Codeforces Round #301 (Div. 2)---D. Bad Luck (probabilistic DP)

The bad Luck are inhabited by three kinds of species:r rocks, s scissors and P papers. At some moments of time both random individuals meet (all pairs of individuals can meet equiprobably), and if they belong t o 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's the probability, this species would 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'll be is the on ly surviving species, respectively. The answer would be a considered correct if the relative or absolute error of each number doesn ' t exceed?-? 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

DP[R][S][P] Indicates the probability that there is a rock,s scissors,p papers of R left.

/************************************************************************* > File name:d.cpp > Author:al ex > Mail: [email protected] > Created time:2015 May 01 Friday 20:06 20 seconds ************************************** **********************************/#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <map>#include <bitset>#include <set>#include <vector>using namespace STD;Const DoublePI =ACOs(-1.0);Const intINF =0x3f3f3f3f;Const DoubleEPS =1e-15;typedef Long LongLL;typedefPair <int,int> PLL;Doubledp[101][101][101];intMain () {intR, S, p; while(Cin>> R >> s >> p) { for(inti =0; I <= R; ++i) { for(intj =0; J <= S; ++J) { for(intK =0; K <= P; ++K) {Dp[i][j][k] =0; }}} Dp[r][s][p] =1; for(inti = r; I >=0; -I.) { for(intj = S; J >=0; --J) { for(intK = p; K >=0; --K) {DoublePP =0;if(i + j + K <1) {Continue; }if(I >1) {pp + = i * (i-1) *1.0/2; }if(J >1) {pp + = J * (J-1) *1.0/2; }if(k >1) {pp + k * (K-1) *1.0/2; }Doublex = ((i + j + k +1) * (i + j + k) *1.0) /2;if(i + j + k >1) {pp/= ((i + j + k) * (i + j + K-1)) /2; }Else{pp =0; }DoubleTMP =0;if(i +1<= r && k >0) {Dp[i][j][k] + = Dp[i +1][J][K] * ((i +1) * k *1.0)/x; }if(j +1<= s && i >0) {Dp[i][j][k] + = Dp[i][j +1][K] * ((j +1) * I *1.0)/x; }if(k +1<= P && J >0) {Dp[i][j][k] + = Dp[i][j][k +1] * ((k +1) * J *1.0)/x; }if(pp! =1.0) {Dp[i][j][k]/= (1.0-PP); }//dp[i][j][k] + = tmp;            //printf ("dp[%d][%d][%d] =%lf\n", I, J, K, Dp[i][j][k]);}            }        }DoubleANS1 =0, Ans2 =0, ANS3 =0; for(inti =1; I <= R; ++i) {ans1 + = dp[i][0][0]; } for(inti =1; I <= s; ++i) {ans2 + = dp[0][i][0]; } for(inti =1; I <= p; ++i) {ans3 + = dp[0][0][i]; }printf("%.12f%.12f%.12f\n", Ans1, Ans2, ANS3); }return 0;}

Codeforces Round #301 (Div. 2)---D. Bad Luck (probabilistic DP)