网站介绍:文章浏览阅读403次。题面在这里首先容斥,把问题转化为求 ∑i=1n∑j=1m[gcd(i,j)=k]⇒∑i=1⌊nk⌋∑j=1⌊mk⌋[gcd(i,j)=1]\sum _{i=1}^n \sum _{j=1}^m [gcd(i,j)=k] \\\Rightarrow \sum _{i=1}^{\lfloor \frac n k \rfloor} \sum _{j=1}^{\lfloor \frac m k \r_bzoj2301 [haoi2011]problem b
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- 链接标题:【容斥+莫比乌斯反演】BZOJ2301 [HAOI2011]Problem b-CSDN博客
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