def is_prime(n):
if n <= M: return sieves[n]
for i in primes:
if n % i == 0: return False
for i in range(max_prime+2, isqrt(n), 2):
if n % i == 0: return False
return True
def is_prime(n):
if n <= M: return sieves[n]
for i in primes:
if n % i == 0: return False
for i in range(max_prime + 2, isqrt(n), 2):
if n % i == 0: return False
return True
def is_prime(n):
if n < MM: return sieves[n]
isqrtn = isqrt(n)
for p in primes:
if p > isqrtn: break
if n % p == 0: return False
return True
def is_triangle_word(s):
num = sum((ord(c)-ord('A')+1) for c in s)
if num in triangle_numbers: return True
if num < triangle_numbers[-1]: return False
n2 = num*8+1
n = isqrt(n2)
return n*n == n2
def is_prime(n):
if n < MM: return sieves[n]
isqrtn = isqrt(n)
for p in primes:
if p > isqrtn: break
if n % p == 0: return False
return True
def is_prime_plus(n):
if n % 2 == 0: return n == 2
if n % 3 == 0: return n == 3
if n < M: return sieves[n]
isqrtn = isqrt(n)
for p in primes:
if p > isqrtn: return True
if n % p == 0: return False
def test(s):
if full_f[s[0]] == 0: return 0
c = full_f[s[-1]]
if c in (0, 2, 3, 7, 8): return 0
d = full_f[s[-2]]
if c == 5 and d != 2: return 0
if c == 6 and d % 2 == 0: return 0
if d % 2 == 1: return 0
n = reduce(lambda x,y: x*10+y, (full_f[c] for c in s))
return n if isqrt(n)**2 == n else 0
def test(s):
if full_f[s[0]] == 0: return 0
c = full_f[s[-1]]
if c in (0, 2, 3, 7, 8): return 0
d = full_f[s[-2]]
if c == 5 and d != 2: return 0
if c == 6 and d % 2 == 0: return 0
if d % 2 == 1: return 0
n = reduce(lambda x, y: x * 10 + y, (full_f[c] for c in s))
return n if isqrt(n)**2 == n else 0
def pe(n, m):
s, t = 0, isqrt(n)
cache = [(n // i % m) for i in range(1, n // t + 1)]
for i in range(t - 1):
k = cache[i]
s += k * (k + 1) * (2 * k + 1) // 6
k = cache[t - 1]
s -= (t - 1) * k * (k + 1) * (2 * k + 1) // 6
s %= m
for i in range(n // t):
s += (i + 1) ** 2 * cache[i]
return s % m
def pe(n, m):
s, t = 0, isqrt(n)
cache = [(n // i % m) for i in range(1, n // t + 1)]
for i in range(t - 1):
k = cache[i]
s += k * (k + 1) * (2 * k + 1) // 6
k = cache[t - 1]
s -= (t - 1) * k * (k + 1) * (2 * k + 1) // 6
s %= m
for i in range(n // t):
s += (i + 1)**2 * cache[i]
return s % m
def pe(M):
s = 0
for m in range(2, isqrt(M / 2)):
mm = (M - 1) // (2 * m)
m2 = m * m
start = 1 if m % 2 == 0 else 2
for n in range(start, min(mm - m + 1, m), 2):
if gcd(m, n) != 1:
continue
n2 = n * n
if (m2 + n2) % (m2 - n2 - 2 * m * n) != 0:
continue
s += mm // (m + n)
return s
def solve2(n):
u = 1
m = isqrt(n)
for f in primes:
if f > m:
u *= 3
break
k = 0
while n % f == 0:
n //= f
k += 1
u *= (2 * k + 1)
if n == 1: break
return u
def pe(M):
s = 0
for m in range(2, isqrt(M/2)):
mm = (M-1) // (2*m)
m2 = m * m
start = 1 if m % 2 == 0 else 2
for n in range(start, min(mm-m+1, m), 2):
if gcd(m, n) != 1:
continue
n2 = n * n
if (m2+n2) % (m2-n2-2*m*n) != 0:
continue
s += mm // (m+n)
return s
def pe10(MAX):
marked = [0] * MAX
value = 3
