r, s, p = map(int, input().split())
species = [[[0] * 110 for i in range(110)] for i in range(110)]
species[r][s][p] = 1.0
for i in range(r, -1, -1):
for j in range(s, -1, -1):
for k in range(p, -1, -1):
if ((i == 0 and j == 0) or (i == 0 and k == 0) or (j == 0 and k == 0)):
continue
sum = (i * j) + (j * k) + (k * i)
if (i > 0 and j > 0):
species[i][j - 1][k] += species[i][j][k] * i * j / sum
if (j > 0 and k > 0):
species[i][j][k - 1] += species[i][j][k] * j * k / sum
if (i > 0 and k > 0):
species[i - 1][j][k] += species[i][j][k] * i * k / sum
rocks = 0
scissors = 0
papers = 0
for i in range(1, r + 1):
rocks += species[i][0][0]
for i in range(1, s + 1):
scissors += species[0][i][0]
for i in range(1, p + 1):
papers += species[0][0][i]
print(rocks, scissors, papers)
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using vld = vector<ld>;
using vvld = vector<vld>;
using vvvld = vector<vvld>;
#define pb push_back
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
vvvld dp1(101,vvld(101,vld(101,-1)));//initialize all to -1 meaning not yet computed
vvvld dp2(101,vvld(101,vld(101,-1)));
vvvld dp3(101,vvld(101,vld(101,-1)));
ld solve1(int r,int s,int p)
{
if (dp1[r][s][p] != -1)return dp1[r][s][p];//returning if dp1[r][s][p] has already been computed
if (r == 0)return dp1[r][s][p] = 0;//Base Case Part 1
if (r > 0 && s == 0 && p == 0)return dp1[r][s][p] = 1;//Base Case Part 2
ld all = 0;//s or p can be 0 and hence all stores the possible meetings for 2 indivuals whose species are non empty
if (r > 0 && s > 0)all += (ld)(r * s);//r and s can meet only when r > 0 and s > 0
if (r > 0 && p > 0)all += (ld)(p * r);
if (s > 0 && p > 0)all += (ld)(p * s);
dp1[r][s][p] = 0;//reinitialize dp1[r][s][p] to 0
if (s > 0)dp1[r][s][p] += ((ld)(r * s) / all) * solve1(r,s - 1,p);// r and s meet
if (p > 0)dp1[r][s][p] += ((ld)(r * p) / all) * solve1(r - 1,s,p);// r and p meet
if (s > 0 && p > 0)dp1[r][s][p] += ((ld)(s * p) / all) * solve1(r,s,p - 1);// s and p meet
// dp1(r,s,p) = probability of any 2 meeting * next case
return dp1[r][s][p];
}
ld solve2(int r,int s,int p)
{
if (dp2[r][s][p] != -1)return dp2[r][s][p];//only the base cases change
if (s == 0)return dp2[r][s][p] = 0;
if (s > 0 && r == 0 && p == 0)return dp2[r][s][p] = 1;
ld all = 0;
if (r > 0 && s > 0)all += (ld)(r * s);
if (r > 0 && p > 0)all += (ld)(p * r);
if (s > 0 && p > 0)all += (ld)(p * s);
dp2[r][s][p] = 0;
if (r > 0 && s > 0)dp2[r][s][p] += ((ld)(r * s) / all) * solve2(r,s - 1,p);
if (r > 0 && p > 0)dp2[r][s][p] += ((ld)(r * p) / all) * solve2(r - 1,s,p);
if (s > 0 && p > 0)dp2[r][s][p] += ((ld)(s * p) / all) * solve2(r,s,p - 1);
return dp2[r][s][p];
}
ld solve3(int r,int s,int p)
{
if (dp3[r][s][p] != -1)return dp3[r][s][p];//only the base cases change
if (p == 0)return dp3[r][s][p] = 0;
if (p > 0 && r == 0 && s == 0)return dp3[r][s][p] = 1;
ld all = 0;
if (r > 0 && s > 0)all += (ld)(r * s);
if (r > 0 && p > 0)all += (ld)(p * r);
if (s > 0 && p > 0)all += (ld)(p * s);
dp3[r][s][p] = 0;
if (r > 0 && s > 0)dp3[r][s][p] += ((ld)(r * s) / all) * solve3(r,s - 1,p);
if (r > 0 && p > 0)dp3[r][s][p] += ((ld)(r * p) / all) * solve3(r - 1,s,p);
if (s > 0 && p > 0)dp3[r][s][p] += ((ld)(s * p) / all) * solve3(r,s,p - 1);
return dp3[r][s][p];
}
int main()
{
int r,s,p;
cin>>r>>s>>p;
cout<<setprecision(12)<<solve1(r,s,p)<<" "<<solve2(r,s,p)<<" "<<solve3(r,s,p);
//do cout<<setprecision(12)<<real;
//solve1 , solve2 , solve3 are the 3 recursive functions to compute the 3 DP tables
return 0;
}
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