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最新推荐文章于 2017-10-25 09:43:54 发布

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最新推荐文章于 2017-10-25 09:43:54 发布

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求翻译。我翻译出来还是只懂个大概。。求解答这题！

Consider two numbers a and b, whose prime factorizations are $2^{a_2} 3^{a_3} ...$ and $2^{b_2} 3^{b_3} ...$ , respectively. Then their LCM is $2^{max(a_2,b_2)} 3^{max(a_3,b_3)} ...$ This is the insight that lets us solve the problem.

So factorize N: $p_1^{n_1} p_2^{n_2} ... p_k^{n_k}$ . Now say that we have a and b such that LCM(a,b) = N. Then for each i, at least one of $a i$ and $b i$ is equal to $n i$ . This would suggest that there are $2 n i + 1$ possibilities for each prime factor, and that if we multiply them all together, we're done. Unfortunately, we run into the issue of overcounting. Thus we need to modify this approach somewhat. We will solve this problem incrementally and recursively.

For a given N, call the smallest prime factor $p 1$ , and say that $n 1$ is the largest power of $p 1$ that divides N. Break the problem into two cases. In case one, we set $a 1 < b 1 = n 1$ . This gives us $n 1$ possibilities for the first number, and for the rest, we can simply take the product of $(2 n i + 1)$ for $i > 1$ . (Verify that this does not count any pairs more than once.) In the other case, we have $a 1 = b 1 = n 1$ . In this case, we just need to solve the same problem, but for $\frac{N}{p_1^{n_1}}$ . (Similarly, verify that these two cases together do not miss any pairs.) The base case of our recursion is when N = 1, in which case we just return 1.