C1. Simple Polygon Embedding——（平面几何）

总结

几何算法，很少遇到，一遇到就死的很惨，没法看

== n为偶数 ==

每次增加2*2，四条边一定在正方形上：
ans = 1.0 / tan(PI / (2 * n))
== n为奇数==
我不会，cos(PI / (4 * n)) / sin(PI / (2 * n))

题目链接

//#pragma GCC optimize(2)
#include<bits/stdc++.h>
//typedef long long ll;
#define ull       unsigned long long
#define int       long long
#define F           first
#define S           second
#define endl        "\n"//<<flush
#define eps         1e-6
#define base        131
#define lowbit(x)   (x&(-x))
#define PI          acos(-1.0)
#define inf         0x3f3f3f3f
#define MAXN        0x7fffffff
#define INF         0x3f3f3f3f3f3f3f3f
#define ferma(a,b)  pow(a,b-2)
#define pb          push_back
#define decimal(x)  cout << fixed << setprecision(x);
#define all(x)      x.begin(),x.end()
#define memset(a,b) memset(a,b,sizeof(a));
#define IOS         ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
void file()
{
#ifdef ONLINE_JUDGE
#else
    freopen("D:/LSNU/codeforces/duipai/data.txt","r",stdin);
    //  freopen("D:/LSNU/codeforces/duipai/WA.txt","w",stdout);
#endif
}
signed main()
{
    IOS;
    file();
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        n*=2;
        double degree=(n-2)*180*1.0/n/2*PI/180;
        decimal(6);
        cout<<tan(degree)<< endl;
    }
    return 0;
}