Problem:
 Score: 600 points    Problem Statement  E869120 is initially standing at the origin (0, 0) in a two-dimensional plane.  He has N engines, which can be used as follows:   When E869120 uses the i -th engine, his X - and Y -coordinate change by x_i and y_i , respectively. In other words, if E869120 uses the i -th engine from coordinates (X, Y) , he will move to the coordinates (X + x_i, Y + y_i) .  E869120 can use these engines in any order, but each engine can be used at most once. He may also choose not to use some of the engines.   He wants to go as far as possible from the origin.
Let (X, Y) be his final coordinates. Find the maximum possible value of \sqrt{X^2 + Y^2} , the distance from the origin.      Constraints   1 \leq N \leq 100  -1 \ 000 \ 000 \leq x_i \leq 1 \ 000 \ 000  -1 \ 000 \ 000 \leq y_i \leq 1 \ 000 \ 000  All values in input are integers.         Input  Input is given from Standard Input in the following format:  N  x_1  y_1  x_2  y_2  :  :  x_N  y_N       Output  Print the maximum possible final distance from the origin, as a real value.
Your output is considered correct when the relative or absolute error from the true answer is at most 10^{-10} .        Sample Input 1 3
0 10
5 -5
-5 -5      Sample Output 1 10.000000000000000000000000000000000000000000000000  The final distance from the origin can be 10 if we use the engines in one of the following three ways:   Use Engine 1 to move to (0, 10) .  Use Engine 2 to move to (5, -5) , and then use Engine 3 to move to (0, -10) .  Use Engine 3 to move to (-5, -5) , and then use Engine 2 to move to (0, -10) .   The distance cannot be greater than 10 , so the maximum possible distance is 10 .       Sample Input 2 5
1 1
1 0
0 1
-1 0
0 -1      Sample Output 2 2.828427124746190097603377448419396157139343750753  The maximum possible final distance is 2 \sqrt{2} = 2.82842... .
One of the ways to achieve it is:   Use Engine 1 to move to (1, 1) , and then use Engine 2 to move to (2, 1) , and finally use Engine 3 to move to (2, 2) .        Sample Input 3 5
1 1
2 2
3 3
4 4
5 5      Sample Output 3 21.213203435596425732025330863145471178545078130654  If we use all the engines in the order 1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 5 , we will end up at (15, 15) , with the distance 15 \sqrt{2} = 21.2132... from the origin.       Sample Input 4 3
0 0
0 1
1 0      Sample Output 4 1.414213562373095048801688724209698078569671875376  There can be useless engines with (x_i, y_i) = (0, 0) .       Sample Input 5 1
90447 91000      Sample Output 5 128303.000000000000000000000000000000000000000000000000  Note that there can be only one engine.       Sample Input 6 2
96000 -72000
-72000 54000      Sample Output 6 120000.000000000000000000000000000000000000000000000000  There can be only two engines, too.       Sample Input 7 10
1 2
3 4
5 6
7 8
9 10
11 12
13 14
15 16
17 18
19 20      Sample Output 7 148.660687473185055226120082139313966514489855137208

Solution (Python):
rr = raw_input
rri = lambda: int(raw_input())
rrm = lambda: map(int, raw_input().split())

from math import atan2, pi

def ccw(p1, p2, p3):
    val = (p2[1] - p1[1]) * (p3[0] - p2[0])
    val -= (p2[0] - p1[0]) * (p3[1] - p2[1])
    return val < 0

def solve(N, A):
    def getangle(i):
        x,y = A[i]
        return atan2(y,x)
    
    def massage(theta):
        if theta < 0:
            theta += 2 * pi
        return theta

    Abase = [(x, y, massage(atan2(y,x))) for x, y in A]
    Abase.sort(key=lambda z: z[2])
    A = Abase[:]
    for x, y, ang in Abase:
        A.append((x, y, ang + 2*pi))

