def main():
limit = 1500000
perimeters = [0]*(limit + 1)
for m in range(2, int(sqrt(limit>>1))):
for n in range(2 if m&1 else 1, m, 2):
if gcd(m, n) == 1:
p = m*(m+n)<<1 # a + b + c
if p <= limit:
for i in range(p, limit, p):
perimeters[i] += 1
return perimeters.count(1)
# for d in range(8, 1000001): # # lower< n/d < 3/7 # fractions = [target] # upper_n = int(d * upper) + 1 # lower_n = int(d * lower) # for n in range(lower_n, upper_n): # if gcd(d, n) == 1: # fractions.append((float(n)/d, str(n)+'/'+str(d))) # if len(fractions) > 1: # fractions.sort() # idx = fractions.index(target) # if fractions[idx-1][0] > lower: # lower = fractions[idx-1][0] # result = fractions[idx-1] # print result # only need to check n = (d*3/7) from Helper import gcd upper = 0 result = 0 for d in range(8, 1000001): n = int(d * 3.0/7) if gcd(d, n) == 1: temp = float(n) / d if temp > upper: upper = temp result = n print result
""" 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? """ from Helper import gcd limit = 12001 count = 0 for d in range(5, limit): lower = d / 3 + 1 upper = d / 2 + 1 for n in range(lower, upper): if gcd(n, d) == 1: count += 1 print count
# Generate pythagorean triple using
# http://en.wikipedia.org/wiki/Pythagorean_triple
# a = m^2 - n^2, b = 2mn, c = m^2 + n^2
# m > n, and m+n is odd and gcd(n,m) = 1
from math import sqrt
from Helper import gcd
Limit = 1500000
mlimit = int(sqrt(Limit / 2)) + 1 # a+b+c<Limit
# list to point how many times the same value
# a + b + c has been caculated
wires = [0] * Limit
count = 0
for m in range(2, mlimit):
for n in range(1, m):
if (m + n) % 2 == 1 and gcd(m, n) == 1:
a = m ** 2 - n ** 2
b = 2 * m * n
c = m ** 2 + n ** 2
length = a + b + c
while length < Limit:
wires[length] += 1
if wires[length] == 1:
count += 1
elif wires[length] == 2:
count -= 1
length += a + b + c
print count
#https://raw.githubusercontent.com/AaronJiang/ProjectEuler/master/py/problem073.py """ 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? """ from Helper import gcd limit = 12001 count = 0 for d in range(5, limit): lower = d / 3 + 1 upper = d / 2 + 1 for n in range(lower, upper): if gcd(n, d) == 1: count += 1 print count
"""
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 2/5 is the fraction immediately to the left of 3/7.
By listing the set of reduced proper fractions for d <= 1,000,000 in
ascending order of size, find the numerator of the fraction immediately
to the left of 3/7.
"""
from Helper import gcd
upper = 0
result = 0
for d in range(8, 1000001):
n = int(d * 3.0 / 7)
if gcd(d, n) == 1:
temp = float(n) / d
if temp > upper:
upper = temp
result = n
print result
