51nod 1220 约数之和

原题链接.

题解：

必要结论：
$σ(i*j)=\sum_{p|i} \sum_{q|j} p *{ j \over q} (gcd(p, q) =1)$
证明：
$\sum_{p|i} \sum_{q|j} p *{ j \over q} (gcd(p, q) =1)$
$=\sum_{p|i} \sum_{q|j} p *q (gcd(p, { j \over q}) =1)$

对每一个质因子分类讨论：

设$i=\sum p_i^{u_i}$,$j=p_i^{v_i}$
$p=\sum p_i^{l_i}$,$q=\sum p_i^{r_i}$

1.若$u_i>=l_i>0$,则$r_i = v_i$,此时可以表示出$p_i^{(v_i+1)->(u_i+v_i)}$。
2.$l_i=0$，则$v_i>=r_i>=0$,此时可以表示出$p_i^{0->v_i}$.

于是可以得出$p*q$一定可以表示出每个质因子$p_i^{0->(u_i+v_i)}$，合起来就包括了所有i*j的所有约数。

∴$σ(i*j)=\sum_{p|i} \sum_{q|j} p *{ j \over q} (gcd(p, q) =1)$

有了gcd就可以反演了。

$Ans = \sum_{i=1}^n \sum_{j=1}^n σ(i*j)$
$=\sum_{i=1}^n \sum_{j=1}^n\sum_{p|i} \sum_{q|j} p *{ j \over q} (gcd(p, q) =1)$
$=\sum_{p=1}^n \sum_{q=1}^n p*{\lfloor {n \over i} \rfloor}*{\lfloor {n \over j} \rfloor}*({\lfloor {n \over j} \rfloor}+1)/2(gcd(p, q) =1)$
$=\sum_{p=1}^n \sum_{q=1}^n q*{\lfloor {n \over i} \rfloor}*{\lfloor {n \over j} \rfloor}*({\lfloor {n \over j} \rfloor}+1)/2 \sum_{d|gcd(p, q)}μ(d)$
$=\sum_{d=1}^n μ(d)*d*\sum_{i=1}^{{\lfloor {n \over d} \rfloor}} \sum_{j=1}^{{\lfloor {n \over d} \rfloor}}i*{\lfloor {n \over d*i} \rfloor}*{\lfloor {n \over d*j} \rfloor}*({\lfloor {n \over d*j} \rfloor}+1)/2$
$=\sum_{d=1}^n μ(d)*d*(\sum_{i=1}^{{\lfloor {n \over d} \rfloor}}i*{\lfloor {n \over d*i} \rfloor})*(\sum_{i=1}^{{\lfloor {n \over d} \rfloor}} {\lfloor {n \over d*i} \rfloor}*({\lfloor {n \over d*i} \rfloor}+1)/2)$

$μ(d)*d$杜教筛，然后对d分块，$\sum_{i=1}^{{\lfloor {n \over d} \rfloor}}i*{\lfloor {n \over d*i} \rfloor})$和$(\sum_{i=1}^{{\lfloor {n \over d} \rfloor}} {\lfloor {n \over d*i} \rfloor}*({\lfloor {n \over d*i} \rfloor}+1)/2)$继续分块求。

时间复杂度:$O(n^{2/3})$

Code:

#include<cmath>
#include<cstdio>
#define ll long long
#define fo(i, x, y) for(ll i = x; i <= y; i ++)
#define min(a, b) ((a) < (b) ? (a) : (b))
using namespace std;

const ll N = 2000000, M = 6516531;
const ll mo = 1e9 + 7, ni_2 = 5e8 + 4;

bool bz[N + 5];
int p[N]; ll mu[N + 5];

ll n, ans;
ll h[M][2];

ll hash(ll x) {
    ll y = x % M;
    while(h[y][0] != 0 && h[y][0] != x) y = (y + 1) % M;
    return y;
}

ll dg(ll n) {
    if(n <= N) return mu[n];
    ll y = hash(n);
    if(h[y][0] == n) return h[y][1];
    h[y][0] = n;
    ll s = 1;
    fo(i, 1, n) {
        ll j = n / (n / i);
        s -= dg(n / i) * ((j - i + 1) * (i + j) / 2 % mo) % mo;
        i = j;
    }
    s = (s % mo + mo) % mo;
    h[y][1] = s;
    return s;
}

ll Solve(ll n) {
    ll p = 0, q = 0;
    fo(i, 1, n) {
        ll j = n / (n / i);
        p += (j - i + 1) * (i + j) / 2 % mo * (n / i);
        q += (n / i) * (n / i + 1) / 2 % mo * (j - i + 1) % mo;
        i = j;
    }
    return p % mo * (q % mo) % mo;
}


int main() {
    mu[1] = 1;
    fo(i, 2, N) {
        if(!bz[i]) p[++ p[0]] = i, mu[i] = -1;
        fo(j, 1, p[0]) {
            ll k = i * p[j];
            if(k > N) break;
            bz[k] = 1;
            if(i % p[j] == 0) {
                mu[k] = 0; break;
            }
            mu[k] = -mu[i];
        }
    }
    fo(i, 1, N) mu[i] = (mu[i] * i + mu[i - 1] + mo) % mo;
    scanf("%lld", &n);
    fo(d, 1, n) {
        ll d_ = n / (n / d);
        ans += Solve(n / d) * (dg(d_) - dg(d - 1) + mo) % mo;
        d = d_;
    }
    printf("%lld", ans % mo);
}