浙工大姗姗杯round2 G - Bad Luck Island CodeForces - 540D

D. 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

2 2 2

output

0.333333333333 0.333333333333 0.333333333333

input

2 1 2

output

0.150000000000 0.300000000000 0.550000000000

input

1 1 3

output

0.057142857143 0.657142857143 0.285714285714

题意：岛上生活着包、剪、锤三种生物，两两相遇遵照石头剪刀布法则杀死对方。现在给你岛上初始包剪锤各自的数量，求包剪锤各自最后的存活率。

思路：概率dp。本来以为有公式可以推到，但是想想不可能。dp[i][j][k]，i,j,k分别表示包剪锤数量，i*j+i*k+j*k是总的相遇情况，又因为两种物种相遇死亡方已知，所以可以用i*j/(i*j+i*k+j*k)来表示概率。接下来就是dp

#include 
#include 
using namespace std;
double dp[128][128][128];
int main(){
	int r, s, p;
	cin >> r >> s >> p;
	for(int i = 1; i <= 127; i++)
        for(int j = 0; j <= 127; j++)
            dp[i][j][0] = 1;
	for(int i = 1; i <= 127; i++)
        for(int j = 1; j <= 127; j++)
            for(int k = 1; k <= 127; k++){
                double p = i*j + j*k + i*k;//总情况数;
                dp[i][j][k] += i*k*dp[i - 1][j][k];
                dp[i][j][k] += i*j*dp[i][j - 1][k];
                dp[i][j][k] += j*k*dp[i][j][k - 1];
                dp[i][j][k] /= p;
            }
    printf("%.12lf %.12lf %.12lf\n", dp[r][s][p], dp[s][p][r], dp[p][r][s]);
	return 0;
}