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Optimal solution in Clear category for Feed Pigeons by veky
def pbs(cond, start=0):
"""last int after start satisfying cond
Precondition: cond(n)=>cond(n-1) & cond(start) & exists n>start not cond(n)
Complexity: max. 2*ld(result-start), calls cond on distinct numbers only"""
step = 1
while cond(start + step): step <<= 1
start += step >> 1
step >>= 2
while step:
if cond(start + step): start += step
step >>= 1
return start
def checkio(n):
k = pbs(lambda k: k**3 - k <= 6*n)
return max(n - (k**3 - k)//6, (k**2 - k)//2)
Feb. 6, 2014
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