LibreOJ 6693 「MtOI2019」幽灵乐团

这题其实是个六合一（
不仅代码写得精神污染，这篇题解写得更加精神污染……
以下分析复杂度时设 $A,B,C$ 同阶，并以 $n$ 代替。

以及 CYJ 题解中对于 $\mathrm{t}\mathrm{y}\mathrm{p}\mathrm{e}=2$ 的枚举 $gcd(i,j)$ 的推导好像有点问题，不过也许是我太菜了（
所以这里直接枚举 $gcd(i,j,k)$。

第一步：

$$\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{\left(\frac{\mathrm{lcm}(i,j)}{gcd(i,k)}\right)}^{f(\text{type})}=\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{\left(\frac{ij}{gcd(i,j)gcd(i,k)}\right)}^{f(\text{type})}$$

所以？
所以分别算

$$\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{i}^{f(\text{type})},\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{gcd(i,j)}^{f(\text{type})}$$

即可。
于是这就成了个六合一（

type = 0

$$\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^{C}i=(A!{)}^{BC}$$

$$\begin{array}{rl}\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^{C}gcd(i,j)& =\prod _{i=1}^A\prod _{j=1}^B{\text{gcd}}^C(i,j)\\ & =\prod _{d=1}^{min(A,B)}\prod _{i=1}^{\lfloor \frac{A}{d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{d}\rfloor }{d}^{[gcd(i,j)=1]C}\\ & =\prod _{d=1}^{min(A,B)}{d}^{C\sum _{i=1}^{\lfloor \frac{A}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{B}{d}\rfloor }[gcd(i,j)=1]}\\ & =\prod _{d=1}^{min(A,B)}{d}^{C\sum _{k=1}^{min\left(\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor \right)}\mu (k)\lfloor \frac{A}{dk}\rfloor \lfloor \frac{B}{dk}\rfloor }\\ & =\prod _{T=1}^{min(A,B)}\prod _{d|T}{d}^{\mu \left(\frac{T}{d}\right)\lfloor \frac{A}{T}\rfloor \lfloor \frac{B}{T}\rfloor C}\\ & =\prod _{T=1}^{min(A,B)}{\left(\prod _{d|T}{d}^{\mu \left(\frac{T}{d}\right)}\right)}^{\lfloor \frac{A}{T}\rfloor \lfloor \frac{B}{T}\rfloor C}\end{array}$$

$O(n\log n)$ 预处理 $\prod _{d|T}{d}^{\mu \left(\frac{T}{d}\right)}$ 的前缀积和前缀积的逆，然后单次询问 $O(\sqrt{n}\log n)$。

type = 1

$$\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{i}^{ijk}={\left(\prod _{i=1}^A{i}^i\right)}^{BC(B+1)(C+1)/4}$$

$$\begin{array}{rl}\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{\text{gcd}}^{ijk}(i,j)& ={\left(\prod _{i=1}^A\prod _{j=1}^B{\text{gcd}}^{ij}(i,j)\right)}^{C(C+1)/2}\\ & ={\left(\prod _{d=1}^{min(A,B)}\prod _{i=1}^{\lfloor \frac{A}{d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{d}\rfloor }{d}^{[gcd(i,j)=1]d^{2}ij}\right)}^{C(C+1)/2}\\ & ={\left(\prod _{d=1}^{min(A,B)}{d}^{d^2\sum _{i=1}^{\lfloor \frac{A}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{B}{d}\rfloor }[gcd(i,j)=1]ij}\right)}^{C(C+1)/2}\\ & ={\left(\prod _{d=1}^{min(A,B)}{d}^{d^2\sum _{k=1}^{min\left(\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor \right)}\mu (k)k^2\lfloor \frac{A}{dk}\rfloor \lfloor \frac{B}{dk}\rfloor \left(\lfloor \frac{A}{dk}\rfloor +1\right)\left(\lfloor \frac{B}{dk}\rfloor +1\right)/4}\right)}^{C(C+1)/2}\\ & ={\left(\prod _{T=1}^{min(A,B)}{\left(\prod _{d|T}{d}^{\mu \left(\frac{T}{d}\right)}\right)}^{T^2\lfloor \frac{A}{T}\rfloor \lfloor \frac{B}{T}\rfloor \left(\lfloor \frac{A}{T}\rfloor +1\right)\left(\lfloor \frac{B}{T}\rfloor +1\right)/4}\right)}^{C(C+1)/2}\end{array}$$

