Daniele Fucini
3a09f3d4f5
Added solutions for problem 66, 67, 68, 69 and 70 in C and python. Also, minor improvements to the code for a few older problems.
56 lines
1.6 KiB
Python
56 lines
1.6 KiB
Python
#!/usr/bin/python3
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# Euler's Totient function, φ(n) [sometimes called the phi function], is used to determine the number of numbers less than n which are
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# relatively prime to n. For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6.
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#
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# n Relatively Prime φ(n) n/φ(n)
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# 2 1 1 2
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# 3 1,2 2 1.5
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# 4 1,3 2 2
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# 5 1,2,3,4 4 1.25
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# 6 1,5 2 3
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# 7 1,2,3,4,5,6 6 1.1666...
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# 8 1,3,5,7 4 2
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# 9 1,2,4,5,7,8 6 1.5
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# 10 1,3,7,9 4 2.5
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#
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# It can be seen that n=6 produces a maximum n/φ(n) for n ≤ 10.
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#
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# Find the value of n ≤ 1,000,000 for which n/φ(n) is a maximum.
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from timeit import default_timer
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from projecteuler import is_prime
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def main():
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start = default_timer()
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N = 1000000
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i = 1
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res = 1
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# Using Euler's formula, phi(n)=n*prod(1-1/p), where p are the distinct
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# primes that divide n. So n/phi(n)=1/prod(1-1/p). To find the maximum
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# value of this function, the denominator must be minimized. This happens
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# when n has the most distinct small prime factor, i.e. to find the solution
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# we need to multiply the smallest consecutive primes until the result is
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# larger than 1000000.
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while res < N:
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i = i + 1
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if is_prime(i):
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res = res * i
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# We need the previous value, because we want i<1000000
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res = res // i
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end = default_timer()
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print('Project Euler, Problem 69')
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print('Answer: {}'.format(res))
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print('Elapsed time: {:.9f} seconds'.format(end - start))
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if __name__ == '__main__':
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main()
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