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Improve solution (#5705)
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@ -18,7 +18,7 @@ Number of numbers between 1 and n that are coprime to n is given by the Euler's
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function, phi(n). So, the answer is simply the sum of phi(n) for 2 <= n <= 1,000,000
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Sum of phi(d), for all d|n = n. This result can be used to find phi(n) using a sieve.
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Time: 3.5 sec
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Time: 1 sec
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"""
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@ -36,8 +36,9 @@ def solution(limit: int = 1_000_000) -> int:
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phi = [i - 1 for i in range(limit + 1)]
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for i in range(2, limit + 1):
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for j in range(2 * i, limit + 1, i):
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phi[j] -= phi[i]
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if phi[i] == i - 1:
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for j in range(2 * i, limit + 1, i):
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phi[j] -= phi[j] // i
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return sum(phi[2 : limit + 1])
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