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arrow_back 欧拉函数

$$ \prod{i=1}^n\prod{j=1}^n\frac{lcm(i,j)}{\gcd(i,j)} \ =\prod{i=1}^n\prod{j=1}^n\frac{ij}{\gcd

zc
2020-03-20 16:33

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

一个点(x,y)和原点之间的点数为gcd(x,y)-1

n<m

$$ 设t=\sum{i=1}^n \sum{j=1}^m gcd(i,j) \ =\sum_{d=1}^n d

zc
2020-03-17 23:04

```cpp

include<bits/stdc++.h>

typedef long long ll;

pragma GCC optimize(3)

define il inline _

zc
2020-03-17 17:16

把题目强行转换为[POI2007]ZAP-Queries

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

zc
2020-03-15 16:05

\overbrace{88...88}^{x\text{个}8}可以表示成:

8\times \frac{(10^x-1)}9

那么$8\times \frac{(10^x-1)}9=kL

zc
2019-12-31 11:31

```cpp

include<bits/stdc++.h>

namespace ZDY{

#pragma GCC optimize(3)
#define il __inline__ 
zc
2019-12-21 19:47

GCD-Extreme top: 0


g(n) = \sum_{i=1}^{n-1} \gcd(i,n)

$$

\sum{i=1}^{n-1} \sum{j=i+1}^n \g

zc
2019-12-21 19:47

$$ \sumi^n \sum{j=i+1}^n gcd(i,j) \\ =\sum_i^n \sum_j^{i-1} gcd(i,j) \\ =\sum_d^n d \sum_i^n \

zc
2019-12-21 19:47
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