def pe169(n, k=None):
if n < 0:
return 0
if n <= 1:
return 1
k2 = int(log(n, 2))
if k is None:
k = k2
elif k < k2 - 1:
return 0
else:
k = min(k, k2)
if k == 0:
return 1 if n < 3 else 0
kmax = 2 ** k
return pe169(n, k-1)+pe169(n-kmax, k-1)+pe169(n-kmax*2, k-1)
def pe169(n, k=None):
if n < 0:
return 0
if n <= 1:
return 1
k2 = int(log(n, 2))
if k is None:
k = k2
elif k < k2 - 1:
return 0
else:
k = min(k, k2)
if k == 0:
return 1 if n < 3 else 0
kmax = 2**k
return pe169(n, k - 1) + pe169(n - kmax, k - 1) + pe169(
n - kmax * 2, k - 1)
#!/usr/bin/python
# -*- coding: utf-8 -*-
#2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
#What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
#Answer:
#232792560
from time import time; t=time()
from mathplus import prime, log
NUM = 20
n = NUM
logn = log(NUM)
ret = 1
for p in prime():
if p > n: break
ret *= p**int(logn/log(p))
print(ret)#, time()-t
#!/usr/bin/python
# -*- coding: utf-8 -*-
#2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder.
#What is the smallest positive number that is evenly divisible by all of the numbers from 1 to 20?
#Answer:
#232792560
from time import time
t = time()
from mathplus import prime, log
NUM = 20
n = NUM
logn = log(NUM)
ret = 1
for p in prime():
if p > n: break
ret *= p**int(logn / log(p))
print(ret) #, time()-t
