Codeforces 540D Bad Luck Island

 1 #include<cstdio>
 2 #include<cmath>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<iostream>
 6 #define dou long double
 7 dou jc[20005];
 8 dou f[105][105][105];
 9 int n,m,K;
10 int read(){
11     int t=0,f=1;char ch=getchar();
12     while (ch<‘0‘||ch>‘9‘){if (ch==‘-‘) f=-1;ch=getchar();}
13     while (‘0‘<=ch&&ch<=‘9‘){t=t*10+ch-‘0‘;ch=getchar();}
14     return t*f;
15 }
16 int main(){
17     n=read();m=read();K=read();
18     f[n][m][K]=1.0;
19     for (int i=n;i>=0;i--)
20      for (int j=m;j>=0;j--)
21       for (int k=K;k>=0;k--)
22        if (f[i][j][k]>0)
23       {
24         int cnt=((int)(i==0))+((int)(j==0))+((int)(k==0));
25         if (cnt>=2) continue;
26         double tot=i*j+j*k+k*i;
27         if (i>0)
28         f[i-1][j][k]+=f[i][j][k]*(i*k)/tot;
29         if (j>0)
30         f[i][j-1][k]+=f[i][j][k]*(i*j)/tot;
31         if (k>0)
32         f[i][j][k-1]+=f[i][j][k]*(k*j)/tot;
33       }
34     double ans1=0,ans2=0,ans3=0;
35     for (int i=1;i<=n;i++)
36      ans1+=(double)f[i][0][0];
37      for (int i=1;i<=m;i++)
38      ans2+=(double)f[0][i][0];
39     for (int i=1;i<=K;i++)
40      ans3+=(double)f[0][0][i];
41     printf("%.9f %.9f %.9f\n",ans1,ans2,ans3);   
42     return 0;
43 }