'''
Created on 2011-7-29
@author: huangkan
'''
from Mytools import primes
p=primes(1000)
def factors(d):
f=[]
for i in p:
if d>1 and d not in p:
if d%i==0:
f.append(i)
while d%i==0:
d=d//i
else:
break
if d>1:f.append(d)
return f
def Totient(n):
n1=n
f=factors(n)
for i in f:
n1=n1*(i-1)//i
return n1
sum=0
for i in range(2,1000000+1):
sum+=Totient(i)
print(sum)
'''
Created on 2011-7-15
@author: huangkan
I delete the right edition mistakenly. This code is to slow to get the answer.
The key point is to measure the rate and length of totient whenever a new factor is found. It will accelerate the whole search.
'''
from Mytools import primes
p=primes(3163)
def factors(d):
f=[]
for i in p:
if d>1 and d not in p:
if d%i==0:
f.append(i)
while d%i==0:
d=d//i
else:
break
if d>1:f.append(d)
return f
def Totient(n):
n1=n
f=factors(n)
for i in f:
n1=n1*(i-1)//i
return n1