$O(n\log n)$ 预处理 ${\left(\prod _{d|T}{d}^{\mu \left(\frac{T}{d}\right)}\right)}^{T^2}$ 的前缀积和前缀积的逆，然后单词询问 $O(\sqrt{n}\log n)$。

type = 2

$$\begin{array}{rl}\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{i}^{gcd(i,j,k)}& =\prod _{d=1}^{min(A,B,C)}\prod _{i=1}^{\lfloor \frac{A}{d}\rfloor }(id{)}^{d\sum _{j=1}^{\lfloor \frac{B}{d}\rfloor }\sum _{k=1}^{\lfloor \frac{B}{d}\rfloor }[gcd(i,j,k)=1]}\\ & =\prod _{d=1}^{min(A,B,C)}\prod _{t=1}^{\lfloor \frac{A}{d}\rfloor }\prod _{i=1}^{\lfloor \frac{A}{td}\rfloor }(itd{)}^{\mu (t)d\lfloor \frac{B}{td}\rfloor \lfloor \frac{C}{td}\rfloor }\\ & =\prod _{T=1}^{min(A,B,C)}\prod _{d|T}{\left(\lfloor \frac{A}{T}\rfloor !T^{\lfloor \frac{A}{T}\rfloor }\right)}^{\mu \left(\frac{T}{d}\right)d\lfloor \frac{B}{T}\rfloor \lfloor \frac{C}{T}\rfloor }\\ & =\prod _{T=1}^{min(A,B,C)}{\left(\lfloor \frac{A}{T}\rfloor !T^{\lfloor \frac{A}{T}\rfloor }\right)}^{\phi (T)\lfloor \frac{B}{T}\rfloor \lfloor \frac{C}{T}\rfloor }\end{array}$$

预处理一下 $T^{\phi (T)}$ 的前缀积即可。

$$\begin{array}{rl}\prod _{i=1}^A\prod _{j=1}^B\prod _{k=1}^C{gcd(i,j)}^{gcd(i,j,k)}& =\prod _{d=1}^{min(A,B,C)}\prod _{i=1}^{\lfloor \frac{A}{d}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{d}\rfloor }(dgcd(i,j){)}^{d\sum _{k=1}^{\lfloor \frac{C}{d}\rfloor }[gcd(i,j,k)=1]}\\ & =\prod _{d=1}^{min(A,B,C)}\prod _{t=1}^{min\left(\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \right)}\prod _{i=1}^{\lfloor \frac{A}{td}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{td}\rfloor }(tdgcd(i,j){)}^{\mu (t)d\lfloor \frac{C}{td}\rfloor }\end{array}$$

把底数拆出来，看 $td$ 的部分：

$$\begin{array}{rl}& \prod _{d=1}^{min(A,B,C)}\prod _{t=1}^{min\left(\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \right)}\prod _{i=1}^{\lfloor \frac{A}{td}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{td}\rfloor }(td{)}^{\mu (t)d\lfloor \frac{C}{td}\rfloor }\\ =& \prod _{d=1}^{min(A,B,C)}\prod _{t=1}^{min\left(\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \right)}(td{)}^{\mu (t)d\lfloor \frac{A}{td}\rfloor \lfloor \frac{B}{td}\rfloor \lfloor \frac{C}{td}\rfloor }\\ =& \prod _{T=1}^{min(A,B,C)}{T}^{\lfloor \frac{A}{T}\rfloor \lfloor \frac{B}{T}\rfloor \lfloor \frac{C}{T}\rfloor \phi (T)}\end{array}$$

是否有一种似曾相识的感觉？
看看上面那个 $i^{gcd(i,j,k)}$ 的式子，发现可以约分掉这部分。
于是乎那个式子只要求

$$\prod _{T=1}^{min(A,B,C)}{\left(\lfloor \frac{A}{T}\rfloor !\right)}^{\phi (T)\lfloor \frac{B}{T}\rfloor \lfloor \frac{C}{T}\rfloor }$$

