F - Roshambo
Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u
Submit Status Practice URAL 1936

Description
Bootstrap: Wondering how it‘s played?
Will: It‘s a game of deception. But your bet includes all the dice, not just your own. What are they wagering?
Bootstrap: Oh, the only thing we have. Years of service.
Will: So any crew member can be challenged?
Bootstrap: Aye. Anyone.
Will: I challenge Davy Jones.
All that the pirates have on the Flying Dutchman is the years of service that are left for them. Every crewman wants to shorten it. That is why gambling is very popular on the ship, the winner have a chance to shorten his years of service significantly.
Pirates often gather to play “Roshambo”, also known as “rock-scissors-paper”. The game consists of several sets. In the beginning of each set players stand in a circle, count to three and show one of three gestures simultaneously, conventionally called as rock, scissors and paper. If everyone shows the same gesture or if each of the three gestures is shown, then nobody leave the game and they play another set. If among the shown gestures there are only two different then only players that chose the victorious gesture play the next set. Scissors beats rock, rock beats paper and paper beats scissors. The game continues until the only one player is left, and that pirate is called the winner. The winner’s time of service is shortened on the number of years that equals the number of the sets played, while the losers get extra years.
Bootstrap Bill decided to try his fortune. You should help him determine the expected value of prize in case of his victory. Pirates don’t know any complicated strategies for this game. So you can suppose that pirates show every gesture equiprobably.

Input
The only line contains integer n that is the number of sailors that are going to play, including Bill (2 ≤ n ≤ 100).

Output
Output the expected amount of years that will be taken off from winner. Absolute or relative error should be no more than 10?6.

Sample Input
input    output

2

    

1.5

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44 
45 //我发现那个kn=1.0然后慢慢减，得的答案特别逗，要kn=0，然后慢慢加，才不逗，这是什么原理
46 #include<cstdio>
47 #include<iostream>
48 #include<cstring>
49 #include<vector>
50 #include<stack>
51 #include<queue>
52 #include<cmath>
53 using namespace std;
54 
55 long double f[111];
56 long double jc[111];
57 long double cf[111];
58 
59 int main()
60 {
61     int i,j;
62     jc[1]=1.0;
63     for(i=2;i<=100;i++)
64         jc[i]=jc[i-1]*i;
65     cf[0]=1.0;
66     for(i=1;i<=100;i++)
67         cf[i]=cf[i-1]*3.0;
68     f[1]=0;
69     f[2]=1.5;
70     for(i=3;i<=100;i++)
71     {
72         f[i]=0;
73         long double kn=0.0;
74         for(j=1;j<i;j++)
75         {
76             long double k=1.0/jc[j]/jc[i-j]/cf[i-1]*jc[i];
77             kn+=k;
78             f[i]+=k*(1.0+f[j]);
79         }
80         f[i]+=1.0-kn;
81         f[i]/=kn;
82         //cout<<i<<". "<<f[i]<<"    kn="<<kn<<endl;
83     }
84     int n;
85     while(scanf("%d",&n)!=EOF)
86         printf("%lf\n",(double)f[n]);
87     return 0;
88 }