    P = [(0,0)]
    sx=sy = 0
    for x, y, th in A:
        sx+=x;sy+=y
        P.append((sx,sy))

    ans = 0
    for i in xrange(N):
        for j in xrange(i, i+N):
            dx = P[j+1][0] - P[i][0]
            dy = P[j+1][1] - P[i][1]
            ans = max(ans, dx*dx+dy*dy)
    return ans**.5
    """
    def reflected(i):
        pt = A[i]
        return (-pt[0], -pt[1])

    def condition(i, j):
        
        return ccw(A[i], A[j], reflected(i))

    print A
    
    j = 0
    ans = 0
    for i in xrange(N):
        
        j = max(j, i)
        while j+1 <= i+N-1 and condition(i, j+1):
            j += 1
        print '!', i,j
        dx = P[j+1][0] - P[i][0]
        dy = P[j+1][1] - P[i][1]
        ans = max(ans, dx*dx+dy*dy)
    return ans**.5
    """

                
N=rri()
A=[rrm() for _ in xrange(N)]
print solve(N, A)

Solution (Python):

def det(p, q):
    return p[0]*q[1] - p[1]*q[0]
def sub(p, q):
    return (p[0]-q[0], p[1]-q[1])
def get_convex_hull(points):
    n = len(points)
    points.sort()
    size_convex_hull = 0
    ch = []
    for i in range(n):
        while size_convex_hull > 1:
            v_cur = sub(ch[-1], ch[-2])
            v_new = sub(points[i], ch[-2])
            if det(v_cur, v_new) > 0:
                break
            size_convex_hull -= 1
            ch.pop()
        ch.append(points[i])
        size_convex_hull += 1

    t = size_convex_hull
    for i in range(n-2, -1, -1):
        while size_convex_hull > t:
            v_cur = sub(ch[-1], ch[-2])
            v_new = sub(points[i], ch[-2])
            if det(v_cur, v_new) > 0:
                break
            size_convex_hull -= 1
            ch.pop()
        ch.append(points[i])
        size_convex_hull += 1

    return ch[:-1]

N = int(input())
XY = [list(map(int, input().split())) for _ in range(N)]
ps = [(0, 0)]
for x, y in XY:
    ps_new = ps[:]
    for x_, y_ in ps:
        ps_new.append((x+x_, y+y_))
    ps = get_convex_hull(ps_new)
ans = 0
for x, y in ps:
    ans = max(ans, x*x+y*y)
print(ans ** 0.5)

Solution (Python):
import sys
import math

stdin = sys.stdin

ni = lambda: int(ns())
na = lambda: list(map(int, stdin.readline().split()))
ns = lambda: stdin.readline().rstrip()  

n = ni()
co = []
for i in range(n):
    t = na()
    if t[0] == 0 and t[1] == 0:
        continue
    co.append(t)

n = len(co)
if n == 0:
    print(0)
    sys.exit(0)

co.sort(key=lambda x: math.atan2(x[1], x[0]))

ans = 0
co = co + co
for i in range(n):
    vx, vy = 0, 0
    for j in range(i, i+n):
        vx += co[j][0]
        vy += co[j][1]
        ans = max(ans, vx*vx+vy*vy)
print("{:.14f}".format(math.sqrt(ans)))

Solution (Python):
import sys
input = sys.stdin.readline
sys.setrecursionlimit(10 ** 7)

"""
適切な回転に対して、x座標最大化の問題になる。
ということは、適当な半平面に入ってるものを全部足す
"""

import numpy as np

N = int(input())
X,Y = np.array([input().split() for _ in range(N)],dtype=np.int64).T

zero = (X==0)&(Y==0)
if(zero.all()):
    print(0)
    exit()

X = X[~zero]
Y = Y[~zero]
N = len(X)


angle = np.arctan2(Y,X)
i = angle.argsort()
X = X[i]; Y = Y[i]; angle = angle[i];
X = np.concatenate([X,X])
Y = np.concatenate([Y,Y])
angle = np.concatenate([angle,angle+2*np.pi])



Xcum = X.cumsum()
Ycum = Y.cumsum()

items = np.arange(1,N+1)[None,:]
L = np.arange(N)[:,None]
dx = Xcum[L+items-1] - Xcum[L] + X[L]
dy = Ycum[L+items-1] - Ycum[L] + Y[L]

answer = (dx*dx+dy*dy).max()**.5
print(answer)