Solution -「洛谷 P3911」最小公倍数之和

(mathcal{Description})

  Link.

  给定 ({a_n})，求：

[sum_{i=1}^nsum_{j=1}^noperatorname{lcm}(a_i,a_j) ]

  (1le n,a_ile5 imes10^4)。

(mathcal{Solution})

  数论题在序列上搞不太现实，记最大值 (m)，有 (c_i) 个 (a_j=i)，推式子：

[egin{aligned} sum_{i=1}^nsum_{j=1}^noperatorname{lcm}(a_i,a_j)&=sum_{i=1}^msum_{j=1}^mfrac{ij}{gcd(i,j)}c_ic_j\ &=sum_{d=1}^msum_{i=1}^{lfloorfrac{m}d floor}sum_{j=1}^{lfloorfrac{m}d floor}[gcd(i,j)=1]dijc_ic_j\ &=sum_{d=1}^msum_{i=1}^{lfloorfrac{m}d floor}sum_{j=1}^{lfloorfrac{m}d floor}dijc_ic_jsum_{D|iland D|j}mu(D)~~~~( ext{Mobius 反演})\ &=sum_{d=1}^mdsum_{D=1}^{lfloorfrac{m}d floor}mu(D)D^2sum_{i=1}^{lfloorfrac{m}{dD} floor}sum_{j=1}^{lfloorfrac{m}{dD} floor}ijc_{idD}c_{jdD}~~~~( ext{交换枚举顺序})\ &=sum_{T=1}^mTsum_{D|T}mu(D)Dsum_{i=1}^{lfloorfrac{m}T floor}sum_{j=1}^{lfloorfrac{m}T floor}ijc_{iT}c_{jT}~~~~( ext{改换枚举}~T=dD)\ &=sum_{T=1}^mTleft(sum_{i=1}^{lfloorfrac{m}T floor}ic_{iT} ight)^2sum_{D|T}mu(D)D end{aligned} ]

  (mathcal O(n+msqrt m)) 算就好啦。

(mathcal{Code})

#include <cmath>
#include <cstdio>

const int MAXN = 5e4;
int n, m, c[MAXN + 5];
int pn, pr[MAXN + 5], mu[MAXN + 5];
bool vis[MAXN + 5];

inline int rint () {
	int x = 0; char s = getchar ();
	for ( ; s < '0' || '9' < s; s = getchar () );
	for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
	return x;
}

inline void sieve ( const int n ) {
	mu[1] = 1;
	for ( int i = 2; i <= n; ++ i ) {
		if ( !vis[i] ) mu[pr[++ pn] = i] = -1;
		for ( int j = 1, t; j <= pn && ( t = i * pr[j] ) <= n; ++ j ) {
			vis[t] = true;
			if ( !( i % pr[j] ) ) break;
			mu[t] = -mu[i];
		}
	}
}

int main () {
	n = rint ();
	for ( int i = 1, a; i <= n; ++ i ) {
		++ c[a = rint ()];
		if ( m < a ) m = a;
	}
	sieve ( m );
	long long ans = 0;
	for ( int i = 1; i <= m; ++ i ) {
		long long a = 0, b = 0;
		for ( int j = 1, t = m / i; j <= t; ++ j ) a += 1ll * j * c[i * j];
		for ( int j = 1, t = sqrt ( i ); j <= t; ++ j ) {
			if ( i % j ) continue;
			b += mu[j] * j;
			if ( j * j < i ) b += mu[i / j] * i / j;
		}
		ans += 1ll * i * a * a * b;
	}
	printf ( "%lld
", ans );
	return 0;
}

(mathcal{Details})

  推的时候把 (ij) 系数搞丢了自闭半天 qaq。