Codeforces Round #301 (Div. 2) D. Bad Luck Island 概率DP

D. Bad Luck Island

 

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

题意：

　　

　　给你剪刀石头布三个种类的数量，每次随机两个，问你最后分别剩下这三个的概率

 

题解：

 

　　假设我们已知数量分别为a,b,c的概率即等于dp[a][b][c]；

　　那么我们可以得到的是dp[a][b][c-1] 、dp[a][b-1][c] 、 dp[a-1][b][c];

　　暴力推下去就得解

#include<bits/stdc++.h>
using namespace std;
const int N = 1e2+20, M = 1e6+10, mod = 1e9+7, inf = 1e9+1000;
typedef long long ll;

double dp[N][N][N];
double a,b,c;
int main() {
    int r,s,p;
    scanf("%d%d%d",&r,&s,&p);
    dp[r][s][p] = 1;
    for(int i=r;i>=0;i--) {
        for(int j=s;j>=0;j--) {
            for(int k=p;k>=0;k--) {
                double all = i*1.0*j+j*1.0*k+k*1.0*i;
                if(all==0) continue;
                if(k-1>=0)dp[i][j][k-1] += dp[i][j][k]*(double)(j*1.0*k)/all;
                if(j-1>=0)dp[i][j-1][k] += dp[i][j][k]*(double)(j*1.0*i)/all;
                if(i-1>=0)dp[i-1][j][k] += dp[i][j][k]*(double)(i*1.0*k)/all;
            }
        }
    }
    for(int i=1;i<=r;i++) a+=dp[i][0][0];
    for(int i=1;i<=s;i++) b+=dp[0][i][0];
    for(int i=1;i<=p;i++) c+=dp[0][0][i];
    printf("%.9f %.9f %.9f\n",a,b,c);
    return 0;
}