99 lines
2.2 KiB
Python
99 lines
2.2 KiB
Python
def parse(file):
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while True:
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a = file.readline()
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if not a:
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break
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b = file.readline()
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pr = file.readline()
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blank = file.readline()
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ax, ay = parse_numbers(a)
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bx, by = parse_numbers(b)
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px, py = parse_numbers(pr)
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yield ax,ay, bx,by, px,py
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import re
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def parse_numbers(s):
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return [int(x) for x in re.findall(r"\d+", s)]
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from fractions import Fraction as F
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import numpy
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def solve(file):
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total = 0
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total2 = 0
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records = list(parse(file))
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for record in records:
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#print(record)
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ax, ay, bx, by, px, py = record
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# Part 1
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# Solve the linear equation:
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# k_a ax + k_b * bx = px
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# k_a ay + k_b * by = py
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M = numpy.array([[ax, bx], [ay, by]])
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V = numpy.array([px, py])
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sol = numpy.linalg.solve(M,V)
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k1 = int(round(sol[0]))
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k2 = int(round(sol[1]))
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if k1*ax + k2*bx == px and k1*ay + k2*by == py:
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print("yes", k1, k2)
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total += 3*k1 + k2
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for record in records:
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ax, ay, bx, by, px, py = record
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# Part 2
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#
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# Can't use numpy because the values are too large for floats
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# So here's an ad-hoc solver
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px += 10000000000000
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py += 10000000000000
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m00, m01 = ax, bx
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m10, m11 = ay, by
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v1 = px
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v2 = py
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# subtract c*eq1 from eq2 to cancel first coefficient
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c = F(m10,m00)
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m10 -= c*m00
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m11 -= c*m01
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v2 -= c*v1
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# scale eq2
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c = F(1,m11)
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v2 *= c
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m11 *= c
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# subtract c*eq2 from eq1 to cancel second coefficient
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c = F(m01,m11)
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m01 -= c*m11
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v1 -= c*v2
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# scale eq1
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c = F(1,m00)
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v1 *= c
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m00 *= c
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# we now have
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# 1*k_a + 0 = v1
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# 0 + 1*k_b = v2
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print(m00, m01, m10, m11, v1, v2)
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if v1.denominator == 1 and v2.denominator == 1 and v1 > 0 and v2 > 0:
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if v1*ax + v2*bx == px and v1*ay + v2*by == py:
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print("yes2", v1, v2)
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total2 += 3*v1 + v2
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print(total)
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print(total2)
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solve(open("sample1.in"))
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solve(open("input"))
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