Solution -「洛谷 P3911」最小公倍數之和
\(\mathcal{Description}\)
Link.
給定 \(\{a_n\}\),求:
\[\sum_{i=1}^n\sum_{j=1}^n\operatorname{lcm}(a_i,a_j) \]
\(1\le n,a_i\le5\times10^4\)。
\(\mathcal{Solution}\)
數論題在序列上搞不太現實,記最大值 \(m\),有 \(c_i\) 個 \(a_j=i\),推式子:
\[\begin{aligned} \sum_{i=1}^n\sum_{j=1}^n\operatorname{lcm}(a_i,a_j)&=\sum_{i=1}^m\sum_{j=1}^m\frac{ij}{\gcd(i,j)}c_ic_j\\ &=\sum_{d=1}^m\sum_{i=1}^{\lfloor\frac{m}d\rfloor}\sum_{j=1}^{\lfloor\frac{m}d\rfloor}[\gcd(i,j)=1]dijc_ic_j\\ &=\sum_{d=1}^m\sum_{i=1}^{\lfloor\frac{m}d\rfloor}\sum_{j=1}^{\lfloor\frac{m}d\rfloor}dijc_ic_j\sum_{D|i\land D|j}\mu(D)~~~~(\text{Mobius 反演})\\ &=\sum_{d=1}^md\sum_{D=1}^{\lfloor\frac{m}d\rfloor}\mu(D)D^2\sum_{i=1}^{\lfloor\frac{m}{dD}\rfloor}\sum_{j=1}^{\lfloor\frac{m}{dD}\rfloor}ijc_{idD}c_{jdD}~~~~(\text{交換列舉順序})\\ &=\sum_{T=1}^mT\sum_{D|T}\mu(D)D\sum_{i=1}^{\lfloor\frac{m}T\rfloor}\sum_{j=1}^{\lfloor\frac{m}T\rfloor}ijc_{iT}c_{jT}~~~~(\text{改換列舉}~T=dD)\\ &=\sum_{T=1}^mT\left(\sum_{i=1}^{\lfloor\frac{m}T\rfloor}ic_{iT}\right)^2\sum_{D|T}\mu(D)D \end{aligned} \]
\(\mathcal O(n+m\sqrt m)\) 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\n", ans ); return 0; }
\(\mathcal{Details}\)
推的時候把 \(ij\) 係數搞丟了自閉半天 qaq。