (t[i][j] < t[k][l]) {
            tmp *= cpos[k][l];
            cant = false;
            break;
          }
        }
        if (cant) tmp = 0;
      }
      sum += tmp;
    }
    cout << sum << endl;
  }

  return 0;
}