正方形内接正多边形做法证明参考：正四边形内接正六边形
具体做法见代码
The statement of this problem is the same as the statement of problem C1. The only difference is that, in problem C1, n is always even, and in C2, n
is always odd.
You are given a regular polygon with 2⋅n
vertices (it's convex and has equal sides and equal angles) and all its sides have length 1. Let's name it as 2n
-gon.
Your task is to find the square of the minimum size such that you can embed 2n
-gon in the square. Embedding 2n-gon in the square means that you need to place 2n-gon in the square in such way that each point which lies inside or on a border of 2n
-gon should also lie inside or on a border of the square.
You can rotate 2n
-gon and/or the square.
Input
The first line contains a single integer T
(1≤T≤200
) — the number of test cases.
Next T
lines contain descriptions of test cases — one per line. Each line contains single odd integer n (3≤n≤199). Don't forget you need to embed 2n-gon, not an n
-gon.
Output
Print T
real numbers — one per test case. For each test case, print the minimum length of a side of the square 2n-gon can be embedded in. Your answer will be considered correct if its absolute or relative error doesn't exceed 10−6
.
Example
Input
Copy
3 3 5 199
Output
Copy
1.931851653 3.196226611 126.687663595
#include<bits/stdc++.h>
#define LOCAL
using namespace std;
typedef unsigned u;
#define ll long long
#define maxn 1350500
#define esp 1e-10
#define pi acos(-1.0)
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
double n;
cin>>n;
double s=round(n/4.0),x=pi/n;
//printf("s=%.5f\n",s);
double a=0.5/sin(x/2.0);
double ans1=a*sqrt(2.0)*sin(x*s);
double ans2=a*sqrt(2.0)*sin(pi/2.0-x*s);
//d*=2.0/sqrt(2.0);
//d+=d*cos(x/2.0);
printf("%.9f\n",ans1+ans2);
}
}
/*
333
3
1.931851653
5
3.196226611
7
4.465702135
9
5.736856623
11
7.008771023
13
8.281093789
15
9.553661305
*/