ygmjsjdboy的博客

门 题目 ∑i=1n∑j=1n(i+j)kgcd(i,j)μ2(gcd(i,j))\sum_{i=1}^n\sum_{j=1}^n(i+j)^kgcd(i,j)\mu^2(gcd(i,j))i=1∑n​j=1∑n​(i+j)kgcd(i,j)μ2(gcd(i,j)) 题解 ∑i=1n∑j=1n(i+j)kgcd(i,j)μ2(gcd(i,j))∑d=1ndk+1μ2(d)∑i=1⌊nd⌋∑j=1⌊nd⌋(i+j)k[gcd(i,j)==1]∑d=1ndk+1μ2(d)∑i=1⌊nd⌋∑j=1⌊nd⌋(i+j

门（洛谷的门不知道什么时候修好） 题目 ∑i=1n∑j=i+1nlcm(ai,aj)\sum_{i=1}^n\sum_{j=i+1}^nlcm(a_i,a_j)i=1∑n​j=i+1∑n​lcm(ai​,aj​) 题解 ∑i=1n∑j=i+1nlcm(ai,aj)∑i=1n∑j=1nlcm(ai,aj)−∑i=1nai2 \sum_{i=1}^n\sum_{j=i+1}^nlcm(a_i,a_j)\\ \frac {\sum_{i=1}^n\sum_{j=1}^nlcm(a_i,a_j)-\sum_{i=1

门 题目 ∑i=1n∑j=1mσ1(gcd(i,j))[σ1(gcd(i,j))≤A]\sum_{i=1}^n\sum_{j=1}^m\sigma_1(gcd(i,j))[\sigma_1(gcd(i,j))\leq A]i=1∑n​j=1∑m​σ1​(gcd(i,j))[σ1​(gcd(i,j))≤A] 题解 先不考虑A ∑i=1n∑j=1mσ1(gcd(i,j))∑d=1upσ1(d)∑i=1⌊nd⌋∑j=1⌊md⌋[gcd(i,j)==1]∑d=1upσ1(d)∑i=1⌊nd⌋∑j=1⌊md⌋∑e∣g

门 题目 ∑i=1n∑j=1nijgcd(i,j)\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j)i=1∑n​j=1∑n​ijgcd(i,j) 题解 ∑i=1n∑j=1nijgcd(i,j)∑d=1nd3∑i=1⌊nd⌋i∑j=1⌊nd⌋j[gcd(i,j)==1]∑d=1nd3∑i=1⌊nd⌋i∑j=1⌊nd⌋j∑e∣gcd(i,j)μ(e)\sum_{i=1}^n\sum_{j=1}^nijgcd(i,j)\\\sum_{d=1}^nd^3\sum_{i=1}^{\lfloor \f

门 题目 ∑i=1A∑j=1B∑k=1Cd(i,j,k)\sum_{i=1}^A\sum_{j=1}^B\sum_{k=1}^Cd(i,j,k)i=1∑A​j=1∑B​k=1∑C​d(i,j,k) 题解 前置：d(i,j,k)=∑a∣i∑b∣j∑c∣k[gcd(a,b)==1][gcd(a,c)==1][gcd(b,c)==1]d(i,j,k)=\sum_{a\mid i}\sum_{b\mid j}\sum_{c\mid k}[gcd(a,b)==1][gcd(a,c)==1][gcd(b,c)==1]d

假装我是目录P5495 [Dirichlet 前缀和](https://www.luogu.com.cn/problem/P5495)P3455 [[POI2007]ZAP-Queries](https://www.luogu.com.cn/problem/P3455)P2257 [YY的GCD](https://www.luogu.com.cn/problem/P2257)P2522 [[HAOI2011]Problem b](https://www.luogu.com.cn/problem/P2522)

P3455 [POI2007]ZAP-Queries 题目 ∑i=1n∑j=1m[gcd(i,j)==k]\sum_{i=1}^n \sum_{j=1}^m[gcd(i,j)==k]i=1∑n​j=1∑m​[gcd(i,j)==k] 题解 莫反入坑作 ∑i=1n∑j=1m[gcd(i,j)==k]∑i=1⌊nk⌋∑j=1⌊mk⌋[gcd(i,j)==1]∑i=1⌊nk⌋∑j=1⌊mk⌋∑d∣gcd(i,j)μ(d)∑d=1min(n,m)∑i=1⌊nkd⌋∑i=1⌊mkd⌋1∑d=1min(n,m)⌊n