而这个式子要求

$$\begin{array}{rl}& \prod _{d=1}^{min(A,B,C)}\prod _{t=1}^{min\left(\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \right)}\prod _{i=1}^{\lfloor \frac{A}{td}\rfloor }\prod _{j=1}^{\lfloor \frac{B}{td}\rfloor }{\text{gcd}}^{\mu (t)d\lfloor \frac{C}{td}\rfloor }(i,j)\\ =& \prod _{d=1}^{min(A,B,C)}\prod _{t=1}^{min\left(\lfloor \frac{A}{d}\rfloor ,\lfloor \frac{B}{d}\rfloor ,\lfloor \frac{C}{d}\rfloor \right)}\prod _{k=1}^{min\left(\lfloor \frac{A}{td}\rfloor ,\lfloor \frac{B}{td}\rfloor ,\lfloor \frac{C}{td}\rfloor \right)}\prod _{l=1}^{min\left(\lfloor \frac{A}{tdk}\rfloor ,\lfloor \frac{B}{tdk}\rfloor ,\lfloor \frac{C}{tdk}\rfloor \right)}(kl{)}^{\mu (t)\mu (l)d\lfloor \frac{C}{td}\rfloor \lfloor \frac{A}{tdkl}\rfloor \lfloor \frac{B}{tdkl}\rfloor }\\ =& \prod _{T=1}^{min(A,B,C)}\prod _{K=1}^{min\left(\lfloor \frac{A}{T}\rfloor ,\lfloor \frac{B}{T}\rfloor ,\lfloor \frac{C}{T}\rfloor \right)}\prod _{d|K}{d}^{\mu \left(\frac{K}{d}\right)\lfloor \frac{A}{TK}\rfloor \lfloor \frac{B}{TK}\rfloor \phi (T)\lfloor \frac{C}{T}\rfloor }\end{array}$$

然后两只数论分块加上 $O(n\log n)$ 预处理出前缀积和前缀积的逆就可以单次询问 $O(n^{3/4}\log n)$ 了。
（在洛谷要开 O2 才能过，不过常数确实还有优化空间）

代码：

#pragma GCC optimize (2)
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 1e5;
int T,mod;
int fac[N + 5],ifac[N + 5],inv[N + 5];
int es[N + 5],s[N + 5];
int vis[N + 5],cnt,prime[N + 5];
int mu[N + 5],emu[N + 5],pemu[N + 5],iemu[N + 5],esmu[N + 5],pesmu[N + 5],iesmu[N + 5];
int phi[N + 5],sphi[N + 5],ephi[N + 5];
int A,B,C;
int ans1 = 1,ans2 = 1;
inline int fpow(int a,int b)
{
    int ret = 1;
    for(;b;b >>= 1)
        (b & 1) && (ret = (long long)ret * a % mod),a = (long long)a * a % mod;
    return ret;
}
inline int f(int A,int B)
{
    int ret = 1;
    for(register int l = 1,r;l <= min(A,B);l = r + 1)
    {
        r = min(A / (A / l),B / (B / l));
        ret = (long long)ret * fpow((long long)pemu[r] * iemu[l - 1] % mod,(long long)(A / l) * (B / l) % (mod - 1)) % mod;
    }
    return ret;
}
int main()
{
    scanf("%d%d",&T,&mod),fac[0] = inv[0] = mu[1] = phi[1] = pemu[0] = es[0] = pesmu[0] = 1;
    for(register int i = 2;i <= N;++i)
    {
        if(!vis[i])
            mu[prime[++cnt] = i] = mod - 2,phi[i] = i - 1;
        for(register int j = 1;j <= cnt && i * prime[j] <= N;++j)
        {
            vis[i * prime[j]] = 1;
            if(!(i % prime[j]))
            {
                phi[i * prime[j]] = (long long)phi[i] * prime[j] % mod;
                break;
            }
            mu[i * prime[j]] = (mod - 1 - mu[i]) % (mod - 1),phi[i * prime[j]] = (long long)phi[i] * (prime[j] - 1) % mod;
        }
    }
    for(register int i = 0;i <= N;++i)
        emu[i] = 1,s[i] = (long long)i * (i + 1) / 2 % (mod - 1);
    for(register int i = 1;i <= N;++i)
        for(register int j = 1;i * j <= N;++j)
            emu[i * j] = (long long)emu[i * j] * fpow(i,mu[j]) % mod;
    for(register int i = 1;i <= N;++i)
        esmu[i] = fpow(emu[i],(long long)i * i % (mod - 1)),
        sphi[i] = (sphi[i - 1] + phi[i]) % (mod - 1);
    for(register int i = 1;i <= N;++i)
        pemu[i] = (long long)pemu[i - 1] * emu[i] % mod,
        fac[i] = (long long)fac[i - 1] * i % mod,
        es[i] = (long long)es[i - 1] * fpow(i,i) % mod,
        pesmu[i] = (long long)pesmu[i - 1] * esmu[i] % mod;
    iemu[N] = fpow(pemu[N],mod - 2),
    ifac[N] = fpow(fac[N],mod - 2),
    iesmu[N] = fpow(pesmu[N],mod - 2);
    for(register int i = N;i;--i)
        iemu[i - 1] = (long long)iemu[i] * emu[i] % mod,
        ifac[i - 1] = (long long)ifac[i] * i % mod,
        inv[i] = (long long)ifac[i] * fac[i - 1] % mod,
        iesmu[i - 1] = (long long)iesmu[i] * esmu[i] % mod;
    for(;T;--T)
    {
        scanf("%d%d%d",&A,&B,&C);

