Solution -「51nod 1584」加权约数和
\(\mathcal{Description}\)
Link.
令 \(\sigma(n)\) 为 \(n\) 的约数之和。求:
\[\sum_{i=1}^n\sum_{j=1}^n\max\{i,j\}\sigma(ij)\bmod(10^9+7)
\]
多测,\(n\le10^6\),数据组数 \(\le5\times10^4\)。
\(\mathcal{Solution}\)
直 接 来 owo!
\[\sum_{i=1}^n\sum_{j=1}^n\max\{i,j\}\sigma(ij)=2\sum_{i=1}^ni\sum_{j=1}^i\sigma(ij)-\sum_{i=1}^ni\sigma(i^2)
\]
先研究一下 \(\sigma(ij)\)。考虑分别枚举 \(i\) 的约数 \(x\) 和 \(j\) 的约数 \(y\),若 \(x\perp\frac{j}y\),则对 \(\sigma(ij)\) 贡献一个 \(xy\),显然贡献不重不漏。即:
\[\sigma(ij)=\sum_{x|i}\sum_{y|j}xy[x\perp\frac{j}y]
\]
考虑原式前一项,记 \(f(n)=n\sum_{i=1}^n\sigma(ni)\),有:
\[\begin{aligned}
f(n)&=n\sum_{i=1}^n\sum_{x|n}\sum_{y|i}xy[x\perp\frac{i}y]\\
&=n\sum_{i=1}^n\sum_{x|n}\sum_{y|i}xy\sum_{d|x\land d|\frac{i}y}\mu(d)\\
&=n\sum_{i=1}^n\sum_{d|n\land d|i}\mu(d)\sum_{x|n\land d|x}\sum_{y|i\land d|y}\frac{ix}y\\
&=n\sum_{i=1}^n\sum_{d|n\land d|i}\mu(d)\sum_{x|\frac{n}d}\sum_{y|\frac{i}d}\frac{ix}y~~~~~~~~(x,y\mbox{ 同时约掉 } d)\\
&=n\sum_{i=1}^n\sum_{d|n\land d|i}\mu(d)\sigma(\frac{n}d)\sum_{y|\frac{i}d}\frac{i}y\\
&=n\sum_{i=1}^n\sum_{d|n\land d|i}\mu(d)\sigma(\frac{n}d)d\sum_{y|\frac{i}d}\frac{\frac{i}y}d\\
&=n\sum_{i=1}^n\sum_{d|n\land d|i}d\mu(d)\sigma(\frac{n}d)\sigma(\frac{i}d)\\
&=n\sum_{d|n}d\mu(d)\sigma(\frac{n}d)\sum_{i=1}^\frac{n}{d}\sigma(i)
\end{aligned}
\]
筛出 \(\mu,\sigma\),枚举 \(d\) 和 \(\frac{n}d\),可以 \(\mathcal O(n\ln n)\) 算出所有 \(f\)。
后一项呢,就是要筛 \(\sigma(n^2)\)。和筛 \(\sigma(n)\) 类似,记录一下当前最小素因子的等比数列求和,就可以 \(\mathcal O(n)\) 算出来。
综上,复杂度 \(\mathcal O(n\ln n+T)\)。
\(\mathcal{Code}\)
/* Clearink */
#include <cstdio>
typedef long long LL;
inline int rint () {
int x = 0, f = 1; char s = getchar ();
for ( ; s < '0' || '9' < s; s = getchar () ) f = s == '-' ? -f : f;
for ( ; '0' <= s && s <= '9'; s = getchar () ) x = x * 10 + ( s ^ '0' );
return x * f;
}
const int MAXN = 1e6, MOD = 1e9 + 7;
int pn, pr[MAXN + 5], mpwr[MAXN + 5], mu[MAXN + 5], sig[MAXN + 5], sigs[MAXN + 5];
int f[MAXN + 5], ans[MAXN + 5];
LL dpwr[MAXN + 5], g[MAXN + 5];
bool vis[MAXN + 5];
inline void init ( const int n ) {
mu[1] = sig[1] = sigs[1] = g[1] = 1;
for ( int i = 2; i <= n; ++ i ) {
if ( !vis[i] ) {
mu[pr[++ pn] = i] = -1;
sig[i] = mpwr[i] = i + 1;
dpwr[i] = g[i] = 1ll * i * i + i + 1;
}
for ( int j = 1, t; j <= pn && ( t = i * pr[j] ) <= n; ++ j ) {
vis[t] = true;
if ( !( i % pr[j] ) ) {
mpwr[t] = mpwr[i] * pr[j] + 1;
dpwr[t] = dpwr[i] * pr[j] * pr[j] + pr[j] + 1;
sig[t] = sig[i] / mpwr[i] * mpwr[t];
g[t] = g[i] / dpwr[i] * dpwr[t];
break;
}
mu[t] = -mu[i];
mpwr[t] = mpwr[pr[j]];
dpwr[t] = dpwr[pr[j]];
sig[t] = sig[i] * sig[pr[j]];
g[t] = g[i] * g[pr[j]];
}
sigs[i] = ( sigs[i - 1] + sig[i] ) % MOD;
}
for ( int i = 1; i <= n; ++ i ) g[i] = i * g[i] % MOD;
for ( int i = 1; i <= n; ++ i ) {
for ( int j = 1, t = n / i; j <= t; ++ j ) {
f[i * j] = ( f[i * j] + 1ll * i * mu[i] * sig[j] % MOD * sigs[j] ) % MOD;
}
}
for ( int i = 1; i <= n; ++ i ) {
f[i] = 1ll * i * ( f[i] + MOD ) % MOD;
ans[i] = ( ( ans[i - 1] + 2ll * f[i] - g[i] ) % MOD + MOD ) % MOD;
}
}
int main () {
init ( MAXN );
for ( int T = rint (), i = 1; i <= T; ++ i ) {
printf ( "Case #%d: %d\n", i, ans[rint ()] );
}
return 0;
}
\(\mathcal{Details}\)
突然觉得推式子好养生啊。(