「MtOI2019」幽灵乐团

Description

求∏i=1A∏j=1B∏k=1C([i,j](i,k))f(type)\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}({[i,j]\over (i,k)})^{f(type)}i=1∏Aj=1∏Bk=1∏C((i,k)[i,j])f(type)
f(0)=1,f(1)=ijk,f(2)=(i,j,k)f(0)=1,f(1)=ijk,f(2)=(i,j,k)f(0)=1,f(1)=ijk,f(2)=(i,j,k)
设T为数据组数
A,B,C<=10^5,T<=70

Solution

被车万骗进来的我做了一天的数学题
三合一差评
公式太多了会跳步，建议手推
所有的除法都是下取整
M表示min(a,b,c)或min(a,b)

显然分成两部分，一部分是∏i=1A∏j=1B∏k=1Cif(type)\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}i^{f(type)}∏i=1A∏j=1B∏k=1Cif(type)
另一部分是∏i=1A∏j=1B∏k=1C(i,j)f(type)\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}(i,j)^{f(type)}∏i=1A∏j=1B∏k=1C(i,j)f(type)
考虑第一部分
type=1直接预处理阶乘快速幂
type=2也是预处理∏i=1nii\prod_{i=1}^{n}i^i∏i=1nii快速幂
type=3推式子：
∏i=1A∏j=1B∏k=1Ci(i,j,k)\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}i^{(i,j,k)}∏i=1A∏j=1B∏k=1Ci(i,j,k)
∏d=1M∏i=1A/d(id)d∑j=1B/d∑k=1C/d[(i,j,k)=1]\prod_{d=1}^{M}\prod_{i=1}^{A/d}(id)^{d\sum_{j=1}^{B/d}\sum_{k=1}^{C/d}[(i,j,k)=1]}∏d=1M∏i=1A/d(id)d∑j=1B/d∑k=1C/d[(i,j,k)=1]
∏x=1M∏d=1M/x∏i=1A/xd(idx)dμ(x)(B/xd)(C/xd)\prod_{x=1}^{M}\prod_{d=1}^{M/x}\prod_{i=1}^{A/xd}(idx)^{d\mu(x)(B/xd)(C/xd)}∏x=1M∏d=1M/x∏i=1A/xd(idx)dμ(x)(B/xd)(C/xd)
∏T=1M∏i=1A/T(iT)∑d∣Tdμ(T/x)(B/T)(C/T)\prod_{T=1}^{M}\prod_{i=1}^{A/T}(iT)^{\sum_{d|T}d\mu(T/x)(B/T)(C/T)}∏T=1M∏i=1A/T(iT)∑d∣Tdμ(T/x)(B/T)(C/T)
∏T=1M∏i=1A/T(iT)φ(T)(B/T)(C/T)\prod_{T=1}^{M}\prod_{i=1}^{A/T}(iT)^{\varphi(T)(B/T)(C/T)}∏T=1M∏i=1A/T(iT)φ(T)(B/T)(C/T)
预处理∏i=1niφ(i)\prod_{i=1}^{n}i^{\varphi(i)}∏i=1niφ(i)+分块即可

考虑第二部分
type=1求∏i=1A∏j=1B(i,j)\prod_{i=1}^{A}\prod_{j=1}^{B}(i,j)∏i=1A∏j=1B(i,j)再快速幂
∏d=1Md∑i=1A/d∑j=1B/d[(i,j)=1]\prod_{d=1}^{M}d^{\sum_{i=1}^{A/d}\sum_{j=1}^{B/d}[(i,j)=1]}∏d=1Md∑i=1A/d∑j=1B/d[(i,j)=1]
∏x=1M∏d=1M/xd(A/xd)(B/xd)μ(x)\prod_{x=1}^{M}\prod_{d=1}^{M/x}d^{(A/xd)(B/xd)\mu(x)}∏x=1M∏d=1M/xd(A/xd)(B/xd)μ(x)
∏T=1M∏d∣Tdμ(T/d)(A/T)(B/T)\prod_{T=1}^{M}\prod_{d|T}d^{\mu(T/d)(A/T)(B/T)}∏T=1M∏d∣Tdμ(T/d)(A/T)(B/T)
n log n预处理+分块
type=2同理，不再过多阐述
type=3
∏i=1A∏j=1B∏k=1C(i,j)(i,j,k)\prod_{i=1}^{A}\prod_{j=1}^{B}\prod_{k=1}^{C}(i,j)^{(i,j,k)}∏i=1A∏j=1B∏k=1C(i,j)(i,j,k)
∏d=1min(A,B)d(∑k=1C(k,d))(∑i=1A/d∑j=1B/d[(i,j)=1)]\prod_{d=1}^{min(A,B)}d^{(\sum_{k=1}^{C}(k,d))(\sum_{i=1}^{A/d}\sum_{j=1}^{B/d}[(i,j)=1)]}∏d=1min(A,B)d(∑k=1C(k,d))(∑i=1A/d∑j=1B/d[(i,j)=1)]
∏x=1min(A,B)∏d=1min(A,B)/xdμ(x)(∑k=1C(k,d))(A/xd)(B/xd)\prod_{x=1}^{min(A,B)}\prod_{d=1}^{min(A,B)/x}d^{\mu(x)(\sum_{k=1}^{C}(k,d))(A/xd)(B/xd)}∏x=1min(A,B)∏d=1min(A,B)/xdμ(x)(∑k=1C(k,d))(A/xd)(B/xd)
考虑∑k=1C(k,d)\sum_{k=1}^{C}(k,d)∑k=1C(k,d)
∑T∣dT∑k=1C/T[(k,d/T)=1]\sum_{T|d}T\sum_{k=1}^{C/T}[(k,d/T)=1]∑T∣dT∑k=1C/T[(k,d/T)=1]
∑x∣dμ(x)∑T∣dxT(C/Tx)\sum_{x|d}\mu(x)\sum_{T|{d\over x}}T(C/Tx)∑x∣dμ(x)∑T∣xdT(C/Tx)
∑i∣d(C/i)∑x∣iμ(x)ix\sum_{i|d}(C/i)\sum_{x|i}\mu(x){i\over x}∑i∣d(C/i)∑x∣iμ(x)xi
∑i∣d(C/i)φ(i)\sum_{i|d}(C/i)\varphi(i)∑i∣d(C/i)φ(i)
原式=∏i=1M∏x=1min(A,B)/i∏d=1min(A,B)/ix(id)(C/i)φ(i)μ(x)(A/idx)(B/idx)\prod_{i=1}^{M}\prod_{x=1}^{min(A,B)/i}\prod_{d=1}^{min(A,B)/ix}(id)^{(C/i)\varphi(i)\mu(x)(A/idx)(B/idx)}∏i=1M∏x=1min(A,B)/i∏d=1min(A,B)/ix(id)(C/i)φ(i)μ(x)(A/idx)(B/idx)
分成两部分
∏i=1Mi(C/i)φ(i)∑x=1min(A,B)/i∑d=1min(A,B)/ixμ(x)(A/idx)(B/idx)\prod_{i=1}^{M}i^{(C/i)\varphi(i)\sum_{x=1}^{min(A,B)/i}\sum_{d=1}^{min(A,B)/ix}\mu(x)(A/idx)(B/idx)}∏i=1Mi(C/i)φ(i)∑x=1min(A,B)/i∑d=1min(A,B)/ixμ(x)(A/idx)(B/idx)
考虑∑x=1min(A,B)/i∑d=1min(A,B)/ixμ(x)(A/idx)(B/idx)\sum_{x=1}^{min(A,B)/i}\sum_{d=1}^{min(A,B)/ix}\mu(x)(A/idx)(B/idx)∑x=1min(A,B)/i∑d=1min(A,B)/ixμ(x)(A/idx)(B/idx)
∑T=1min(A,B)/i(A/Ti)(B/Ti)∑d∣Tμ(d)\sum_{T=1}^{min(A,B)/i}(A/Ti)(B/Ti)\sum_{d|T}\mu(d)∑T=1min(A,B)/i(A/Ti)(B/Ti)∑d∣Tμ(d)
后面那个显然只有T=1为1
所以这一部分就是∏i=1Mi(A/i)(B/i)(C/i)φ(i)\prod_{i=1}^{M}i^{(A/i)(B/i)(C/i)\varphi(i)}∏i=1Mi(A/i)(B/i)(C/i)φ(i)，直接分块
∏i=1M∏x=1min(A,B)/i∏d=1min(A,B)/ixd(C/i)φ(i)μ(x)(A/idx)(B/idx)\prod_{i=1}^{M}\prod_{x=1}^{min(A,B)/i}\prod_{d=1}^{min(A,B)/ix}d^{(C/i)\varphi(i)\mu(x)(A/idx)(B/idx)}∏i=1M∏x=1min(A,B)/i∏d=1min(A,B)/ixd(C/i)φ(i)μ(x)(A/idx)(B/idx)
∏i=1M∏T=1min(A,B)/i∑d∣Tdμ(T/d)(A/iT)(B/iT)(C/i)φ(i)\prod_{i=1}^{M}\prod_{T=1}^{min(A,B)/i}\sum_{d|T}d^{\mu(T/d)(A/iT)(B/iT)(C/i)\varphi(i)}∏i=1M∏T=1min(A,B)/i∑d∣Tdμ(T/d)(A/iT)(B/iT)(C/i)φ(i)，分块套分块即可

如果写错了什么那也是无可奈何啊(摊手

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
using namespace std;typedef long long ll;const int N=1e5+5;int q,Mo,p[N],mu[N],phi[N];
bool vis[N];
ll f1[N],t1[N],f2[N],t2[N],s[N],f3[N],sp[N],Inv_t1[N],Inv_t2[N],a,b,c;ll pwr(ll x,ll y) {y%=Mo-1;if (y<0) y+=Mo-1;ll z=1;for(;y;y>>=1,x=x*x%Mo)if (y&1) z=z*x%Mo;return z;
}ll Inv(ll x) {return pwr(x,Mo-2);}ll S(ll x) {return x*(x+1)/2%(Mo-1);}void pre(int N) {mu[1]=phi[1]=1;fo(i,2,N) {if (!vis[i]) p[++p[0]]=i,mu[i]=-1,phi[i]=i-1;fo(j,1,p[0]) {int k=i*p[j];if (k>N) break;vis[k]=1;if (!(i%p[j])) {phi[k]=phi[i]*p[j];break;}mu[k]=-mu[i];phi[k]=phi[i]*(p[j]-1);}}f1[0]=f2[0]=f3[0]=1;fo(i,1,N) {f1[i]=f1[i-1]*i%Mo;f2[i]=f2[i-1]*pwr(i,i)%Mo;f3[i]=f3[i-1]*pwr(i,phi[i])%Mo;}fo(i,0,N) t1[i]=t2[i]=1;fo(d,1,N) fo(x,1,N/d) {int now=pwr(d,mu[x]);t1[d*x]=t1[d*x]*now%Mo;t2[d*x]=t2[d*x]*now%Mo;}Inv_t1[0]=Inv_t2[0]=1;fo(i,1,N) {t1[i]=t1[i-1]*t1[i]%Mo;t2[i]=t2[i-1]*pwr(t2[i],(ll)i*i)%Mo;Inv_t1[i]=Inv(t1[i]);Inv_t2[i]=Inv(t2[i]);sp[i]=(sp[i-1]+phi[i])%(Mo-1);}
}namespace Type_1{ll calc(ll n,ll m) {if (n>m) swap(n,m);ll ans=1;for(ll l=1,r=0;l<=n;l=r+1) {r=min(n/(n/l),m/(m/l));ans=ans*pwr(t1[r]*Inv_t1[l-1]%Mo,(n/l)*(m/r))%Mo;}return ans;}
}namespace Type_2{ll calc(ll n,ll m) {if (n>m) swap(n,m);ll ans=1;for(ll l=1,r=0;l<=n;l=r+1) {r=min(n/(n/l),m/(m/l));ans=ans*pwr(t2[r]*Inv_t2[l-1]%Mo,S(n/l)*S(m/l))%Mo;}return ans;}
}namespace Type_3{ll calc(ll a,ll b) {if (a>b) swap(a,b);ll ans=1;for(ll l=1,r=0;l<=a;l=r+1) {r=min(a/(a/l),b/(b/l));ans=ans*pwr(t1[r]*Inv_t1[l-1]%Mo,(a/l)*(b/l))%Mo;}return ans;}ll calc_1(ll a,ll b,ll c) {ll ans=1;int m=min(min(a,b),c);for(ll l=1,r=0;l<=m;l=r+1) {r=min(min(a/(a/l),b/(b/l)),c/(c/l));ll ret=(sp[r]-sp[l-1])%(Mo-1);ans=ans*pwr(f1[a/l],(b/l)*(c/l)%(Mo-1)*ret)%Mo;}return ans;}ll calc_2(ll a,ll b,ll c) {ll ans=1;int m=min(min(a,b),c);for(ll l=1,r=0;l<=m;l=r+1) {r=min(min(a/(a/l),b/(b/l)),c/(c/l));ans=ans*pwr(calc(a/l,b/l),(sp[r]-sp[l-1])*(c/l))%Mo;}return ans;}
}int main() {scanf("%d%d",&q,&Mo);pre(1e5);for(;q;q--) {scanf("%lld%lld%lld",&a,&b,&c);ll ans=pwr(f1[a],b*c)%Mo*pwr(f1[b],a*c)%Mo;ans=ans*Inv(pwr(Type_1::calc(a,b),c))%Mo;ans=ans*Inv(pwr(Type_1::calc(a,c),b))%Mo;printf("%lld ",ans);ans=pwr(f2[a],S(b)*S(c))*pwr(f2[b],S(a)*S(c))%Mo;ans=ans*Inv(pwr(Type_2::calc(a,b),S(c)))%Mo;ans=ans*Inv(pwr(Type_2::calc(a,c),S(b)))%Mo;printf("%lld ",ans);ans=Type_3::calc_1(a,b,c)*Type_3::calc_1(b,a,c)%Mo;ans=ans*Inv(Type_3::calc_2(a,b,c))%Mo;ans=ans*Inv(Type_3::calc_2(a,c,b))%Mo;printf("%lld\n",ans);}return 0;
}