P9308 「DTOI-5」#1f1e33 题解

思路#

没有脑子怎么办，使用纯套路解决这道题。

$$\begin{array}{rl}& =\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^n[i+j+k=n]\mathrm{lcm}(i,gcd(j,k))\\ & =\sum _{i=1}^n\sum _{j=1}^n\sum _{k=1}^n[i+j+k=n]\sum _{d=1}^n\mathrm{lcm}(i,d)[gcd(j,k)=d]& (\text{枚举 gcd})\\ & =\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^{\frac{n}{d}}\sum _{k=1}^{\frac{n}{d}}[i+jd+kd=n]\mathrm{lcm}(i,d)[gcd(j,k)=1]\\ & =\sum _{d=1}^n\sum _{i=1}^n\sum _{j=1}^{\frac{n}{d}}\sum _{k=1}^{\frac{n}{d}}[i+jd+kd=n]\mathrm{lcm}(i,d)\sum _{p|gcd(j,k)}\mu (p)\\ & =\sum _{d=1}^n\sum _{p=1}^n\mu (p)\sum _{j=1}^{\frac{n}{dp}}\sum _{k=1}^{\frac{n}{dp}}[j+k\le \frac{n}{dp}]\mathrm{lcm}(n-jdp-kdp,d)\end{array}$$

令 $T=dp$。

$$\begin{array}{rl}& =\sum _{T=1}^n\sum _{d|T}\sum _{j=1}^{\frac{n}{T}}\sum _{k=1}^{\frac{n}{T}}[j+k\le \frac{n}{T}]\mathrm{lcm}(n-jT-kT,d)\mu (\frac{T}{d})\\ & =\sum _{T=1}^n\sum _{d|T}\sum _{j=1}^{\frac{n}{T}}\sum _{k=1}^{j-1}\mathrm{lcm}(n-jT,d)\mu (\frac{T}{d})& (\text{枚举 j+k 和 j})\\ & =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\sum _{j=1}^{\frac{n}{T}}\mathrm{lcm}(n-jT,d)(j-1)\\ & =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\sum _{j=1}^{\frac{n}{T}}\frac{(n-jT)d(j-1)}{gcd(n-jT,d)}& (\text{拆开 lcm})\end{array}$$

注意到 $d|T$，所以有 $gcd(n-jT,d)=gcd(n,d)$。

$$\begin{array}{rl}& =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\sum _{j=1}^{\frac{n}{T}}\frac{(n-jT)d(j-1)}{gcd(n,d)}\\ & =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})d\sum _{j=1}^{\frac{n}{T}}\frac{(n-jT)(j-1)}{gcd(n,d)}\\ & =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\frac{d}{gcd(n,d)}\sum _{j=1}^{\frac{n}{T}}(n-jT)(j-1)\end{array}$$

令 $S(x)=1+\cdots +x,S2(x)=1^2+\cdots +x^2$。

$$\begin{array}{rl}& =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\frac{d}{gcd(n,d)}\sum _{j=1}^{\frac{n}{T}}(n-jT)(j-1)\\ & =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\frac{d}{gcd(n,d)}\sum _{j=1}^{\frac{n}{T}}nj-n-j^{2}T+jT\\ & =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\frac{d}{gcd(n,d)}F(n,T)\end{array}$$

其中：

$$\begin{array}{rl}F(n,T)& =nS(\frac{n}{T})-n(\frac{n}{T})-TS2(\frac{n}{T})+TS(\frac{n}{T})\end{array}$$

