P6070 [RC-02] GCD [杜教筛，莫比乌斯反演]

没啥好说的，杜教筛板子题。

\[\sum_{i=1}^{N} \sum_{j=1}^{N}\sum_{p=1}^{\lfloor \frac{N}{j} \rfloor}\sum_{q=1}^{\lfloor \frac{N}{j} \rfloor} [\gcd(i,j)==1][\gcd(p,q)==1] \]

容易发现，我们枚举 \(j\) 其实是相当于枚举 \(\gcd\)
才不是枚举题目

然后式子可以变成

\[\sum_{i=1}^{N}\sum_{p=1}^{N}\sum_{q=1}^{N} [\gcd(i,p,q)==1] \]

然后套路式的枚举 \(gcd\) 依旧不是枚举题目

\[\sum_{d=1}^{N}\sum_{i=1}^{N}\sum_{p=1}^{N}\sum_{q=1}^{N}[\gcd(i,p,q)==d] \]

熟悉的形式，其实就等于

\[\sum_{d=1}^{N}\sum_{i=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{p=1}^{\lfloor \frac{N}{d} \rfloor} \sum_{q=1}^{\lfloor \frac{N}{d} \rfloor} \mu(d) \]

\[\sum_{d=1}^{N} \mu(d) \lfloor \frac{N}{d}\rfloor^3 \]

然后整除分块就完了，由于 \(N\) 比较大，大力杜教筛就完事了，话说我好像是这题除掉出题人的最优解

// powered by c++11
// by Isaunoya
#include<bits/stdc++.h>
#define rep(i , x , y) for(register int i = (x) ; i <= (y) ; ++ i)
#define Rep(i , x , y) for(register int i = (x) ; i >= (y) ; -- i)
using namespace std ;
using db = double ;
using ll = long long ;
using uint = unsigned int ;
#define int long long
using pii = pair < int , int > ;
#define ve vector
#define Tp template
#define all(v) v.begin() , v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
#define fir first
#define sec second
// the cmin && cmax
Tp < class T > void cmax(T & x , const T & y) { if(x < y) x = y ; }
Tp < class T > void cmin(T & x , const T & y) { if(x > y) x = y ; }
// sort , unique , reverse
Tp < class T > void sort(ve < T > & v) { sort(all(v)) ; }
Tp < class T > void unique(ve < T > & v) { sort(all(v)) ; v.erase(unique(all(v)) , v.end()) ; }
Tp < class T > void reverse(ve < T > & v) { reverse(all(v)) ; }
const int SZ = 0x191981 ;
struct FILEIN {
	~ FILEIN () {} char qwq[SZ] , * S = qwq , * T = qwq , ch ;
	char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq , 1 , SZ , stdin) , S == T) ? EOF : * S ++ ; }
	FILEIN & operator >> (char & c) { while(isspace(c = GETC())) ; return * this ; }
	FILEIN & operator >> (string & s) {
		while(isspace(ch = GETC())) ; s = ch ;
		while(! isspace(ch = GETC())) s += ch ; return * this ;
	}
	Tp < class T > void read(T & x) {
		bool sign = 1 ; while((ch = GETC()) < 0x30) if(ch == 0x2d) sign = 0 ;
		x = (ch ^ 0x30) ; while((ch = GETC()) > 0x2f) x = x * 0xa + (ch ^ 0x30) ;
		x = sign ? x : -x ;
	}
	FILEIN & operator >> (int & x) { return read(x) , * this ; }
	FILEIN & operator >> (signed & x) { return read(x) , * this ; }
	FILEIN & operator >> (unsigned & x) { return read(x) , * this ; }
} in ;
struct FILEOUT { const static int LIMIT = 0x114514 ;
	char quq[SZ] , ST[0x114] ; signed sz , O ;
	~ FILEOUT () { sz = O = 0 ; }
	void flush() { fwrite(quq , 1 , O , stdout) ; fflush(stdout) ; O = 0 ; }
	FILEOUT & operator << (char c) { return quq[O ++] = c , * this ; }
	FILEOUT & operator << (string str) {
		if(O > LIMIT) flush() ; for(char c : str) quq[O ++] = c ; return * this ;
	}
	Tp < class T > void write(T x) {
		if(O > LIMIT) flush() ; if(x < 0) { quq[O ++] = 0x2d ; x = -x ; }
		do { ST[++ sz] = x % 0xa ^ 0x30 ; x /= 0xa ; } while(x) ;
		while(sz) quq[O ++] = ST[sz --] ; return ;
	}
	FILEOUT & operator << (int x) { return write(x) , * this ; }
	FILEOUT & operator << (signed x) { return write(x) , * this ; }
	FILEOUT & operator << (unsigned x) { return write(x) , * this ; }
} out ;

const int maxn = 5e5 ;
int mu[maxn + 10] ;
const int mod = 998244353 ;
map < int , int > _mu ;
int getmu(int x) {
	if(x <= maxn) return mu[x] ;
	if(_mu[x]) return _mu[x] ;
	int ans = 1 ;
	int l = 2 , r = 0 ;
	for( ; l <= x ; l = r + 1) {
		r = x / (x / l) ;
		ans -= getmu(x / l) * (r - l + 1) ;
		ans = (ans + mod) % mod ;
	}
	return _mu[x] = ans ;
}
signed main() {
#ifdef _WIN64
	freopen("testdata.in" , "r" , stdin) ;
#else
	ios_base :: sync_with_stdio(false) ;
	cin.tie(nullptr) , cout.tie(nullptr) ;
#endif
// code begin.
	mu[1] = 1 ;
	for(int i = 1 ; i <= maxn ; i ++)
		for(int j = i + i ; j <= maxn ; j += i) 
			mu[j] -= mu[i] ;
	for(int i = 2 ; i <= maxn ; i ++)
		mu[i] = (mu[i] + mu[i - 1]) % mod ;
	int n ;
	in >> n ;
	int l = 1 , r = 0 ;
	int ans = 0 ;
	for( ; l <= n ; l = r + 1) {
		r = n / (n / l) ;
		int qwq = (n / l) * (n / l) % mod * (n / l) % mod ;
		ans = (ans + (getmu(r) - getmu(l - 1) + mod) % mod * qwq % mod) % mod ;
	}
	out << ans << '\n' ;
	return out.flush() , 0 ;
// code end.
}