Exemplo n.º 1
0
def solution2(a, b, n=100000, max_y=None, min_y=0):
    max_y = max([f(random.uniform(a, b)) for i in xrange(10 * n)]) if not None else max_y
    print max_y
    coordinates = [(random.uniform(a, b), random.uniform(min_y, max_y)) for i in xrange(n)]

    # file2_b = open('test2_b.txt', 'r+')
    k = 0
    for x, y in coordinates:
        # print >> file2_b, (x, y)
        if f(x) >= y:
            k += 1
    # file2_b.close()

    return (max_y - min_y) * (b - a) * (float(k) / n)
Exemplo n.º 2
0
def solution2(a, b, n=100000, max_y=None, min_y=0):
    max_y = max([f(random.uniform(a, b))
                 for i in xrange(10 * n)]) if not None else max_y
    print max_y
    coordinates = [(random.uniform(a, b), random.uniform(min_y, max_y))
                   for i in xrange(n)]

    # file2_b = open('test2_b.txt', 'r+')
    k = 0
    for x, y in coordinates:
        # print >> file2_b, (x, y)
        if f(x) >= y:
            k += 1
    # file2_b.close()

    return (max_y - min_y) * (b - a) * (float(k) / n)
Exemplo n.º 3
0
# Python3

from solution1 import startupName as f

qa = [(['coolcompany', 'nicecompany', 'legendarycompany'], ['e', 'l']),
      (['nameone', 'nametwo', 'namethree'], ['o', 't']),
      (['heh', 'hah', 'funny'], ['h']),
      (['amazon', 'ebay', 'aliexpress'], ['e']), (['abc', 'def', 'hijk'], []),
      (['cool', 'looc', 'loco'], [])]

for *q, a in qa:
    for i, e in enumerate(q):
        print('input{0}: {1}'.format(i + 1, e))
    ans = f(*q)
    if ans != a:
        print('  [failed]')
        print('    output:', ans)
        print('    expected:', a)
    else:
        print('  [ok]')
        print('    output:', ans)
    print()
Exemplo n.º 4
0
                      ['Five', 'Four', 'One', 'Six', 'Two'],
                      ['Five', 'Four', 'One', 'Three', 'Two'],
                      ['Five', 'Four', 'Seven', 'Six', 'Three'],
                      ['Five', 'Four', 'Seven', 'Six', 'Two'],
                      ['Five', 'Four', 'Seven', 'Three', 'Two'],
                      ['Five', 'Four', 'Six', 'Three', 'Two'],
                      ['Five', 'One', 'Seven', 'Six', 'Three'],
                      ['Five', 'One', 'Seven', 'Six', 'Two'],
                      ['Five', 'One', 'Seven', 'Three', 'Two'],
                      ['Five', 'One', 'Six', 'Three', 'Two'],
                      ['Five', 'Seven', 'Six', 'Three', 'Two'],
                      ['Four', 'One', 'Seven', 'Six', 'Three'],
                      ['Four', 'One', 'Seven', 'Six', 'Two'],
                      ['Four', 'One', 'Seven', 'Three', 'Two'],
                      ['Four', 'One', 'Six', 'Three', 'Two'],
                      ['Four', 'Seven', 'Six', 'Three', 'Two'],
                      ['One', 'Seven', 'Six', 'Three', 'Two']])]

for *q, a in qa:
    for i, e in enumerate(q):
        print('input{0}: {1}'.format(i + 1, e))
    ans = [*map(list, f(*q))]
    if ans != a:
        print('  [failed]')
        print('    output:', ans)
        print('    expected:', a)
    else:
        print('  [ok]')
        print('    output:', ans)
    print()
Exemplo n.º 5
0
    (['abs',
      'math.sin',
      'lambda x: 3 * x / 2'],
     3.1415, 1),
    (['math.sin',
      'math.cos',
      'lambda x: x * 2',
      'lambda x: x ** 2'],
     1, -0.404239153852),
    (['lambda z: z',
      'lambda z: 1.0 * z / 13'],
     -1000, -76.9230769231),
    (['float'],
     1000, 1000),
    (['abs'],
     -20, 20)
]

for *q, a in qa:
    for i, e in enumerate(q):
        print('input{0}: {1}'.format(i + 1, e))
    ans = round(f(*q), 6)
    if ans != round(a, 6):
        print('  [failed]')
        print('    output:', ans)
        print('    expected:', a)
    else:
        print('  [ok]')
        print('    output:', ans)
    print()
Exemplo n.º 6
0
# Python3

from solution1 import rangeFloat as f

qa = [
    ([5], [0, 1, 2, 3, 4]),
    ([0.5, 7.5], [0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5]),
    ([-1.1, 2.4, 0.3], [-1.1, -0.8, -0.5, -0.2, 0.1, 0.4, 0.7, 1, 1.3, 1.6, 1.9, 2.2]),
    ([5.6, 3.2], []),
    ([-3.2, -123, 10], []),
    ([-4], []),
    ([10.4, 3.2, -1.2], [10.4, 9.2, 8, 6.8, 5.6, 4.4, 3.2]),
    ([10, 10, 3], []),
    ([33.3], [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33])
]

for *q, a in qa:
    for i, e in enumerate(q):
        print('input{0}: {1}'.format(i + 1, e))
    ans = [ round(i, 5) for i in f(*q) ]
    if ans != a:
        print('  [failed]')
        print('    output:', ans)
        print('    expected:', a)
    else:
        print('  [ok]')
        print('    output:', ans)
    print()
Exemplo n.º 7
0
from solution1 import kthPermutation as f

qa = [
    ([1, 2, 3, 4, 5], 4,
     [1, 2, 4, 5, 3]),
    ([1, 2], 1,
     [1, 2]),
    ([11, 22, 31, 43, 56], 120,
     [56, 43, 31, 22, 11]),
    ([14, 25, 27, 29, 30, 40, 55, 89, 100, 239], 238,
     [14, 25, 27, 29, 40, 239, 100, 55, 89, 30]),
    ([14, 25, 27, 29, 30, 40, 55, 89, 100, 239], 3628800,
     [239, 100, 89, 55, 40, 30, 29, 27, 25, 14]),
    ([50, 100, 123, 789], 15,
     [123, 100, 50, 789])
]

for *q, a in qa:
    for i, e in enumerate(q):
        print('input{0}: {1}'.format(i + 1, e))
    ans = list(f(*q))
    if ans != a:
        print('  [failed]')
        print('    output:', ans)
        print('    expected:', a)
    else:
        print('  [ok]')
        print('    output:', ans)
    print()