Created
November 8, 2015 02:51
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Problem 3 "Planet Mining" Solution
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from random import uniform | |
numberOfCases = 1000 | |
print(numberOfCases + 1) | |
# an edge case to test for | |
print('0.1 0.1 1 1') | |
for i in range(numberOfCases): | |
starting_concentration = uniform(0.25, 0.5) | |
ending_concentration = uniform(0.001, 0.01) | |
volume = uniform(1, 10) | |
outflow_rate = uniform(1, 10) | |
print('{0:0.3f} {1:0.3f} {2:0.3f} {3:0.3f}'.format(starting_concentration, ending_concentration, volume, outflow_rate)) |
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from math import log | |
N = int(input().strip()) | |
for i in range(N): | |
starting_concentration, ending_concentration, volume, outflow_rate = map(float, input().strip().split()) | |
# the formula for model concentration is exponential, and can be written like this: | |
# concentration(t) = concentration(0) * exp(-outflow_rate * t / volume) | |
# If we know concentration(t) = ending_concentration, we can solve for t. This is the formula you get | |
ending_time = volume / outflow_rate * log(starting_concentration / ending_concentration) | |
# now pretty print | |
print('{0:0.3f}h'.format(ending_time)) |
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