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

Div. 2) D. Bad Luck Island"/>

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

原题链接：

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.

Sample test(s) 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：有三种人，r，s，p，其中r可以杀死s，s可以杀死p，p可以杀死r，任意时间，只要两种不同的人相遇就会杀死其中一方，已知初始r，s，p的人数，求在无限长时间以后分别只剩r，s，p类人的概率。

思路：设某一时间的人数状态为（i,j,k），则下一状态可以为(i-1,,j,k),(i,j-1,k) , (i,j,k-1)，若dp[i][j][k]表示这一状态出现的概率，则以下一状态为(i-1,j,k)为例，其中任意不同类两人相遇的情况为（i*j+j*k+i*k），而i死去一个，则为i和k相遇的情况(i*k)，即dp[i-1][j][k]+=dp[i][j][k]*(i*k)/(i*j+j*k+i*k)。有一些边界条件要注意，还有输出误差要求是10-9。

代码：

#include "stdio.h"
#include "iostream"
#include "string.h"
#include "stdlib.h"
#include "algorithm"
#include "math.h"
#include "map"
#include "queue"
#include "stack"
using namespace std;
const int INF=0x3f3f3f3f;
const int MAXN=100005;
typedef long long LL;int a,b,c;
double dp[101][101][101];int cc[3][3]={-1,0,0,0,-1,0,0,0,-1};void getdp(int i,int j,int k,int w)
{double s=i*j+j*k+i*k;int a1,a2,a3;a1=i+cc[w][0];a2=j+cc[w][1];a3=k+cc[w][2];if(a1<0||a2<0||a3<0||a1+a2+a3<=0||s==0)return;if(w==0){dp[a1][a2][a3]+=dp[i][j][k]*(i*k)/s;}else if(w==1){dp[a1][a2][a3]+=dp[i][j][k]*(i*j)/s;}else{dp[a1][a2][a3]+=dp[i][j][k]*(j*k)/s;}
}int main()
{memset(dp,0,sizeof(dp));scanf("%d%d%d",&a,&b,&c);dp[a][b][c]=1;double ans1=0,ans2=0,ans3=0;for(int i=a;i>=0;i--){for(int j=b;j>=0;j--){for(int w=c;w>=0;w--){for(int q=0;q<3;q++){getdp(i,j,w,q);}}}}for(int i=1;i<=a;i++)ans1+=dp[i][0][0];for(int i=1;i<=b;i++)ans2+=dp[0][i][0];for(int i=1;i<=c;i++)ans3+=dp[0][0][i];printf("%.10lf %.10lf %.10lf\n",ans1,ans2,ans3);return 0;
}