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## logistic_regression.ipynb

      
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                xeonqq
                / logistic_regression.ipynb
            
            
              Created
              August 4, 2015 13:24
            
          
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## Codility_MinMaxDivision.py
def blockPossible(A, bound, K):
    s = A[0]
    blocks = 1;
    for a in A[1:]:
        s += a
        if s>bound:
            blocks+=1
            s=a
        if blocks > K:
            return False

## codility_NumberSolitaire.py
# you can use print for debugging purposes, e.g.
# print "this is a debug message"

def solution(A):
    n = len(A)
    MIN_INT = -n * 20000
    v = [A[0]]+(n-1)*[MIN_INT]
    for i in xrange(1, n):
        for j in xrange(1, 7):
            if j <= i:

## MaxNonoverlappingSegments.py
def solution(A, B):
    # write your code in Python 2.7
    pre = -1
    N = len(A)
    ans=0
    for i in xrange(N):
        if A[i] > pre:
            ans+=1
            pre = B[i]
    return ans

## Triangle.py
def solution(A):
    # write your code in Python 2.7
    A.sort()
    for i in xrange(len(A)-2):
        if (((A[i] + A[i+1]) > A[i+2]) and ((A[i+2] + A[i+1]) > A[i]) and ((A[i+2] + A[i]) > A[i+1])):
            return 1
    return 0

## NumberOfDiscIntersections.py
def solution(A):
    # write your code in Python 2.7
    B = []
    for i in xrange(len(A)):
        B.append((i-A[i],i+A[i]))
    B = sorted(B, key=lambda x:x[0])
    cnt = 0
    for i in xrange(len(A)-1):
        right = B[i][1]
        start = i+1

## MaxDoubleSliceSum.py
def solution(A):
    # write your code in Python 2.7
    n = len(A)
    if n <=3:
        return 0
    leftMax = [0]*n
    rightMax = [0]*n
    for i in xrange(2,n-1):
        leftMax[i] = max(0,leftMax[i-1]+A[i-1])
    for i in xrange(n-3,0,-1):

## Equi.py
def solution(A):
    # write your code in Python 2.7
    s = []
    tmp = 0
    n = len(A)
    for a in A:
        tmp = a +  tmp
        s.append(tmp)
    s.append(0)

## try_bluetooth_function.c
void tryBT()
{

	delay(20000);
	led_green_flash();
	long baudrates[5] = {9600, 19200, 38400, 57600, 115200};
	String inputString="";
	for (int i =0; i< 5; i++)
	{
		DebugSerial.begin(baudrates[i]); // debug output

## MinPerimeterRectangle.py
from math import sqrt

def solution(N):
    # write your code in Python 2.7
    minperimeter = 2*(N + 1)

    if N == 1:
        return minperimeter

    denominator = 2
	def blockPossible(A, bound, K):
	s = A[0]
	blocks = 1;
	for a in A[1:]:
	s += a
	if s>bound:
	blocks+=1
	s=a
	if blocks > K:
	return False
	# you can use print for debugging purposes, e.g.
	# print "this is a debug message"

	def solution(A):
	n = len(A)
	MIN_INT = -n * 20000
	v = [A[0]]+(n-1)*[MIN_INT]
	for i in xrange(1, n):
	for j in xrange(1, 7):
	if j <= i:
	def solution(A, B):
	# write your code in Python 2.7
	pre = -1
	N = len(A)
	ans=0
	for i in xrange(N):
	if A[i] > pre:
	ans+=1
	pre = B[i]
	return ans
	def solution(A):
	# write your code in Python 2.7
	A.sort()
	for i in xrange(len(A)-2):
	if (((A[i] + A[i+1]) > A[i+2]) and ((A[i+2] + A[i+1]) > A[i]) and ((A[i+2] + A[i]) > A[i+1])):
	return 1
	return 0
	def solution(A):
	# write your code in Python 2.7
	B = []
	for i in xrange(len(A)):
	B.append((i-A[i],i+A[i]))
	B = sorted(B, key=lambda x:x[0])
	cnt = 0
	for i in xrange(len(A)-1):
	right = B[i][1]
	start = i+1
	def solution(A):
	# write your code in Python 2.7
	s = []
	tmp = 0
	n = len(A)
	for a in A:
	tmp = a + tmp
	s.append(tmp)
	s.append(0)
	void tryBT()
	{

	delay(20000);
	led_green_flash();
	long baudrates[5] = {9600, 19200, 38400, 57600, 115200};
	String inputString="";
	for (int i =0; i< 5; i++)
	{
	DebugSerial.begin(baudrates[i]); // debug output
	from math import sqrt

	def solution(N):
	# write your code in Python 2.7
	minperimeter = 2*(N + 1)

	if N == 1:
	return minperimeter

	denominator = 2