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

D. Bad Luck

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'll be considered correct if the relative or absolute error of each number doesn ' t exceed -9.

Examples input

2 2 2

Output

0.333333333333 0.333333333333 0.333333333333

Test Instructions:Give you a pair of scissors stone cloth Three kinds of number, each time randomly two, ask you finally respectively left these three probabilityThe following:Assuming that we known quantity the probability of a,b,c is equal to dp[a][b][c]; then we can get dp[a][b][c-1], Dp[a][b-1][c], dp[a-1][b][c]; The violence pushes down, it's solved.

#include <bits/stdc++.h>using namespacestd;Const intN = 1e2+ -, M = 1e6+Ten, mod = 1e9+7, INF = 1e9+ +; typedefLong Longll;DoubleDp[n][n][n];Doublea,b,c;intMain () {intr,s,p; scanf ("%d%d%d",&r,&s,&p); DP[R][S][P]=1;  for(inti=r;i>=0; i--) {         for(intj=s;j>=0; j--) {             for(intk=p;k>=0; k--) {                Doubleall = 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(intI=1; i<=r;i++) a+=dp[i][0][0];  for(intI=1; i<=s;i++) b+=dp[0][i][0];  for(intI=1; i<=p;i++) c+=dp[0][0][i]; printf ("%.9f%.9f%.9f\n", A,b,c); return 0;}

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