[51 Nod 1584] 加权约数和

题意

求$\large \sum_{i=1}^N\sum_{j=1}^Nmax(i,j)\cdot\sigma_1(ij)$
其中
$1\le N\le10^6$
$1\le T\le5\cdot10^4$
$\sigma_1(n)$表示$n$的约数和

题目分析

令$A=\sum_{i=1}^n\sum_{j=1}^ii\cdot\sigma_1(ij),B=\sum_{i=1}^ni\cdot\sigma_1(i^2)$,则$Ans=2A-B$

且有
$\large \begin{aligned} A&=\sum_{i=1}^ni\cdot\sum_{j=1}^i\sum_{x|i}\sum_{y|j}x\cdot j/y[(x,y)=1]\\ &=\sum_{i=1}^ni\cdot\sum_{j=1}^i\sum_{x|i}\sum_{y|j}x\cdot j/y\sum_{d|(x,y)}\mu(d)\\ &=\sum_{d=1}^n\mu(d)\sum_{d|x}x\sum_{x|i}i\sum_{d|y}\sum_{y|j}^i\frac jy\\ &=\sum_{d=1}^n\mu(d)\sum_{x=1}^{\lfloor\frac nd\rfloor}dx\sum_{x|i}^{\lfloor\frac nd\rfloor}di\sum_{y=1}^{\lfloor\frac nd\rfloor}\sum_{y|j}^i\frac{dj}{dy}\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y=1}^{\lfloor\frac nd\rfloor}\sum_{y|j}^iy\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)\sum_{j=1}^{i}\sigma_1(j)\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)\cdot S_{\sigma_1}(i)\\\end{aligned}$
$\large \begin{aligned} B&=\sum_{i=1}^ni\sum_{x|i}\sum_{y|i}x\cdot i/y[(x,y)=1]\\ &=\sum_{i=1}^ni\cdot\sum_{x|i}\sum_{y|i}x\cdot i/y\sum_{d|(x,y)}\mu(d)\\ &=\sum_{d=1}^n\mu(d)\sum_{d|i}i\sum_{d|x|i}\sum_{d|y|i}x\cdot \frac iy\\ &=\sum_{d=1}^n\mu(d)\sum_{i=1}^{\lfloor\frac nd\rfloor}di\sum_{x|i}\sum_{y|i}dx\cdot \frac {di}{dy}\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y|i}\frac {i}{y}\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\sum_{x|i}x\sum_{y|i}y\\ &=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)^2\\ \end{aligned}$
$\large \begin{aligned} \therefore Ans(n)&=2A-B\\&=\sum_{d=1}^n\mu(d)d^2\sum_{i=1}^{\lfloor\frac nd\rfloor}i\cdot\sigma_1(i)(2\cdot S_{\sigma_1}(i)-\sigma_1(i)) \end{aligned}$

发现$Ans(n)$可以递推：
$Ans(n)=Ans(n-1)+\sum_{d|n}\mu(d)d^2{\lfloor\frac nd\rfloor}\cdot\sigma_1({\lfloor\frac nd\rfloor})(2\cdot S_{\sigma_1}({\lfloor\frac nd\rfloor})-\sigma_1({\lfloor\frac nd\rfloor}))$
那么就可以$O(n\log n)$预处理了。

然后每次$O(1)$回答就完了。

CODE

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 1000005;
const int mod = 1000000007;
int Cnt, Prime[N], mu[N];
LL ans[N], sd[N], a[N], d[N], f[N];
bool IsnotPrime[N];

void init() {
  d[1] = 1, a[1] = 1, mu[1] = 1;
  for(int i = 2; i < N; ++i) {
    if(!IsnotPrime[i])
      Prime[++Cnt] = i, d[i] = a[i] = 1+i, mu[i] = -1;
    for(int j = 1, k; j <= Cnt && i * Prime[j] < N; ++j) {
      IsnotPrime[k = i * Prime[j]] = 1;
      if(i % Prime[j] == 0) {
        mu[k] = 0;
        a[k] = a[i] * Prime[j] + 1;
        d[k] = d[i] / a[i] * a[k];
        break;
      }
      mu[k] = -mu[i];
      a[k] = 1 + Prime[j];
      d[k] = d[i] * a[k];
    }
  }
  for(int i = 1; i < N; ++i) {
    sd[i] = (sd[i-1] + (d[i]%=mod)) % mod;
    f[i] = i*d[i]%mod*(2*sd[i]-d[i]+mod)%mod;
  }
  for(int i = 1; i < N; ++i) if(mu[i])
    for(int j = i; j < N; j += i)
      ans[j] = (ans[j] + 1ll*mu[i]*i*i%mod*f[j/i]%mod + mod) % mod;
  for(int i = 1; i < N; ++i) ans[i] = (ans[i-1] + ans[i]) % mod;
}
int main () {
  init();
  int T, ks = 0, N; scanf("%d", &T);
  while(T--) scanf("%d", &N), printf("Case #%d: %lld\n", ++ks, ans[N]);
}