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 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 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
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
#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
print(sum(1 for i in p if i == 1))#, time()-t
#!/usr/bin/python
# -*- coding: utf-8 -*-
#Consider the fraction, n/d, where n and d are positive integers. If n<d and HCF(n,d)=1, it is called a reduced proper fraction.
#If we list the set of reduced proper fractions for d ≤ 8 in ascending order of size, we get:
#1/8, 1/7, 1/6, 1/5, 1/4, 2/7, 1/3, 3/8, 2/5, 3/7, 1/2, 4/7, 3/5, 5/8, 2/3, 5/7, 3/4, 4/5, 5/6, 6/7, 7/8
#It can be seen that there are 3 fractions between 1/3 and 1/2.
#How many fractions lie between 1/3 and 1/2 in the sorted set of reduced proper fractions for d ≤ 12,000?
#Note: The upper limit has been changed recently.
#Answer:
#7295372
from time import time; t=time()
from mathplus import gcd
D=12000
cnt = 0
for n in range(5, D+1):
for m in range(n//3+1, (n+1)//2):
if gcd(n, m) == 1: cnt += 1
print(cnt)#, time()-t
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
ss = 0
MM = M/2+1
p = [0] * MM
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
ss = 0
MM = M/2+1
p = [0] * MM