这样的式子已经可以获得 $20\text{pts}$ 的高分了。

$$\begin{array}{rl}& =\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\frac{d}{gcd(n,d)}F(n,T)\\ & =\sum _{p=1}^n\sum _{T=1}^n\sum _{d|T}\mu (\frac{T}{d})\frac{d}{p}[gcd(n,d)=p]F(n,T)& (\text{传统艺能})\\ & =\sum _{p=1}^n[p|n]\sum _{T=1}^n\sum _{d|T,p|d}\mu (\frac{T}{d})\frac{d}{p}[gcd(n,d)=1]F(n,T)\\ & =\sum _{p=1}^n[p|n]\sum _{T=1}^n\sum _{d|T,p|d}\mu (\frac{T}{d})\frac{d}{p}\sum _{k|gcd(n,d)}\mu (k)F(n,T)\\ & =\sum _{p=1}^n\sum _{k=1}^n\mu (k)[pk|n]\sum _{T=1}^n\sum _{d|T,pk|d}\mu (\frac{T}{d})\frac{d}{p}F(n,T)\end{array}$$

令 $H=pk$。

$$\begin{array}{rl}& =\sum _{H|n}\sum _{k|H}\mu (k)\sum _{T=1}^n\sum _{d|T,H|d}\mu (\frac{T}{d})\frac{d}{p}F(n,T)\\ & =\sum _{H|n}\sum _{k|H}\mu (k)\sum _{T=1}^{\frac{n}{H}}\sum _{d|T}\mu (\frac{TH}{dH})\frac{dH}{p}F(n,TH)\\ & =\sum _{H|n}\sum _{k|H}\mu (k)\sum _{T=1}^{\frac{n}{H}}\sum _{d|T}\mu (\frac{T}{d})dkF(n,TH)\\ & =\sum _{H|n}\sum _{T=1}^{\frac{n}{H}}F(n,TH)\sum _{k|H}\mu (k)\sum _{d|T}\mu (\frac{T}{d})dk\\ & =\sum _{H|n}\sum _{T=1}^{\frac{n}{H}}F(n,TH)\sum _{k|H}\mu (k)k\sum _{d|T}\mu (\frac{T}{d})d\end{array}$$

令 $G1(x)=\sum _{d|x}\mu (\frac{x}{d})d=\phi (x),G2(x)=\sum _{d|x}\mu (x)x$。

$$\begin{array}{rl}& =\sum _{H|n}\sum _{T=1}^{\frac{n}{H}}F(n,TH)\sum _{k|H}\mu (k)k\sum _{d|T}\mu (\frac{T}{d})d\\ & =\sum _{H|n}\sum _{T=1}^{\frac{n}{H}}F(n,TH)G1(T)G2(H)\end{array}$$

这个式子就非常的简洁，可以获得 $40\text{pts}$ 的高分了。

$$\begin{array}{rl}& =\sum _{H|n}\sum _{T=1}^{\frac{n}{H}}F(n,TH)G1(T)G2(H)\\ & =\sum _{H|n}G2(H)\sum _{T=1}^{\frac{n}{H}}F(n,TH)G1(T)\\ & =\sum _{H|n}G2(H)\sum _{T=1}^{\frac{n}{H}}((n+TH)S(\frac{n}{TH})-n(\frac{n}{TH})-THS2(\frac{n}{TH}))G1(T)\\ & =\sum _{H|n}G2(H)\sum _{T=1}^{\frac{n}{H}}H((\frac{n}{H}+T)S(\frac{n}{TH})-\frac{n}{H}(\frac{n}{TH})-TS2(\frac{n}{TH}))G1(T)\\ & =\sum _{H|n}G2(H)H\sum _{T=1}^{\frac{n}{H}}((\frac{n}{H}+T)S(\frac{n}{TH})-\frac{n}{H}(\frac{n}{TH})-TS2(\frac{n}{TH}))G1(T)\\ & =\sum _{H|n}G2(H)HG(\frac{n}{H})\end{array}$$

其中：

$$\begin{array}{r}G(n)=\sum _{T=1}^n((n+T)S(\frac{n}{T})-n(\frac{n}{T})-TS2(\frac{n}{T}))G1(T)\end{array}$$

