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Re:我有个证明，大家可以讨论一下。。In Reply To:Re:我有个证明，大家可以讨论一下。。 Posted by:MyTalent at 2013-05-17 15:14:01 > 蒽，这个应该不难想到。关键是怎样快速给出N,M条件的所有 > gcd(x1,x2,..xn,M)=1 > 的方法数呢 呃啊不好意思刚才太愚钝了。直接用总数减gcd不为1的好像就行 Followed by: Post your reply here: |
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