Project Euler:problem Ordered Fractions

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, Quarter, 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 are 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 O f the fraction immediately to the left of 3/7.

p/q < 3/7 ==> 7p<3q ==> 7p<=3q-1 ==> p=floor ((3q-1)/7)

Save the current score nearest 3/7 to r/s if r/s <p/q is R*q < p*s assigns p,q to R,s respectively

Import matha=3b=7r=0s=1for Q in Range (2,1000000):    P=math.floor ((a*q-1)/b)    if r*q < s*p:        r=p        s= Qprint (' r = ', r) print (' s = ', s)

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Project Euler:problem Ordered Fractions