#include <bits/stdc++.h>
using namespace std;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef vector<vector<long long>> VVL;
typedef vector<string> VS;
typedef pair<int, int> P;
typedef tuple<int,int,int> tpl;

#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REVERSE(c) reverse((c).begin(),(c).end())
#define LB(a,x) lower_bound((a).begin(), (a).end(), x) - (a).begin()
#define UB(a,x) upper_bound((a).begin(), (a).end(), x) - (a).begin()

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define RFOR(i,a,b) for(int i=(a)-1;i>=(b);--i)
#define RREP(i,n) RFOR(i,n,0)

#define en "\n"

constexpr double EPS = 1e-9;
constexpr double PI  = 3.1415926535897932;
constexpr int INF = 2147483647;
constexpr long long LINF = 1LL<<60;
constexpr long long MOD = 1000000007; // 998244353;

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

struct V {
    double x,y;
    int id;
    V(double x=0, double y=0): x(x), y(y) {}

    V& operator += (const V& v) { x += v.x; y += v.y; return *this;}
    V operator + (const V& v) const { return V(*this) += v;}
    V& operator -= (const V& v) { x -= v.x; y -= v.y; return *this;}
    V operator - (const V& v) const { return V(*this) -= v;}
    V& operator *= (double s) { x *= s; y *= s; return *this;}
    V operator * (double s) const { return V(*this) *= s;}
    V& operator /= (double s) { x /= s; y /= s; return *this;}
    V operator / (double s) const { return V(*this) /= s;}
    bool operator <(const V &v) const {
		return x < v.x || (x == v.x && y < v.y);
	}

    double dot(const V& v) const { return x*v.x + y*v.y;}
    double norm2() const { return x*x + y*y;}
    double norm() const { return sqrt(norm2());}
    V& rotate(double theta) { x = x*cos(theta)-y*sin(theta); y = x*sin(theta)+y*cos(theta); return *this;}
    V rotate90() { return V(-y, x);}
};

// 3D cross product of OA and OB vectors, (i.e z-component of their "2D" cross product, but remember that it is not defined in "2D").
// Returns a positive value, if OAB makes a counter-clockwise turn,
// negative for clockwise turn, and zero if the points are collinear.
double cross(const V &O, const V &A, const V &B)
{
	return (A.x - O.x) * (B.y - O.y) - (A.y - O.y) * (B.x - O.x);
}

// Returns a list of points on the convex hull in counter-clockwise order.
// Note: the last point in the returned list is the same as the first one.
vector<V> convex_hull(vector<V> pt){
	size_t n = pt.size(), k = 0;
    if (n <= 3) return pt;
	vector<V> H(2*n);

	// Sort points lexicographically
    sort(pt.begin(), pt.end());

	// Build lower hull
	for (size_t i = 0; i < n; ++i) {
		while (k >= 2 && cross(H[k-2], H[k-1], pt[i]) <= 0) k--;
		H[k++] = pt[i];
	}

	// Build upper hull
	for (size_t i = n-1, t = k+1; i > 0; --i) {
		while (k >= t && cross(H[k-2], H[k-1], pt[i-1]) <= 0) k--;
		H[k++] = pt[i-1];
	}

	H.resize(k-1);
	return H;
}

void Main(){
    int N; cin >> N;
    vector<V> pt(N); REP(i,N){cin >> pt[i].x >> pt[i].y; pt[i].id = i;}

    if(N == 2){
        vector<double> ans{0.5,0.5};
        REP(i,N) cout << ans[i] << en;
        return;
    }

    vector<V> H = convex_hull(pt);
    int K = H.size();
    VI idx; REP(i,K) idx.push_back(H[i].id);
    vector<double> ans(N,0);

    REP(i,K){
        V p1 = (pt[idx[i]] - pt[idx[(i+1)%K]]).rotate90();
        V p2 = (pt[idx[(i+2)%K]] - pt[idx[(i+1)%K]]).rotate90().rotate90().rotate90();
        double t1 = atan2(p1.y, p1.x), t2 = atan2(p2.y, p2.x);
        double theta = abs(t1-t2);
        if(theta > PI) theta = 2*PI-theta;
        ans[idx[(i+1)%K]] = theta / PI / 2;
    }

    REP(i,N) cout << ans[i] << en;
    return;
}

int main(void){
    cin.tie(0);ios_base::sync_with_stdio(0);cout<<fixed<<setprecision(15);
    int t=1; //cin>>t;
    REP(_,t){
        Main();
    }
    return 0;
}

Measured memory_bytes: 512