Daniele Fucini
7c055b749e
Added solutions for problem 71, 72, 73, 74 and 75 in C and python and made a few corrections in the code for other problems.
55 lines
1.4 KiB
Python
55 lines
1.4 KiB
Python
#!/usr/bin/python3
|
|
|
|
# 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 math import gcd
|
|
|
|
from timeit import default_timer
|
|
|
|
def main():
|
|
start = default_timer()
|
|
|
|
N = 1000000
|
|
|
|
max_ = 0
|
|
|
|
# For each denominator q, we need to find the biggest numerator p for which
|
|
# p/q<a/b, where a/b is 3/7 for this problem. So:
|
|
# pb<aq
|
|
# pb<=aq-1
|
|
# p<=(aq-1)/b
|
|
# So we can set p=(3*q-1)/7 (using integer division).
|
|
for i in range(2, N+1):
|
|
j = (3 * i - 1) // 7
|
|
|
|
if j / i > max_:
|
|
n = j
|
|
d = i
|
|
max_ = n / d
|
|
|
|
# Reduce the fractions if it's not already reduced.
|
|
if gcd(i, j) > 1:
|
|
n /= gcd(i, j)
|
|
d /= gcd(i, j)
|
|
|
|
max_n = n
|
|
|
|
end = default_timer()
|
|
|
|
print('Project Euler, Problem 71')
|
|
print('Answer: {}'.format(max_n))
|
|
|
|
print('Elapsed time: {:.9f} seconds'.format(end - start))
|
|
|
|
if __name__ == '__main__':
|
|
main()
|