s = 2
limit = isqrt(MAX)+1
while value < limit:
if marked[value] == 0:
s += value
for i in range(value*value, MAX, value*2):
marked[i] = 1
value += 2
for i in range(limit//2*2+1, MAX, 2):
if marked[i] == 0:
s += i
return s
def solve2(n):
u = 1
m = isqrt(n)
for f in primes:
if f > m:
u *= 3
break
k = 0
while n % f == 0:
n //= f
k += 1
u *= 2 * k + 1
if n == 1:
break
return u
def get_root(n, m):
#assert 0 < n < 100
root = [0] * m
root[0] = isqrt(n)
n -= root[0]**2
if n == 0: return [root[0]]
a2 = root[0] * 20
for i in range(1, m):
n *= 100
for b in range(n // a2, -1, -1):
k = (a2 + b) * b
if k < n:
n -= k
break
root[i] = b
a2 = 10 * (a2 + 2 * b)
return root
def get_root(n, m):
#assert 0 < n < 100
root = [0]*m
root[0] = isqrt(n)
n -= root[0]**2
if n == 0: return [root[0]]
a2 = root[0]*20
for i in range(1, m):
n *= 100
for b in range(n//a2, -1, -1):
k = (a2+b)*b
if k < n:
n -= k
break
root[i] = b
a2 = 10*(a2+2*b)
return root
def pe171(n, m):
up = isqrt(n * 81)
uplimit = up ** 2 + 1
d = [[[0, 0] for j in range(uplimit)] for i in range(n)]
for k in range(10):
d[0][k * k] = [1, k]
for i in range(1, n):
ii = 10 ** i if i < m else 0
for j in range(uplimit):
for k in range(10):
kk = k * k
if j < kk:
break
p, q = d[i - 1][j - kk]
d[i][j][0] += p
d[i][j][1] += q + ii * k * p
# print [(k, d[n-1][k*k]) for k in range(up+1)]
return str(sum(d[n - 1][k * k][1] for k in range(1, up + 1)))[-m:]
def pe171(n, m):
up = isqrt(n * 81)
uplimit = up**2 + 1
d = [[[0, 0] for j in range(uplimit)] for i in range(n)]
for k in range(10):
d[0][k * k] = [1, k]
for i in range(1, n):
ii = 10**i if i < m else 0
for j in range(uplimit):
for k in range(10):
kk = k * k
if j < kk:
break
p, q = d[i - 1][j - kk]
d[i][j][0] += p
d[i][j][1] += q + ii * k * p
# print [(k, d[n-1][k*k]) for k in range(up+1)]
return str(sum(d[n - 1][k * k][1] for k in range(1, up + 1)))[-m:]
def func(n):
global primes, sieves, SIZE, PRIME_CNT
factors = 1
while n % 2 == 0:
factors += 1
n /= 2
if factors >= 2: factors -= 1
isqrtn = isqrt(n)
while isqrtn >= PRIME_CNT:
SIZE = 10*SIZE
primes, sieves = get_primes_by_sieve(SIZE)
PRIME_CNT = len(primes)
for i in range(1,isqrtn):
cnt = 1
p = primes[i]
while n % p == 0:
cnt += 1
n /= p
factors *= cnt
if n == 1: break
return factors
def func(n):
global primes, sieves, SIZE, PRIME_CNT
factors = 1
while n % 2 == 0:
factors += 1
n /= 2
if factors >= 2: factors -= 1
isqrtn = isqrt(n)
while isqrtn >= PRIME_CNT:
SIZE = 10 * SIZE
primes, sieves = get_primes_by_sieve(SIZE)
PRIME_CNT = len(primes)
for i in range(1, isqrtn):
cnt = 1
p = primes[i]
while n % p == 0:
cnt += 1
n /= p
factors *= cnt
if n == 1: break
return factors
def pe(M):
rad = [0] * M
rad[1] = 1
for p in range(2, M):
if rad[p] > 0: continue
isqrtp = isqrt(p)
flag = True
for q in range(2, isqrtp + 1):
if p % q != 0: continue
pp = p
while pp % q == 0:
pp //= q
qq = q * rad[pp]
pp = p
while pp < M:
rad[pp] = qq
pp *= q
flag = (q > isqrtp)
break
if flag:
