Codeforces B. Bad Luck Island（概率dp）

题目描述：

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

思路：

这道题刚看的时候觉得好神奇啊，一定要用到好高大上的概率知识，但是我的数学已经丢完了，只能跳过了。知道了解法之后更觉得神奇，啊~神奇的概率dp。

#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;
}