# Solution of;
# Project Euler Problem 363: Bézier Curves
# https://projecteuler.net/problem=363
# 
# A cubic Bézier curve is defined by four points: $P_0, P_1, P_2,$ and $P_3$. 
# The curve is constructed as follows:On the segments $P_0 P_1$, $P_1 P_2$, 
# and $P_2 P_3$ the points $Q_0, Q_1,$ and $Q_2$ are drawn such that 
# $\dfrac{P_0 Q_0}{P_0 P_1} = \dfrac{P_1 Q_1}{P_1 P_2} = \dfrac{P_2 Q_2}{P_2 
# P_3} = t$, with $t$ in $[0, 1]$. On the segments $Q_0 Q_1$ and $Q_1 Q_2$ the 
# points $R_0$ and $R_1$ are drawn such that$\dfrac{Q_0 R_0}{Q_0 Q_1} = 
# \dfrac{Q_1 R_1}{Q_1 Q_2} = t$ for the same value of $t$. On the segment $R_0 
# R_1$ the point $B$ is drawn such that $\dfrac{R_0 B}{R_0 R_1} = t$ for the 
# same value of $t$. The Bézier curve defined by the points $P_0, P_1, P_2, 
# P_3$ is the locus of $B$ as $Q_0$ takes all possible positions on the 
# segment $P_0 P_1$. (Please note that for all points the value of $t$ is the 
# same. )From the construction it is clear that the Bézier curve will be 
# tangent to the segments $P_0 P_1$ in $P_0$ and $P_2 P_3$ in $P_3$. A cubic 
# Bézier curve with $P_0 = (1, 0), P_1 = (1, v), P_2 = (v, 1),$ and $P_3 = (0, 
# 1)$ is used to approximate a quarter circle. The value $v \gt 0$ is chosen 
# such that the area enclosed by the lines $O P_0, OP_3$ and the curve is 
# equal to $\dfrac{\pi}{4}$ (the area of the quarter circle). By how many 
# percent does the length of the curve differ from the length of the quarter 
# circle?That is, if $L$ is the length of the curve, calculate $100 \times 
# \dfrac{L - \frac{\pi}{2}}{\frac{\pi}{2}}$Give your answer rounded to 10 
# digits behind the decimal point.
# 
# by lcsm29 http://github.com/lcsm29/project-euler
import timed


def dummy(n):
    pass


if __name__ == '__main__':
    n = 1000
    i = 10000
    prob_id = 363
    timed.caller(dummy, n, i, prob_id)
