CodeForces - 540D Bad Luck Island (概率dp)

最新推荐文章于 2019-09-19 13:30:53 发布

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本文介绍了一种使用概率动态规划的方法来计算三种物种（石头、剪刀、布）在相互作用过程中的长期存活概率。通过定义状态和转移方程，解决了给定初始数量时，每种物种最终成为唯一幸存者的概率。

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：有r个石头，s个剪刀，p个布，求在足够长一段时间后，石头、剪刀、布分别存活下来的概率。

思路：概率dp，定义dp[i][j][k],表示有i个石头，j个剪刀，k个布情况下的概率。

状态转移方程：dp[i-1][j][k]=dp[i-1][j][k]+dp[i][j][k]*i*k/(i*j+i*k+j*k); 表示布包石头的情况，石头-1。

ansr表示石头最终存活下来的概率。对所有dp[i][j][0]求和即可，即布已为0，石头一定存活。

代码：

#include<bits/stdc++.h>
using namespace std;

const int N=105;
double dp[N][N][N];

int main(){
	int r,s,p;
	double ansr=0,anss=0,ansp=0;
	scanf("%d%d%d",&r,&s,&p);
	memset(dp,0,sizeof(dp));
	dp[r][s][p]=1;
	for(int i=r;i>=1;i--){
		for(int j=s;j>=1;j--){
			for(int k=p;k>=1;k--){
				int temp=i*j+i*k+j*k;
				dp[i-1][j][k]+=dp[i][j][k]*i*k/temp;
				dp[i][j-1][k]+=dp[i][j][k]*j*i/temp;
				dp[i][j][k-1]+=dp[i][j][k]*k*j/temp;
			}
		}
	} 
	for(int i=r;i>=0;i--){
		for(int j=s;j>=0;j--){
			ansr+=dp[i][j][0];
		}
	}
	printf("%.12f ",ansr);
	
	for(int i=s;i>=0;i--){
		for(int j=p;j>=0;j--){
			anss+=dp[0][i][j];
		}
	}
	printf("%.12f ",anss);
	
	for(int i=r;i>=0;i--){
		for(int j=p;j>=0;j--){
			ansp+=dp[i][0][j];
		}
	}
	printf("%.12f\n",ansp);
	
	return 0;
}