// Original C++ solution (slower):
#include<bits/stdc++.h>

using namespace std;



#define FOR(i,a,b) for (int i = (a); i < (b); i++)

#define RFOR(i,b,a) for (int i = (b) - 1; i >= (a); i--)

#define ITER(it, a) for(__typeof(a.begin()) it = a.begin(); it != a.end(); it++)

#define FILL(a, value) memset(a, value, sizeof(a))



#define SZ(a) (int)a.size()

#define ALL(a) a.begin(), a.end()

#define PB push_back

#define MP make_pair



typedef long long LL;

typedef vector<int> VI;

typedef pair<int, int> PII;



const double PI = acos(-1);

const int INF = 1000 * 1000 * 1000 + 7;

const LL LINF = INF * (LL)INF;



struct point

{

	LL x, y;

	point(){}

	point(LL x1, LL y1)

	{

		x = x1;

		y = y1;

	}



	LL d2() const

	{

		return x * x + y * y;

	}



	point operator -(const point& a) const

	{

		return point(x - a.x, y - a.y);

	}



	LL operator *(const point& a) const

	{

		return x * a.y - a.x * y;

	}



	bool operator <(const point& a) const

	{

		if (x == a.x) return y < a.y;

		return x < a.x;

	}



};



const int MAX = 100 + 47;

double const EPS = 1e-8;



point A[MAX];

double ANS[MAX];

int n;

point a, b;



double getAng(point a, point b)

{

	double a1 = atan2(a.y, a.x);

	if (a1 < 0) a1 += 2 * PI;



	double a2 = atan2(b.y, b.x);

	if (a2 < 0) a2 += 2 * PI;



	double alpha = abs(a1 - a2);

	if (alpha > PI) alpha = 2 * PI - alpha;

	return PI - alpha;

}



bool isOk(point a, point b)

{

	bool ccw = false;

	bool cw = false;



	FOR(i, 0, n)

	{

		LL val = (b - a) * (A[i] - a);

		if (val > 0) ccw = true;



		if (val < 0) cw = true;

	}



	if (ccw && cw) return false;



	return true;

}



bool inside(point a, point b, point c)

{

	if (a.x < b.x && b.x < c.x) return true;

	if (a.y < b.y && b.y < c.y) return true;



	return false;

}



bool inOneLine()

{

	FOR(i, 2, n)

	{

		if ((A[i] - A[i - 1]) * (A[i - 1] - A[i - 2]) != 0) return false;

	}



	return true;

}



int main()

{

	//freopen("in.txt", "r", stdin);

	ios::sync_with_stdio(false); cin.tie(0);

	cin >> n;

	FOR(i, 0, n)

	{

		cin >> A[i].x >> A[i].y;

	}



	if (inOneLine())

	{

		int mn = 0, mx = 0;

		FOR(i, 0, n)

		{

			if (A[i] < A[mn]) mn = i;

			if (A[mx] < A[i]) mx = i;

		}



		ANS[mn] = ANS[mx] = 0.5;

	}

	else

	{

		FOR(i, 0, n)

		{

			point a = A[i];



			FOR(j, 0, n)

			{

				if (i == j) continue;

				point b = A[j];



				if (!isOk(a, b)) continue;

				FOR(k, 0, n)

				{

					if (k == i || k == j) continue;

					point c = A[k];

					if (!isOk(b, c)) continue;

					//cout << i + 1 << " " << j + 1 << " " << k + 1 << endl;



					if ((a - b) * (c - b) == 0) continue;

					double ans = getAng(a - b, c - b) / (2 * PI);

					ANS[j] = ans;

				}

			}

		}

	}



	FOR(i, 0, n)

	{

		cout << fixed << setprecision(12) << ANS[i] << "\n";

	}

	return 0;

}


// Optimized solution to the same problem (faster):
#include <cstdio>

#include <cstring>

#include <algorithm>

#include <cmath>

#define ld long double

#define MN 205

#define eps 1e-9

using namespace std;

const ld pi=acos(-1);

struct vec

{

	ld x,y; int pos;

	friend vec operator-(const vec& a,const vec& b) {return (vec){a.x-b.x,a.y-b.y};}

	friend ld operator/(const vec& a,const vec& b) {return a.x*b.y-a.y*b.x;}

	friend ld operator*(const vec& a,const vec& b) {return a.x*b.x+a.y*b.y;}

	friend ld abs(const vec& a) {return sqrt(a.x*a.x+a.y*a.y);}

}a[MN];

ld ans[MN];

int q[MN];

int n,tp;



inline int read()

{

	int n=0,f=1; char c=getchar();

	while (c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}

	while (c>='0' && c<='9') {n=n*10+c-'0'; c=getchar();}

	return n*f;

}



bool cmp1(const vec& a,const vec& b) {return a.x<b.x || a.x==b.x && a.y<b.y;}

bool cmp2(const vec& A,const vec& B)

{

	ld lt=(A-a[1])/(B-a[1]);

	if (fabs(lt)>eps) return lt>0;

	else return abs(A-a[1])<abs(B-a[1]);

}



int main()

{

//	freopen("data.in","r",stdin);

//	freopen("data.out","w",stdout);

	register int i;

	n=read();

//	printf("%d\n",n);	

	for (i=1;i<=n;++i)

		a[i].x=read(),a[i].y=read(),a[i].pos=i;

	if (n==1) return 0*printf("1");

	else if (n==2) return 0*printf("0.5000000\n0.5000000");

	sort(a+1,a+n+1,cmp1);

	sort(a+2,a+n+1,cmp2);

	q[tp=1]=1;

	for (i=2;i<=n;++i)

	{

		while (tp>1&&(a[q[tp]]-a[q[tp-1]])/(a[i]-a[q[tp-1]])<eps) --tp;

		q[++tp]=i;

	}

	q[tp+1]=q[1]; q[0]=q[tp];

//	for (i=1;i<=tp;++i) printf("%lld,%lld\n",a[q[i]].x,a[q[i]].y);	

//	for (i=1;i<=tp;++i) printf("%d ",a[q[i]].pos); puts("");

//	printf("%d\n",tp);

//	printf("%d %d\n",n,tp);

	for (i=1;i<=tp;++i)

	

	{

	

		ans[a[q[i]].pos]=

		(pi-acos((double)(

		((a[q[i-1]]-a[q[i]])*

		(a[q[i+1]]-a[q[i]]))

		/(abs(a[q[i-1]]-a[q[i]])*

		  abs(a[q[i+1]]-a[q[i]])))))/(pi*2);

//	printf("%.5lf\n",acos(

//		((a[q[i-1]]-a[q[i]])*

//		(a[q[i+1]]-a[q[i]]))

//		/(abs(a[q[i-1]]-a[q[i]])*

//		  abs(a[q[i+1]]-a[q[i]]))));

	}

	

	for (i=1;i<=n;++i) printf("%.10lf\n",(double)ans[i]);

}

// Fastest known solution for this problem: