LibreOJ 6693 「MtOI2019」幽灵乐团

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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 type=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： A∏i=1B∏j=1C∏k=1(lcm(i,j)gcd(i,k))f(type)=A∏i=1B∏j=1C∏k=1(ijgcd(i,j)gcd(i,k))f(type)

所以？
所以分别算 A∏i=1B∏j=1C∏k=1if(type),A∏i=1B∏j=1C∏k=1gcd(i,j)f(type)

即可。
于是这就成了个六合一（

type = 0

A∏i=1B∏j=1C∏k=1i=(A!)BC

A∏i=1B∏j=1C∏k=1gcd(i,j)=A∏i=1B∏j=1gcdC(i,j)=min(A,B)∏d=1⌊Ad⌋∏i=1⌊Bd⌋∏j=1d[gcd(i,j)=1]C=min(A,B)∏d=1dC⌊Ad⌋∑i=1⌊Bd⌋∑j=1[gcd(i,j)=1]=min(A,B)∏d=1dCmin(⌊Ad⌋,⌊Bd⌋)∑k=1μ(k)⌊Adk⌋⌊Bdk⌋=min(A,B)∏T=1∏d|Tdμ(Td)⌊AT⌋⌊BT⌋C=min(A,B)∏T=1(∏d|Tdμ(Td))⌊AT⌋⌊BT⌋C

O(nlogn) 预处理 ∏d|Tdμ(Td) 的前缀积和前缀积的逆，然后单次询问 O(√nlogn)。

type = 1

A∏i=1B∏j=1C∏k=1iijk=(A∏i=1ii)BC(B+1)(C+1)/4

A∏i=1B∏j=1C∏k=1gcdijk(i,j)=(A∏i=1B∏j=1gcdij(i,j))C(C+1)/2=(min(A,B)∏d=1⌊Ad⌋∏i=1⌊Bd⌋∏j=1d[gcd(i,j)=1]d2ij)C(C+1)/2=(min(A,B)∏d=1dd2⌊Ad⌋∑i=1⌊Bd⌋∑j=1[gcd(i,j)=1]ij)C(C+1)/2=(min(A,B)∏d=1dd2min(⌊Ad⌋,⌊Bd⌋)∑k=1μ(k)k2⌊Adk⌋⌊Bdk⌋(⌊Adk⌋+1)(⌊Bdk⌋+1)/4)C(C+1)/2=(min(A,B)∏T=1(∏d|Tdμ(Td))T2⌊AT⌋⌊BT⌋(⌊AT⌋+1)(⌊BT⌋+1)/4)C(C+1)/2

O(nlogn) 预处理 (∏d|Tdμ(Td))T2 的前缀积和前缀积的逆，然后单词询问 O(√nlogn)。

type = 2

A∏i=1B∏j=1C∏k=1igcd(i,j,k)=min(A,B,C)∏d=1⌊Ad⌋∏i=1(id)d⌊Bd⌋∑j=1⌊Bd⌋∑k=1[gcd(i,j,k)=1]=min(A,B,C)∏d=1⌊Ad⌋∏t=1⌊Atd⌋∏i=1(itd)μ(t)d⌊Btd⌋⌊Ctd⌋=min(A,B,C)∏T=1∏d|T(⌊AT⌋!T⌊AT⌋)μ(Td)d⌊BT⌋⌊CT⌋=min(A,B,C)∏T=1(⌊AT⌋!T⌊AT⌋)φ(T)⌊BT⌋⌊CT⌋

预处理一下 Tφ(T) 的前缀积即可。

A∏i=1B∏j=1C∏k=1gcd(i,j)gcd(i,j,k)=min(A,B,C)∏d=1⌊Ad⌋∏i=1⌊Bd⌋∏j=1(dgcd(i,j))d⌊Cd⌋∑k=1[gcd(i,j,k)=1]=min(A,B,C)∏d=1min(⌊Ad⌋,⌊Bd⌋,⌊Cd⌋)∏t=1⌊Atd⌋∏i=1⌊Btd⌋∏j=1(tdgcd(i,j))μ(t)d⌊Ctd⌋

把底数拆出来，看 td 的部分：

min(A,B,C)∏d=1min(⌊Ad⌋,⌊Bd⌋,⌊Cd⌋)∏t=1⌊Atd⌋∏i=1⌊Btd⌋∏j=1(td)μ(t)d⌊Ctd⌋=min(A,B,C)∏d=1min(⌊Ad⌋,⌊Bd⌋,⌊Cd⌋)∏t=1(td)μ(t)d⌊Atd⌋⌊Btd⌋⌊Ctd⌋=min(A,B,C)∏T=1T⌊AT⌋⌊BT⌋⌊CT⌋φ(T)

是否有一种似曾相识的感觉？
看看上面那个 igcd(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 min(A,B,C)∏T=1(⌊AT⌋!)φ(T)⌊BT⌋⌊CT⌋

而这个式子要求 min(A,B,C)∏d=1min(⌊Ad⌋,⌊Bd⌋,⌊Cd⌋)∏t=1⌊Atd⌋∏i=1⌊Btd⌋∏j=1gcdμ(t)d⌊Ctd⌋(i,j)=min(A,B,C)∏d=1min(⌊Ad⌋,⌊Bd⌋,⌊Cd⌋)∏t=1min(⌊Atd⌋,⌊Btd⌋,⌊Ctd⌋)∏k=1min(⌊Atdk⌋,⌊Btdk⌋,⌊Ctdk⌋)∏l=1(kl)μ(t)μ(l)d⌊Ctd⌋⌊Atdkl⌋⌊Btdkl⌋=min(A,B,C)∏T=1min(⌊AT⌋,⌊BT⌋,⌊CT⌋)∏K=1∏d|Kdμ(Kd)⌊ATK⌋⌊BTK⌋φ(T)⌊CT⌋

然后两只数论分块加上 O(nlogn) 预处理出前缀积和前缀积的逆就可以单次询问 O(n3/4logn) 了。
（在洛谷要开 O2 才能过，不过常数确实还有优化空间）

代码： 

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#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;
    }
}

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