P3768 简单的数学题 [杜教筛，莫比乌斯反演]

\[\sum_{i=1}^{n}\sum_{j=1}^{n} ij\gcd(i,j)
\]

\[=\sum_{d=1}^{n} d \sum_{i=1}^{n}\sum_{j=1}^{n} ij[\gcd(i,j)==d]
\]

\[=\sum_{d=1}^{n} d^3 \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{n}{d}\rfloor} ij[\gcd(i,j)==1]
\]

\[=\sum_{d=1}^{n} d^3 \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{n}{d}\rfloor} ij[\gcd(i,j)==1]
\]

因为 \(\sum_{t|n} \mu(t) = [n == 1]\)

所以

\[=\sum_{d=1}^{n} d^3 \sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{n}{d}\rfloor} ij \sum_{t|\gcd(i,j)} \mu(t)
\]

\[=\sum_{d=1}^{n} d^3 \sum_{t=1}^{\lfloor\frac{n}{d}\rfloor} \mu(t) t^2 \sum_{i=1}^{\lfloor\frac{n}{td}\rfloor} \sum_{j=1}^{\lfloor\frac{n}{td}\rfloor} ij
\]

设 \(sum_x = \sum_{i=1}^{x}\)

则原式等同于

\[=\sum_{d=1}^{n} d^3 \sum_{t=1}^{\lfloor\frac{n}{d}\rfloor} \mu(t) t^2 (sum_{\lfloor \frac{n}{td} \rfloor})^2
\]

设 \(T=td\)

\[=\sum_{T=1}^{n} T^2 (sum_{\lfloor \frac{n}{T} \rfloor})^2 \sum_{d|t} \mu (\frac{T}{d}) d
\]

因为

\[\sum_{d|n} \frac{\mu(d)}{d} = \frac{\varphi(n)}{n}
\]

根据狄利克雷卷积

\[id(x) = x , id * \mu = \varphi
\]

那么这个柿子变成了

\[\sum_{T=1}^{n}T^2 \varphi(T) (sum_{\lfloor \frac{n}{T} \rfloor})^2
\]

然后我们本身知道 形如 \(\lfloor \frac{n}{T} \rfloor\) 之类的玩意可以数论分块

然后 \(f_x = x^2 \varphi(x)\) 的前缀和

设 \(s_x\) 是 \(f_x\) 的前缀和，然后考虑 \(g\) 函数

然后我们知道

\[\sum_{d|n} \varphi(d) = n
\]

考虑 \(g\) 函数为 \(g_i = i^2\)

\[(g*f)(i) = \sum_{d|i} d^2 \varphi (d) \frac{i^2}{d^2} = \sum_{d|i}\varphi (d) i^2 = i^3
\]

\[s_n = \sum_{i=1}^{n} i^3 - \sum_{i=2}^{n} i^2 s_{\frac{n}{i}}
\]

因为

\[\sum_{i=1}^{n}i^3 = (\sum_{i=1}^{n}i)^2
\]

\[\sum_{i=1}^{n}i^2 = \frac{n(n+1)(2n+1)}{6}
\]

Q.E.D.

杜教筛一手，这题没了。

// 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 ;

int n , p ;
const int maxn = 5e6 + 10 ;
int phi[maxn + 5] ;
int iv2 , iv6 ;
int qpow(int x , int y) {
	int res = 1 ;
	for( ; y ; y >>= 1 , x = x * x % p)
		if(y & 1)
			res = res * x % p ;
	return res ;
}
int sum(int x) {
	x = x - (x / p) * p ;
	return (x + 1) * x % p * iv2 % p ;
}
int sum2(int x) {
	x = x - (x / p) * p ;
	return (x + 1) * x % p * (x + x + 1) % p * iv6 % p ;
}

unordered_map < int , int > _phi ;
int s(int x) {
	if(x <= maxn) return phi[x] ;
	if(_phi[x]) return _phi[x] ;
	int ans = sum(x) ;
	ans = ans * ans % p ;
	for(int l = 2 , r ; l <= x ; l = r + 1) {
		r = x / (x / l) ;
		ans = (ans - (sum2(r) - sum2(l - 1) + p) % p * s(x / l) % p + p) % p ;
	}
	return _phi[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.
	in >> p >> n ;
	iv2 = qpow(2 , p - 2) ;
	iv6 = qpow(6 , p - 2) ;
	phi[1] = 1 ;
	for(int i = 2 ; i <= maxn; i ++) {
		if(phi[i])
			continue ;
		for(int j = i ; j <= maxn ; j += i) {
			if(! phi[j])
				phi[j] = j ;
			phi[j] = phi[j] / i * (i - 1) ;
		}
	}
	for(int i = 1 ; i <= maxn ; i ++)
		phi[i] = (phi[i - 1] + 1ll * i * i % p * phi[i] % p) % p ;
	int ans = 0 ;
	for(int l = 1 , r ; l <= n ; l = r + 1) {
		r = n / (n / l) ;
		int k = sum(n / l) ;
		k = k * k % p ;
		ans = (ans + (s(r) - s(l - 1) + p) % p * k % p) % p ;
	}
	out << ans << '\n' ;
	return out.flush() , 0 ;
// code end.
}