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Most searched keywords: Whether you are looking for Ac-pink – This is a free and comprehensive report about.The domain is currently hosted on a server located in Japan with the IP address … … Summary of article content: Articles about Ac-pink – This is a free and comprehensive report about.75 AC PINK ideas in 2022 | pink aesthetic, pastel pink aesthetic, pink.For N < 200 it is slower than pure Python method (the exact value of boundary may depend on machine, on mine it was 200, its best to check it yourself): # method #4 For N > 200 it is faster than pure Python method. Print('Faster, pure python:', (time.time()-start))įaster, pure python: 0.0009944438934326172ĮDIT2: Forss noticed that numpy fast method can be optimized by using x*x instead of x**2. Result3 = sum(x*x * ysum for x, ysum in enumerate(itertools.accumulate(range(n+1))))
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Turns out NumPy has no advantage here, because it has no vectorized code for np.objects. Yes, for large N's, but already at N=100 it is like 8 times faster: start=time.time()ĮDIT: Even faster method (by KellyBundy, the Pumpkin) is by using pure python. Your fast solution is significantly faster than the slow one. It will now give you correct results (checked it up to N=100000). In a fast NumPy method you need to specify dtype=np.object so that NumPy does not convert Python int to its own dtypes ( np.int64 or others). (fastest methods, 3 and 4, are at the end) Print(polynomial((x, naive(x)) for x in range(6))) Return reduce(lambda r, i: r * i // math.gcd(r, i), ints)Ī = for x in xs]Ĭoeffs = / A for i in range(n)]ĭenominator = lcm(c.denominator for c in coeffs)Ĭoeffs = Return sum(x**2 * sum(range(x+1)) for x in range(n+1)) I did it similarly to this, but my matrix A is left-right mirrored compared to that, and I called my y-vector b. Compute the polynomial from the six equations with six unknowns (the polynomial coefficients).Compute six values with a naive implementation.Look up formulas for sum(x**4) and sum(x**3).So the whole thing becomes sum(x**2 * x*(x+1)//2 for x in range(n+1)). Rewrite the inner sum as the well-known x*(x+1)//2.
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