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$$ \sum{i=1}^n\sum{j=1}^n(i+j)^k\mu^2(\gcd(i,j))\gcd(i,j) \ =\sum_{d=1}^n \mu^2(d) d

zc
2020-03-20 12:57

经过一番化简后变成了:

$$ \sum{i=1}^{\sqrt{2n}} \sum{x=L}^{R} [\gcd(i,x)=1]
\ L=Max(1,i-\lfloor {n\over i

zc
2020-03-19 23:44

$$ \sum{i=1}^A\sum{j=1}^B [\gcd(i,j)=d] \ =\sum_{i=1}^{\left \lfloor \frac Ad \right \rfloor}\sum

zc
2020-03-19 20:28

首先L=\left \lceil \frac LK\right \rceil,H=\left \lceil \frac HK\right \rceil

问题转化为在[L,H]中取 N个$

zc
2020-03-19 17:10

\sum_{i=1}^n\sum_{j=1}^n lcm(A_i,A_j)

a_i = \sum_{j=1}^n [A_j=i]

那么

$$ ans = \sum_i^n \sum_j

zc
2020-03-19 15:44

前置知识:

  1. 杜教筛(包括[狄利克雷卷积](http://blog.zcmimi.top/posts
zc
2020-03-18 02:07

```cpp

include<bits/stdc++.h>

typedef long long ll;

pragma GCC optimize(3)

define il inline _

zc
2020-03-17 17:16

$$ f=\sum{i=1}^{a}\sum{j=1}^{b}[gcd(i,j)=d] \ f=\sum_{i=1}^{\left \lfloor \frac ad \right \rfloor

zc
2020-03-15 16:07

把题目强行转换为[POI2007]ZAP-Queries

$Ans((1,b),(1,d))-Ans(

zc
2020-03-15 16:05

n<m $$ \prod{i=1}^n \prod{j=1}^m f{\gcd(i,j)} \ =\prod{d=1}^n{fd}^{\sum{i=1}^n \sum_{j=1}^

zc
2020-03-15 15:48
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