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question71.py
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question71.py
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#!/usr/bin/env python3
# -*- coding: utf-8 -*-
# Ordered fractions
#Problem 71
#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.
import time
import math
import fractions
from fractions import gcd
s1=time.time()
def find_frac(limit):
lower = 2.0/5
upper = 3.0/7
#limit = 8
res= 2
for den in range(limit,1,-1):
#print(den)
nlower = math.floor(lower*den)
nupper = math.ceil(upper*den)
for num in range(nlower,nupper):
if gcd(num,den)==1:
if num/(1.0*den)<upper:
#print(num,den,num/den)
if num/(1.0*den)>lower:
lower = num/(1.0*den)
res = num
return(res)
#test = count_fracs2(8)
#test = count_fracs2(1000000)
print(find_frac(1000000))
#print(test)
print("{}s".format(time.time() - s1))
# 428570
#2.031975269317627s