#
# Using Heron's formula from Lubomir Markov's extension of Goehl's Method
#
M = []
from time import time
st = time()
for m in xrange(1, 1001):
#print "m:",m,
psum = 0
#u = 2*m
fd0 = divisors(2 * m) # Factoring u = 2m
for u in fd0:
ul = u * 3**.5 # choosing factors up to u*sqrt(3)
mxul = int(ul) + 1
for v in xrange(1, mxul):
if gcd(u, v) != 1: continue # choosing v coprime to u
f = 4 * m * m * (u * u + v * v) # = delta1 * delta2
fd = divisors(f) # obtaining factors for d1,d2
ul2 = int(
2 * m *
((u * u + v * v)**.5)) # obtaining sqrt to avoid repetitions
#print "!",m,u,v # showing m,u,v for curiosity's sake
for d1 in fd:
if ((2 * m) * (v**2 - u**2) / u
) <= d1 <= ul2: #choosing range of delta1 to ensure A=mP
if (d1 + 2 * m * u) % v: continue
d2 = f / d1
if (d2 + 2 * m * u) % v:
print "oooooooooooooops:", m, d1, d2, P
continue
a = (d1 + 2 * m * u) / v + (2 * m * v) / u
def notcoprime(cp, Y):
for yy in Y:
if gcd(yy, cp) != 1:
return yy
return -1
def notcoprime(cp,Y):
for yy in Y:
if gcd(yy,cp)!=1:
return yy
return -1
summ = 0
#
# Using Heron's formula from Lubomir Markov's extension of Goehl's Method
#
M = []
for m in xrange(1, 1001):
print "m:", m,
psum = 0
# u = 2*m
fd0 = divisors(2 * m) # Factoring u = 2m
for u in fd0:
ul = u * 3 ** 0.5 # choosing factors up to u*sqrt(3)
mxul = int(ul) + 1
for v in xrange(1, mxul):
if gcd(u, v) != 1:
continue # choosing v coprime to u
f = 4 * m * m * (u * u + v * v) # = delta1 * delta2
fd = divisors(f) # obtaining factors for d1,d2
ul2 = 2 * m * ((u * u + v * v) ** 0.5) # obtaining sqrt to avoid repetitions
# print "!",m,u,v # showing m,u,v for curiosity's sake
for d1 in fd:
if ((2 * m) * (v ** 2 - u ** 2) / u) <= d1 <= ul2: # choosing range of delta1 to ensure A=mP
d2 = f / d1
a = (d1 + 2 * m * u) / v + (2 * m * v) / u
b = (d2 + 2 * m * u) / v + (2 * m * v) / u
c = (d1 + d2 + 4 * m * u) / v
A = ta(a, b, c)
P = a + b + c
R = A / float(P) # paranoid check to ensure ratios match
if R == int(R): # Making sure Ratio is integer,probably not necessary
#
#
# Euler Problem 283
#
#
from Functions3 import GCD,gcd
summ = 0
ctr,ctr1 = 0,0
for m in xrange(2, 4000):
for n in xrange(1, m):
if gcd(m,n)!=1 or (m-n)%2==0: continue
a = (m*m - n*n)
b = 2*m*n
c = (m*m + n*n)
#if GCD((a,b,c))>1:continue
#print m,n,":",a,b,c
#if a**2 +b**2 == c**2:
A0=a*b/2;P0=a+b+c;R=float(A0)/P0
ctr+=1
#print m,n,":",a,b,c,":",A0,P0,R
if R < 1001:
#print a,b,c,":",A0,P0,R
#summ += P0
i=1
while 1:
A = A0*i**2
P = P0*i
R = float(A)/P
#
#
# Euler Problem 283
#
#
from Functions3 import GCD, gcd
summ = 0
ctr, ctr1 = 0, 0
for m in xrange(2, 4000):
for n in xrange(1, m):
if gcd(m, n) != 1 or (m - n) % 2 == 0: continue
a = (m * m - n * n)
b = 2 * m * n
c = (m * m + n * n)
#if GCD((a,b,c))>1:continue
#print m,n,":",a,b,c
#if a**2 +b**2 == c**2:
A0 = a * b / 2
P0 = a + b + c
R = float(A0) / P0
ctr += 1
#print m,n,":",a,b,c,":",A0,P0,R
if R < 1001:
#print a,b,c,":",A0,P0,R
#summ += P0
i = 1
while 1:
A = A0 * i**2
#
# Using Heron's formula from Lubomir Markov's extension of Goehl's Method
#
#M=[]
from time import time
st = time()
for m in xrange(1,1001):
print "m:",m,
psum = 0 # used to determine P sums for each m
#u = 2*m
fd0 = divisors(2*m) # Factoring u = 2m
for u in fd0:
ul = u*3**.5 # choosing factors up to u*sqrt(3)
mxul = int(ul)+1
for v in xrange(1,mxul):
if gcd(u,v)!=1:continue # choosing v coprime to u
f = 4*m*m*(u*u +v*v) # = delta1 * delta2
fd = divisors(f) # obtaining factors for d1,d2
ul2 = int(2*m*((u*u+v*v)**.5)) # obtaining sqrt to avoid repetitions
#print "!",m,u,v # showing m,u,v for curiosity's sake
for d1 in fd:
if ((2*m)*(v**2-u**2)/u)<= d1 <=ul2: #choosing range of delta1 to ensure A=mP
if (d1+2*m*u)%v:continue
d2 = f/d1
if (d2+2*m*u)%v:continue
a = (d1+2*m*u)/v + (2*m*v)/u
b = (d2+2*m*u)/v + (2*m*v)/u
c = (d1+d2+4*m*u)/v
#A = ta(a,b,c)
P = a+b+c