        ans1 = (long long)fpow(fac[A],(long long)B * C % (mod - 1)) * fpow(fac[B],(long long)A * C % (mod - 1)) % mod;
        for(register int l = 1,r;l <= min(A,B);l = r + 1)
        {
            r = min(A / (A / l),B / (B / l));
            ans2 = (long long)ans2 * fpow((long long)pemu[r] * iemu[l - 1] % mod,(long long)(A / l) * (B / l) % (mod - 1) * C % (mod - 1)) % mod;
        }
        for(register int l = 1,r;l <= min(A,C);l = r + 1)
        {
            r = min(A / (A / l),C / (C / l));
            ans2 = (long long)ans2 * fpow((long long)pemu[r] * iemu[l - 1] % mod,(long long)(A / l) * (C / l) % (mod - 1) * B % (mod - 1)) % mod;
        }
        printf("%d ",(int)((long long)ans1 * fpow(ans2,mod - 2) % mod)),ans1 = ans2 = 1;

        ans1 = (long long)fpow(es[A],(long long)s[B] * s[C] % (mod - 1)) * fpow(es[B],(long long)s[A] * s[C] % (mod - 1)) % mod;
        for(register int l = 1,r;l <= min(A,B);l = r + 1)
        {
            r = min(A / (A / l),B / (B / l));
            ans2 = (long long)ans2 * fpow((long long)pesmu[r] * iesmu[l - 1] % mod,(long long)s[A / l] * s[B / l] % (mod - 1) * s[C] % (mod - 1)) % mod;
        }
        for(register int l = 1,r;l <= min(A,C);l = r + 1)
        {
            r = min(A / (A / l),C / (C / l));
            ans2 = (long long)ans2 * fpow((long long)pesmu[r] * iesmu[l - 1] % mod,(long long)s[A / l] * s[C / l] % (mod - 1) * s[B] % (mod - 1)) % mod;
        }
        printf("%d ",(int)((long long)ans1 * fpow(ans2,mod - 2) % mod)),ans1 = ans2 = 1;

        for(register int l = 1,r;l <= min(A,min(B,C));l = r + 1)
        {
            r = min(A / (A / l),min(B / (B / l),C / (C / l)));
            ans1 = (long long)ans1 * fpow(fac[A / l],(long long)(sphi[r] - sphi[l - 1] + mod - 1) * (B / l) % (mod - 1) * (C / l) % (mod - 1)) % mod;
        }
        for(register int l = 1,r;l <= min(A,min(B,C));l = r + 1)
        {
            r = min(A / (A / l),min(B / (B / l),C / (C / l)));
            ans1 = (long long)ans1 * fpow(fac[B / l],(long long)(sphi[r] - sphi[l - 1] + mod - 1) * (A / l) % (mod - 1) * (C / l) % (mod - 1)) % mod;
        }
        for(register int l = 1,r;l <= min(A,min(B,C));l = r + 1)
        {
            r = min(A / (A / l),min(B / (B / l),C / (C / l)));
            ans2 = (long long)ans2 * fpow(f(A / l,B / l),(long long)(sphi[r] - sphi[l - 1] + mod - 1) * (C / l) % (mod - 1)) % mod;
        }
        for(register int l = 1,r;l <= min(A,min(B,C));l = r + 1)
        {
            r = min(A / (A / l),min(B / (B / l),C / (C / l)));
            ans2 = (long long)ans2 * fpow(f(A / l,C / l),(long long)(sphi[r] - sphi[l - 1] + mod - 1) * (B / l) % (mod - 1)) % mod;
        }
        printf("%d\n",(int)((long long)ans1 * fpow(ans2,mod - 2) % mod)),ans1 = ans2 = 1;
    }
}