Codeforces C2. Not So Simple Polygon Embedding (數學 / 計算幾何) (Round 87 Rated for Div.2)

傳送門

題意： 找到能放下正n(n爲奇數)邊形的最小正方形的邊長。

思路：

• 啊啊啊！其實我也不會，看了兩位大佬的博客也是似懂非懂的，感覺可比C1難多了
• 大佬博客1 ， 大佬博客2

代碼實現：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cctype>
#include <cstring>
#include <iostream>
#include <sstream>
#include <string>
#include <list>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <algorithm>
#include <functional>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int  inf = 0x7fffffff;
const double Pi = acos(-1.0);
const double eps = 1e-6;
const ll   mod = 1e9 + 7;
const int  N = 2e5 + 5;

int t, n;

signed main()
{
    cin >> t;
    while(t --){
        cin >> n;
        double ans = cos(Pi / (4 * n)) / sin(Pi / (2 * n));
        printf("%.9f\n", ans);
    }

    return 0;
}