DZY Loves Math IV（杜教筛）

title

solution

这道题是多么的妙啊，完全不是我能推出来的式子呢！

观察数据范围，有点奇怪欸，在暗示我？？

考虑暴力枚举$n$
$$S(n,m)=\sum _{i=1}^m\phi (n\times i)$$
神奇的操作来了，将$n$质因数分解，并把不同的质因数分别拿出一个
$n=\prod p_i^{e_i}$
$q=\prod p_i$
$p=\prod p_i^{e_i-1}$
则有$p\times q=n$

1. 若$i\%j=0$，则$\phi (ij)=\phi (i)\times j$
2. 若$(i,j)=1$，则$\phi (ij)=\phi (i)\times \phi (j)$

$$S(n,m)=\sum _{i=1}^m\phi (n\times i)$$$$=p\cdot \sum _{i=1}^m\phi (q\times i)$$$$=p\cdot \sum _{i=1}^m\phi (\frac{q}{gcd(q,i)}\times i\times gcd(q,i))$$$$=p\cdot \sum _{i=1}^m\phi (\frac{q}{gcd(q,i)})\phi (i\times gcd(q,i))$$$$=p\cdot \sum _{i=1}^m\phi (\frac{q}{gcd(q,i)})\phi (i)gcd(q,i)$$$$=p\sum _{i=1}^m\phi (\frac{q}{gcd(q,i)})\phi (i)\sum _{d|gcd(q,i)}\phi (d)$$$$=p\sum _{i=1}^m\phi (i)\sum _{d|i,d|q}\phi (\frac{q}{d})$$$$=p\sum _{d|q}\phi (\frac{q}{d})\sum _{i=1}^{\lfloor \frac{m}{d}\rfloor }\phi (i\times d)$$$$=p\sum _{d|q}\phi (\frac{q}{d})S(d,\lfloor \frac{m}{d}\rfloor )$$

$\phi$用杜教筛，应该是老熟人了
$S(n,m)$记忆化一下，应该就没了

code

#include <cstdio>
#include <vector>
#include <map>
using namespace std;
#define mod 1000000007
#define int long long
#define maxn 200000
map < int, int > mp, s[maxn];
int cnt;
int minp[maxn + 5]; //minp[i]:i的最大质因子
int prime[maxn], phi[maxn + 5]; //phi[i]:1~i的phi的前缀和 
bool vis[maxn + 5];
 
void init() {
    
    
	phi[1] = 1;
	for( int i = 2;i <= maxn;i ++ ) {
    
    
		if( ! vis[i] ) prime[++ cnt] = i, minp[i] = i, phi[i] = i - 1;
		for( int j = 1;j <= cnt && i * prime[j] <= maxn;j ++ ) {
    
    
			vis[i * prime[j]] = 1, minp[i * prime[j]] = prime[j];
			if( i % prime[j] == 0 ) {
    
    
				phi[i * prime[j]] = phi[i] * prime[j] % mod;//与式子推导的第二步为什么p能直接从φ里面拿出来呼应
				break;
			}
			else
				phi[i * prime[j]] = phi[i] * ( prime[j] - 1 ) % mod;
		}
	}
	for( int i = 1;i <= maxn;i ++ ) phi[i] = ( phi[i] + phi[i - 1] ) % mod;
}

int Phi( int n ) {
    
    
	if( n <= maxn ) return phi[n];
	if( mp[n] ) return mp[n];
	int ans = n * ( n + 1 ) / 2 % mod;
	for( int i = 2, r;i <= n;i = r + 1 ) {
    
    
		r = n / ( n / i );
		ans = ( ans - ( r - i + 1 ) * Phi( n / i ) % mod + mod ) % mod;
	}
	return mp[n] = ans;
}

int solve( int n, int m ) {
    
    
	if( ! m ) return 0;
	if( s[n][m] ) return s[n][m];
	if( n == 1 ) return s[n][m] = Phi( m );
	if( m == 1 ) return s[n][m] = ( Phi( n ) - Phi( n - 1 ) + mod ) % mod;
	vector < int > g;
	int p = 1, q = 1, N = n, x;
	while( N > 1 ) {
    
    
		x = minp[N], q *= x, N /= x, g.push_back( x );
		while( N % x == 0 ) N /= x, p *= x;
	}
	int len = g.size(), ans = 0;
	for( int i = 0;i < ( 1 << len );i ++ ) {
    
     //枚举q的所有质因子(状压) 
		int d = 1;
		for( int j = 0;j < len;j ++ )
			if( i & ( 1 << j ) ) d = d * g[j]; //二进制位为1则有该质因子
		ans = ( ans + ( Phi( q / d ) - Phi( q / d - 1 ) + mod ) % mod * solve( d, m / d ) % mod ) % mod;
	}
	return s[n][m] = ans * p % mod;
}

signed main() {
    
    
	int n, m;
	scanf( "%lld %lld", &n, &m );
	init();
	int ans = 0;
	for( int i = 1;i <= n;i ++ ) 
		ans = ( ans + solve( i, m ) ) % mod;
	printf( "%lld\n", ans );
	return 0;
}