pp = p
while pp < M:
rad[pp] = p
pp *= p
s = 0
for c in range(3, M):
cc = (c - 1) // rad[c]
if rad[c - 1] <= cc:
s += c
if cc < 6: continue
if c % 2 == 0 and cc < 15: continue
if c % 3 == 0 and cc < 10: continue
for a in range(2, c // 2):
b = c - a
if rad[a] > cc or rad[b] > cc: continue
if rad[a] * rad[b] <= cc and gcd(a, b) == 1:
s += c
return s
#840
from time import time
t = time()
from mathplus import isqrt, gcd
M = 1000
p = [0] * (M // 2 + 1)
q = [0] * (M // 2 + 1)
for n in range(1, M // 2 + 1):
for k in range(1, n):
if n % k != 0: continue
m = n // k
if q[m] == 0:
for i in range(isqrt(m // 2) + 1, isqrt(m) + 1):
if m % i != 0: continue
u, v = i, m // i - i
if (u + v) % 2 != 0 and gcd(u, v) == 1:
q[m] += 1
#if n == 60:
#print k, u, v, k*(u*u-v*v), 2*k*u*v, k*(u*u+v*v)
p[n] += q[m]
print(max((x, i) for i, x in enumerate(p))[1] * 2) #, time()-t
'''
from time import time; t=time()
from mathplus import isqrt, gcd
M = 1000
from time import time
t = time()
from mathplus import isqrt
M = 100
def get_root(n, m):
#assert 0 < n < 100
root = [0] * m
root[0] = isqrt(n)
n -= root[0]**2
if n == 0: return [root[0]]
a2 = root[0] * 20
for i in range(1, m):
n *= 100
for b in range(n // a2, -1, -1):
k = (a2 + b) * b
if k < n:
n -= k
break
root[i] = b
a2 = 10 * (a2 + 2 * b)
return root
print(sum(sum(get_root(n, 100)) for n in range(M)
if n != isqrt(n)**2)) #, time()-t
#!/usr/bin/python # -*- coding: utf-8 -*- #The smallest number expressible as the sum of a prime square, prime cube, and prime fourth power is 28. In fact, there are exactly four numbers below fifty that can be expressed in such a way: #28 = 22 + 23 + 24 #33 = 32 + 23 + 24 #49 = 52 + 23 + 24 #47 = 22 + 33 + 24 #How many numbers below fifty million can be expressed as the sum of a prime square, prime cube, and prime fourth power? #Answer: #1097343 from time import time; t=time() from mathplus import get_primes_by_sieve, isqrt, pow M = 50000000 sqrtM = isqrt(M) primes, _ = get_primes_by_sieve(sqrtM) p2 = [p*p for p in primes] p3 = [p**3 for p in primes[:int(pow(M, 1.0/3))]] p4 = [p**4 for p in primes[:isqrt(sqrtM)+1]] pool = set(p+q for p in p3 for q in p4 if p+q < M) pool = set(p+q for p in pool for q in p2 if p+q < M) print(len(pool))#, time()-t
from time import time
t = time()
from mathplus import isqrt, reduce, op
M = 12000
n = M
D = [2 * i for i in range(n + 1)]
def mul(l):
return reduce(op.mul, l, 1)
p = [[i] for i in range(2, isqrt(2 * n) + 1)]
while p:
q = []
for l in p:
m, s = mul(l), sum(l) - len(l)
r = l[-1]
mr = m * r
while mr <= 2 * n:
nn = mr - s - r + 1
if nn > n: break
q.append(l + [r])
D[nn] = min(D[nn], mr)
r += 1
mr += m
p = q
# -*- coding: utf-8 -*-
#Triangle, pentagonal, and hexagonal numbers are generated by the following formulae:
#Triangle Tn=n(n+1)/2 1, 3, 6, 10, 15, ...
#Pentagonal Pn=n(3n−1)/2 1, 5, 12, 22, 35, ...
#Hexagonal Hn=n(2n−1) 1, 6, 15, 28, 45, ...
#It can be verified that T285 = P165 = H143 = 40755.