只需要计算 $G(n)$ 即可。

$$\begin{array}{rl}G(n)& =\sum _{T=1}^n((n+T)S(\frac{n}{T})-n(\frac{n}{T})-TS2(\frac{n}{T}))G1(T)\\ & =\sum _{T=1}^n-n\phi (i)(\frac{n}{i})+n\phi (i)S(\frac{n}{i})+i\phi (i)S(\frac{n}{i})-i\phi (i)S2(\frac{n}{i})\\ & =\sum _{T=1}^n\frac{(3(\frac{n}{i}{)}^2-3(\frac{n}{i}))n-(2(\frac{n}{i}{)}^3-2(\frac{n}{i}))i}{6}\end{array}$$

容易发现，预处理 $\phi (i),i\phi (i)$ 前缀和，可以使用整除分块做到 $O(n\sqrt{n})$。

可以获得 $60\text{pts}$ 的高分。

但实际上在算 $1\sim n$ 所有的整除分块的时候，我们有更加优秀的做法。

具体来说。

枚举 $i$ 与 $d=n/i$。

那么对应的 $n$ 要满足 $id\le n\le i(d+1)-1$。

这样就可以直接用前缀和计算。

至此，所有的东西都可以用最多调和计数的复杂度计算。

时间复杂度：$O(n\ln n)$。

Code#

/*
  ! 如果没有天赋，那就一直重复
  ! Created: 2024/05/24 10:38:16
*/
#include <bits/stdc++.h>
using namespace std;

#define fro(i, x, y) for (int i = (x); i <= (y); i++)
#define pre(i, x, y) for (int i = (x); i >= (y); i--)

const int N = 1e6 + 10;
const int mod = 998244353;

using i64 = long long;

int n, ct;
int vs[N], pr[N], mu[N];
i64 sm[N], fm[N], g1[N], g2[N], p1[N], p2[N], d2[N], d3[N], ans;
vector<int> to[N];

signed main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n, mu[1] = 1;
  fro(i, 2, n) {
    if (!vs[i]) pr[++ct] = i, mu[i] = -1;
    for (int j = 1; j <= ct && pr[j] * i <= n; j++) {
      vs[i * pr[j]] = 1;
      if (i % pr[j] == 0) {
        mu[i * pr[j]] = 0;
        break;
      }
      mu[i * pr[j]] = -mu[i];
    }
  }
  fro(i, 1, n) {
    for (int j = i, k = 1; j <= n; j += i, k++) {
      g1[j] += i * mu[k];
      g2[j] += i * mu[i];
    }
    g1[i] = (g1[i] % mod + mod) % mod;
    g2[i] = (g2[i] % mod + mod) % mod;
    d2[i] = 1ll * i * i % mod;
    d3[i] = 1ll * i * d2[i] % mod;
  }
  fro(i, 1, n) {
    for (i64 k = 1, l = i, r = 2 * i; l <= n; l += i, r += i, k++) {
      if (r > n) r = n + 1;
      p1[l] += g1[i] * (3 * d2[k] - 3 * k);
      p1[r] -= g1[i] * (3 * d2[k] - 3 * k);
      p2[l] += g1[i] * (2 * d3[k] - 2 * k) * i;
      p2[r] -= g1[i] * (2 * d3[k] - 2 * k) * i;
    }
  }
  fro(i, 1, n) {
    p1[i] = (p1[i] + p1[i - 1]) % mod;
    p2[i] = (p2[i] + p2[i - 1]) % mod;
    sm[i] = (p1[i] * i - p2[i]) % mod;
    sm[i] = (sm[i] % mod + mod) % mod;
    sm[i] = (sm[i] * 166374059) % mod;
  }
  fro(i, 1, n) {
    for (int j = i, k = 1; j <= n; j += i, k++)
      (fm[j] += i * g2[i] % mod * sm[k]) %= mod;
    cout << fm[i] % mod << " \n"[i == n];
  }
  return 0;
}