p072.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 21 elements in this set.
#
#How many elements would be contained in the set of reduced proper fractions for d <= 1,000,000?

import logging

from prime import PrimeNumberPool
import permute
import time

def HCF(n, d, prime):
    for p in prime.getPrimeFactor(n):
        if (d % p == 0):
            return 0
    return 1

def Phi(n, prime):
    fn = prime.getPrimeFactor(n)
    num = n
    for i in range(1, len(fn)+1):
        fni = permute.select(i, fn)
        for pni in fni:
            pp = 1
            for p in pni:
                pp *= p
            if (i % 2):
               num -= n//pp
            else:
                num += n//pp
    return num

def main(args):
    if args.test:
        m = 10
    else:
        m = 1000*1000

    ts1 = time.time()
    prime = PrimeNumberPool(m//2)
    ts2 = time.time()
    logging.debug("prepare primes takes {} seconds".format(ts2-ts1))
    
    s = 0
    for d in range(2, m+1):
        nrf = Phi(d, prime)
        s += nrf

    logging.info("answer: {}".format(s))