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[数论] 有序数对的个数

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chudengshuxue posted 2018-9-5 16:02 |Read mode
Last edited by hbghlyj 2025-3-19 18:05已知 $a, b \inZ++$,且 $[a, b]=2^5 \times 3^7 \times 7^2 \times 11$,则有序数对 $(a, b)$ 的个数是

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kuing posted 2018-9-5 16:50
(2*6-1)(2*8-1)(2*3-1)(2*2-1)=2475

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isee posted 2018-9-5 20:19
与解题无关,建立楼主看一看 论坛代码极简入门

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realnumber posted 2018-9-6 15:24
这样考虑,特殊到一般
先考虑[a,b]=2,
[a,b]=$2^2$等找找规律

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kuing posted 2018-9-6 15:40
这还用找规律吗,稍微想想就知道了……

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realnumber posted 2018-9-6 16:25
回复 5# kuing


    恩,在这个问题上,我假设1楼搞晕了

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isee posted 2018-9-6 16:45
我记得似乎楼上两位中某位把这个定理 证明了的,且有过程。。。

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isee posted 2018-9-6 17:00
回复 5# kuing

这个只能说你对这些算法无比的熟悉。。。

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kuing posted 2018-9-6 19:07
回复 8# isee

没什么熟悉不熟悉的,我以前也没做过这类题,这个实在是很容易想到。
就看 2 的次数,公倍数是 5 次,所以两数中有一个是 5 次,另一个则不多于 5 次,所以可以是 (5,0),(5,1),...,(5,5),对称的也一样,再减去 (5,5) 重复计算的,所以这里就有 2*6-1 种可能,其余同理,再由乘法原理乘起来就是 2# 的结果。

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