(mathcal{Description})
Link.
令 (sigma(n)) 为 (n) 的约数之和。求：
[sum_{i=1}^nsum_{j=1}^nmax{i,j}sigma(ij)mod(10^9+7)
]
多测，(nle10^6)，数据组数 (le5 imes10^4)。
(mathcal{Solution})
直 接 来 owo！
[sum_{i=1}^nsum_{j=1}^nmax{i,j}sigma(ij)=2sum_{i=1}^nisum_{j=1}^isigma(ij)-sum_{i=1}^nisigma(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 (xperpfrac{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[xperpfrac{j}y]
]
考虑原式前一项，记 (f(n)=nsum_{i=1}^nsigma(ni))，有：
[egin{aligned}
f(n)&=nsum_{i=1}^nsum_{x|n}sum_{y|i}xy[xperpfrac{i}y]\
&=nsum_{i=1}^nsum_{x|n}sum_{y|i}xysum_{d|xland d|frac{i}y}mu(d)\
&=nsum_{i=1}^nsum_{d|nland d|i}mu(d)sum_{x|nland d|x}sum_{y|iland d|y}frac{ix}y\
&=nsum_{i=1}^nsum_{d|nland d|i}mu(d)sum_{x|frac{n}d}sum_{y|frac{i}d}frac{ix}y~~~~~~~~(x,ymbox{ 同时约掉 } d)\
&=nsum_{i=1}^nsum_{d|nland d|i}mu(d)sigma(frac{n}d)sum_{y|frac{i}d}frac{i}y\
&=nsum_{i=1}^nsum_{d|nland d|i}mu(d)sigma(frac{n}d)dsum_{y|frac{i}d}frac{frac{i}y}d\
&=nsum_{i=1}^nsum_{d|nland d|i}dmu(d)sigma(frac{n}d)sigma(frac{i}d)\
&=nsum_{d|n}dmu(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(nln 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(nln 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
", i, ans[rint ()] );
}
return 0;
}
(mathcal{Details})
突然觉得推式子好养生啊。（