「 51nod1639 」 shoelaces (Probability

 

1639 benchmark time limit for binding shoelaces: 1 second space limit: 131072 kb score: 20 difficulty: 3 levels of algorithm questions favorites followed by N shoelaces mixed together, now repeat n times the following operations: randomly take two shoes to take the lead and tie them together. As you can imagine, after N times, there will be no independent shoes taking the lead, and N shoelaces will be tied into some rings. So what is the probability that all these shoelaces form only one ring? Input

Only one row contains an integer N (2 <=n <= 1000 ).

Output

Output a line, indicating the probability of just forming a ring.

Input example

2

Output example

0.666667

Question

Considering that $ I $ has been completed,

So there are $2 * N-2 * I $ shoes at the moment,

Where you hold one, there are $2 * N-2 * I-1 $ shoes to take the lead,

Only one of these shoes (on the same rope you hold) cannot be knotted,

So this step can hit a qualified knot probability $ \ frac {2 * N-2 * I-2} {2 * N-2 * I-1} $.

$ I $ loops from $0 $ to $ N-2 $, stops when $ n-1 $ knot is hit (because $ n-1 $ knot is already a chain ).

The answer is the result of multiplication.

 1 /* 2 C++ 3 15 ms 4 2108 KB 5 Accepted 6 2018/10/26 7 17:01:37 8 */ 9 #include<iostream>10 #include<cstdio>11 using namespace std;12 int main()13 {14     //freopen("a.in","r",stdin);15     int n;16     scanf("%d",&n);17     double ans=1;18     for(int i=0;i<n-1;++i)19     {20         ans*=(double)(2*n-(2*i)-2)/(2*n-(2*i)-1);21     }22     cout<<ans;23     return 0;24 }

 

「 51nod1639 」 shoelaces (Probability