Codeforces B. Bad Luck Island (probability dp)

Subject description:

Bad Luck Island

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

The Bad Luck Island is inhabited by three kinds of species: r rocks, s scissors and p papers. At some moments of time two random individuals meet (all pairs of individuals can meet equiprobably), and if they belong to different species, then one individual kills the other one: a rock kills scissors, scissors kill paper, and paper kills a rock. Your task is to determine for each species what is the probability that this species will be the only one to inhabit this island after a long enough period of time.

Input

The single line contains three integers r, s and p (1 ≤ r, s, p ≤ 100) — the original number of individuals in the species of rock, scissors and paper, respectively.

Output

Print three space-separated real numbers: the probabilities, at which the rocks, the scissors and the paper will be the only surviving species, respectively. The answer will be considered correct if the relative or absolute error of each number doesn't exceed 10 - 9.

Examples

Input

Copy

2 2 2

Output

Copy

0.333333333333 0.333333333333 0.333333333333

Input

Copy

2 1 2

Output

Copy

0.150000000000 0.300000000000 0.550000000000

Input

Copy

1 1 3

Output

Copy

0.057142857143 0.657142857143 0.285714285714

Ideas:

This question is just look at the time feel good magic ah, be sure to use the knowledge to good probability tall, but my math has been lost over, it can only be skipped. Once you know the law better understand the magical feel, ah ~ magic probability dp.

Definition of DP [i] [j] [k] is the i-th species remaining rock, scissors remaining j-th, k th probability happens cloth left. dp beginning [r] [s] [p] = 1, now consider a state transition.

When a pair of scissors will be less of it? It encountered a stone, the probability of this happening is it? \ (\ {C_S FRAC * C_r ^ ^. 1. 1} + {R & lt C_ {P} S ^ 2} + \) , at least one fabric Similarly it? It encountered a pair of scissors, the probability of it? \ (\ FRAC {C_p ^. 1 * ^. 1 C_S} {C_ {R & lt + S + P} ^ 2} \) , at least one probability stones \ (\ frac {C_p ^ 1 * C_r ^ 1} {C_ { } S + P + R & lt ^ 2} \) , then the state transition equation is

\ [Sum = source + Q + clear \]

\[dp[i-1][j][k]+=dp[i][j][k]*\frac{C_s^1*C_r^1}{sum}\]

\[dp[i][j-1][k]+=dp[i][j][k]*\frac{C_p^1*C_r^1}{C_{sum}^2}\]

\[dp[i][j][k-1]+=dp[i][j][k]*\frac{C_p^1*C_s^1}{C_{sum}^2}\]

Recursive to start from scratch.

Statistics last answer when statistics dp [i] [j] [0] This answer, because the answer to this outcome has been set, there is only one species

Code:

#include <iostream>
#include <iomanip>
#define max_n 105
using namespace std;
int r,s,p;
double dp[max_n][max_n][max_n];
int main()
{
    cin >> r >> s >> p;
    dp[r][s][p] = 1.0;
    for(int i = r;i>0;i--)
    {
        for(int j = s;j>0;j--)
        {
            for(int k = p;k>0;k--)
            {
                double sum = i*j+i*k+j*k;
                dp[i-1][j][k] += dp[i][j][k]*(double)(i*k)/sum;
                dp[i][j-1][k] += dp[i][j][k]*(double)(i*j)/sum;
                dp[i][j][k-1] += dp[i][j][k]*(double)(j*k)/sum;
            }
        }
    }
    double ans1 = 0;
    double ans2 = 0;
    double ans3 = 0;
    for(int i = 0;i<=100;i++)
    {
        for(int j = 0;j<=100;j++)
        {
            ans1 += dp[i][j][0];
            ans2 += dp[0][i][j];
            ans3 += dp[i][0][j];
        }
    }
    cout.setf(ios_base::fixed,ios_base::floatfield);
    cout << setprecision(12) << ans1 << " " << ans2 << " " << ans3 << endl;
    return 0;
}