#Find the next triangle number that is also pentagonal and hexagonal.
#Answer:
#1533776805
from time import time
t = time()
from mathplus import isqrt
n = 143
while True:
n += 1
hn = n * (2 * n - 1)
vpn = hn * 6
vn = isqrt(vpn) + 1
if vn % 3 == 0 and vn * (vn - 1) == vpn:
print(hn) #, time()-t
break
# http://mathworld.wolfram.com/HexagonalPentagonalNumber.html
#9 3 7 48073
#For d = 0 to 9, the sum of all S(4, d) is 273700.
#Find the sum of all S(10, d).
#Answer:
#612407567715
from time import time
t = time()
from mathplus import get_primes_by_sieve, isqrt, product, permutations
M = 10
EM = 10**(M - 1)
MM = isqrt(10**M)
primes, sieves = get_primes_by_sieve(MM)
def is_prime(n):
if n < MM: return sieves[n]
isqrtn = isqrt(n)
for p in primes:
if p > isqrtn: break
if n % p == 0: return False
return True
def gen(l, m, i):
others = list(range(10))
def pe233():
# Unreadable... = =||| #FML
primes, _ = get_primes_by_sieve(N // (5**3 * 13**2) + 1, odd_only=True)
p4k1 = []
p4k3 = []
for p in primes:
if p % 4 == 1:
p4k1.append(p)
else:
p4k3.append(p)
len_c = N // (5**3 * 13**2 * 17) + 1
choices = [False] * len_c
choices[1] = True
for p in p4k3:
if p >= len_c: break
choices[p] = True
for i in range(3, len_c, 2):
if choices[i]: continue
l = isqrt(i)
for p in primes:
if p > l:
break
if i % p == 0:
if p % 4 == 3:
choices[i] = choices[i // p]
break
for i in range(1, (len_c + 1) // 2):
choices[i * 2] = choices[i]
cs = [0] * len_c
ss = 0
for i, f in enumerate(choices):
if f: ss += i
cs[i] = ss
s = 0
Ndd = N // (5 * 5 * 13)
Nd5 = N // 5
for p1 in p4k1:
k = p1**3
if k > Ndd: break
for p2 in p4k1:
if p2 == p1: continue
k2 = k * p2**2
if k2 > Nd5: break
for p3 in p4k1:
if p3 == p1 or p3 == p2: continue
k3 = k2 * p3
if k3 > N: break
s += k3 * cs[N // k3]
for p1 in p4k1:
k = p1**7
if k > Nd5: break
for p2 in p4k1:
if p2 == p1: continue
k2 = k * p2**3
if k2 > N: break
s += k2 * cs[N // k2]
for p1 in p4k1:
k = p1**10
if k > Nd5: break
for p2 in p4k1:
if p2 == p1: continue
k2 = k * p2**2
if k2 > N: break
s += k2 * cs[N // k2]
return s
def is_pentagonal(n):
n2 = n * 24 + 1
n = isqrt(n2)
return n * n == n2 and (n + 1) % 6 == 0
#120 cm: (30,40,50), (20,48,52), (24,45,51)
#Given that L is the length of the wire, for how many values of L ≤ 1,500,000 can exactly one integer sided right angle triangle be formed?
#Note: This problem has been changed recently, please check that you are using the right parameters.
#Answer:
#161667
from time import time; t=time()
from mathplus import gcd, isqrt
M = 1500000
q = [0]*(M+1)
for m in range(2, isqrt(M/2)+1):
for n in range(m % 2 + 1, min(m, M//(2*m)-m+1), 2):
#length = 2*m*(m+n)
#if length > M: break
#x = m*m-n*n
#y = 2*m*n
#if gcd(x,y)==1:
if gcd(m, n)==1:
length = 2*m*(m+n)
q[length] += 1
p = [0]*(M+1)
for m, c in enumerate(q):
if c:
for n in range(m, M+1, m):
p[n] += c
M = 2000000
def tria(n):
return n * (n + 1)
def is_tria(n):
n4 = n * 8 + 1
return isqrt(n4)**2 == n4
#n*(n+1)*m*(m+1) = M*4
cnt = 0
while True:
cnt += 1
for k in (M - cnt, M + cnt):
sn = 0
n = 1
while True:
sn += n
n += 1
if sn > k: break
if k % sn != 0 or not is_tria(k // sn): continue
m = (isqrt(k / sn * 8 + 1) - 1) // 2
print((n - 1) * m) #, time()-t, n-1, m
import sys
sys.exit()
def is_tria(n):
n4 = n * 8 + 1
return isqrt(n4)**2 == n4
def test(u, v):
w = u * u + v * v
return w == isqrt(w)**2
#It was proposed by Christian Goldbach that every odd composite number can be written as the sum of a prime and twice a square.
