CF1354C2 - Not So Simple Polygon Embedding （数学，二分）

Description

思路

大体思路是二分边缩短（\(\Delta L\)）的长度，判断多边形能否通过旋转塞进正方形当中（就是判断左上角的边会不会超出边界）。

#include <iostream>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <map>
#include <set>
#include <vector>
#include <cstring>
#include <string>
#include <deque>
#include <cmath>
#include <iomanip>
#include <cctype>
 
#define endl '\n'
#define IOS std::ios::sync_with_stdio(0); 
#define FILE freopen("..//data_generator//in.txt","r",stdin),freopen("res.txt","w",stdout)
#define FI freopen("..//data_generator//in.txt","r",stdin)
#define FO freopen("res.txt","w",stdout)
#define md make_pair
#define pb push_back
#define mp make_pair
#define seteps(N) fixed << setprecision(N) 
 
typedef long long ll;
 
ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;}
inline ll qmul(ll a, ll b, ll m) {
    ll res = 0;
    while(b) {
        if(b & 1) res = (res + a) % m;
        a = (a << 1) % m;
        b = b >> 1;
    }
    return res;
}
inline ll qpow(ll a, ll b, ll m) {
    ll res = 1;
    while(b) {
        if(b & 1) res = (res * a) % m;
        a = (a * a) % m;
        b = b >> 1;
    }
    return res;
}
inline ll inv(ll x, ll q) {
    return qpow(x, q - 2, q);
}
 
using namespace std;
/*-----------------------------------------------------------------*/
 
#define INF 0x3f3f3f3f
 
const int N = 3e6 + 10;
const double eps = 1e-10;
 
char s[N];
 
const long double PI = 3.14159265358979323846;
 
 
 
int main() {
    IOS;
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        long double o = PI / n;
        long double L = (long double)0.5 / sin(o / 2);
        long double l = 0, r = L - cos(o / 2) * L;
        while(r - l > eps) {
            long double deL = (l + r) / 2;
            long double deo = acos((L - deL) / L);
            if(L * cos(o / 2 - deo) <= L - deL) l = deL;
            else r = deL;
        }
        cout << seteps(10) << 2 * (L - l) << endl;
    }
}