#9 = 7 + 2×12
#15 = 7 + 2×22
#21 = 3 + 2×32
#25 = 7 + 2×32
#27 = 19 + 2×22
#33 = 31 + 2×12
#It turns out that the conjecture was false.
#What is the smallest odd composite that cannot be written as the sum of a prime and twice a square?
#Answer:
#5777
from time import time
t = time()
from mathplus import sieve, isqrt
M = 10000
p = sieve(M)
for n in range(35, M, 2):
if p[n]: continue
for i in range(1, isqrt((n - 3) / 2) + 1):
if p[n - 2 * i * i]: break
else:
print(n) #, time()-t
break
#!/usr/bin/python # -*- coding: utf-8 -*- #There are some prime values, p, for which there exists a positive integer, n, such that the expression n3 + n2p is a perfect cube. #For example, when p = 19, 83 + 82×19 = 123. #What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred. #How many primes below one million have this remarkable property? #Answer: #173 #https://github.com/skydark/project-euler/blob/master/131.py from time import time; t=time() from mathplus import isqrt, get_primes_by_sieve M = 10**6 primes, sieves = get_primes_by_sieve(M) m = isqrt(M/3.) ps = set(3*x*x+3*x+1 for x in range(m)) print(len(ps.intersection(primes)))#, time()-t
#There are ten composites below thirty containing precisely two, not necessarily distinct, prime factors: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26.
#How many composite integers, n < 108, have precisely two, not necessarily distinct, prime factors?
#Answer:
#17427258
from time import time
t = time()
from mathplus import isqrt, get_primes_by_sieve
from gen_primes import get_primes
#M = 30
M = 10**8
sqrtM = isqrt(M)
primes = get_primes(M // 2, odd_only=False)
#print time()-t
sieves = [0] * (sqrtM + 1)
pnext = 2
pnext_index = 0
p = 0
for i in range(sqrtM + 1):
if i == pnext:
pnext_index += 1
pnext = primes[pnext_index]
p += 1
sieves[i] = p
#print time()-t
ss = sieves[sqrtM]
s = ss * (ss + 1) // 2
#!/usr/bin/python
# -*- coding: utf-8 -*-
#Find the number of integers 1 < n < 107, for which n and n + 1 have the same number of positive divisors. For example, 14 has the positive divisors 1, 2, 7, 14 while 15 has 1, 3, 5, 15.
#Answer:
#986262
from time import time; t=time()
from mathplus import isqrt
M=10**7
sieves = [2]*M
sieves[0] = 0
sieves[1] = 1
for n in range(2, isqrt(M)):
nn = n*n
for m in range(nn, M, n): sieves[m] += 2
sieves[nn] -= 1
s = sum(1 for n in range(3, M) if sieves[n] == sieves[n-1])
print(s)#, time()-t
#Let N(n) be the number of t ≤ 1000000 such that t is type L(n); for example, N(15) = 832.
#What is ∑ N(n) for 1 ≤ n ≤ 10?
#Answer:
#209566
from time import time
t = time()
from mathplus import isqrt
M = 1000000
N = range(1, 11) #(15,)#
M //= 4
facts = [1] * (M + 1)
facts[0] = 0
for i in range(1, isqrt(M + 1)):
facts[i * i] -= 1
for i in range(1, M + 1):
for j in range(i, M + 1, i):
facts[j] += 1
facts[i] //= 2
s = [0] * len(N)
for i, n in enumerate(N):
s[i] = facts.count(n)
print(sum(s)) #, time()-t
