[SDOI2017]数字表格 & [MtOI2019]幽灵乐团
作者：互联网
P3704 [SDOI2017]数字表格
首先根据题意写出答案的表达式
\[\large\prod_{i=1}^n\prod_{j=1}^mf_{\gcd(i,j)} \]按常规套路改为枚举 \(d=\gcd(i,j)\)
（不妨设 \(n\le m\) ）
指数上的式子很熟悉了，单独拿出来推一下
\[\begin{aligned} \sum_{i=1}^n\sum_{j=1}^m[(i,j)=d] &=\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}[(i,j)=1]\\ &=\sum_{i=1}^{n/d}\sum_{j=1}^{m/d}\sum_{t\mid (i,j)}\mu(t)\\ &=\sum_{t=1}^{n/d}\mu(t)\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}[t|i][t|j]\\ &=\sum_{t=1}^{n/d}\mu(t)(n/dt)(m/dt) \end{aligned}\]代回原式
\[\large\prod_{d=1}^n{f_d}^{\sum_{t=1}^{n/d}\mu(t)(n/dt)(m/dt)} \]设 \(k=dt\) ，改为枚举 \(k,d\)
\[\large\prod_{k=1}^n\prod_{d\mid k}{f_d}^{\mu(k/d)(n/k)(m/k)} \]\[\large\prod_{k=1}^n\left(\prod_{d\mid k}{f_d}^{\mu(k/d)}\right)^{(n/k)(m/k)} \]\[\large\prod_{k=1}^n\left(\prod_{d\mid k}{f_{k/d}}^{\mu(d)}\right)^{(n/k)(m/k)} \]括号里的东西可以 \(O(n\log n)\) 预处理，然后整除分块计算就可以了
时间复杂度 \(O(n\log n+T\sqrt n\log n)\)
带 \(\log\) 是因为用到了快速幂
到这里已经可以 AC 了
但还是介绍一下 \(O(n\log\log n+T\sqrt n\log n)\) 的做法
可以理解为 dp ，设：
\[\large s_{i,n}=\prod_{d\mid n,d 只含前 i 种质因子}{f_{n/d}}^{\mu(d)} \]\[\large inv_{i,n}={s_{i,n}}^{-1}=\prod_{d\mid n,d 只含前 i 种质因子}{f_{n/d}}^{-\mu(d)} \]枚举质数 \(p_i\)
当 \(p_i\nmid n\) 时，显然 \(s_{i,n}=s_{i-1,n},inv_{i,n}=inv_{i-1,n}\)
当 \(p_i\mid n\) 时，需要考虑 \(d\) 中含因子 \(p_i\) 的个数
\(p_i\nmid d\) 的情况对 \(s_{i,n}\) 的贡献即为 \(s_{i-1,n}\)
\({p_i}^2\mid d\) 时 \(\mu(d)=0\) ，因此这时不产生贡献
若 \(p_i\mid d\) 且 \({p_i}^2\nmid d\) ，设 \(d=p_id'\) ，则贡献为
\[\prod_{d'\mid (n/p_i)}{f_{n/p_id'}}^{\mu(p_id')} \]\(p_i\) 与 \(d'\) 互质，因此 \(\mu(p_id')=\mu(p_i)\mu(d')=-\mu(d')\)
代回之后发现贡献就是 \(inv_{i-1,n/p_i}\)
故 \(s_{i,n}=s_{i-1,n}\times inv_{i-1,n/p_i}\)
显然 \(inv_{i,n}=inv_{i-1,n}\times s_{i-1,n/p_i}\)
综上，转移方程为
\[s_{i,n}=\begin{cases}s_{i-1,n}&(p_i\nmid n)\\ s_{i-1,n}\times invs_{i-1,n/p_i} & (p_i\mid n)\end{cases} \]\[inv_{i,n}=\begin{cases}inv_{i-1,n}&(p_i\nmid n)\\ inv_{i-1,n}\times s_{i-1,n/p_i} &(p_i\mid n)\end{cases} \]初始状态为 \(s_{0,n}=f_n,inv_{0,n}={f_n}^{-1}\)
但是直接求逆元会导致复杂度升到 \(O(n\log P)\) ，就前功尽弃了
所以还要用一下线性求逆元的方法
#include<stdio.h>
const int N=1000010,P=1e9+7; int T,n,m,ans,t1,t2;
int t,cnt,suf,I,p[100000],s[N],inv[N],v[N],pre[N];
inline int min(int x,int y) { return x<y?x:y; }
inline int power(int x,int y) {
int s=1;
while (y) (y&1)&&(s=1ll*s*x%P),x=1ll*x*x%P,y>>=1;
return s;
}
int main() {
s[1]=pre[0]=pre[1]=inv[0]=suf=1;
for (int i=2; i<N; ++i) { //线性筛、初始化 s
v[i]||(p[++cnt]=i);
for (int j=1; j<=cnt&&(t=p[j]*i)<N; ++j) {
v[t]=1;
if (i%p[j]==0) break;
}
(s[i]=s[i-2]+s[i-1])>=P&&(s[i]-=P);
pre[i]=1ll*pre[i-1]*s[i]%P;
}
I=power(pre[N-1],P-2);
for (int i=N-1; i; --i) //初始化 inv
inv[i]=1ll*pre[i-1]*suf%P*I%P,
suf=1ll*suf*s[i]%P;
for (int i=1,j; i<=cnt; ++i) // dp
for (j=(N-1)/p[i],t=j*p[i]; j; --j,t-=p[i])
s[t]=1ll*s[t]*inv[j]%P,
inv[t]=1ll*inv[t]*s[j]%P;
for (int i=2; i<N; ++i) s[i]=1ll*s[i]*s[i-1]%P;
for (int i=2; i<N; ++i) inv[i]=1ll*inv[i]*inv[i-1]%P;
//存储 s 和 inv 的前缀积，便于整除分块计算
for (scanf("%d",&T); T; --T) {
scanf("%d%d",&n,&m),ans=1;
for (int l=1,r,mn=min(n,m); l<=mn; l=r+1) //整除分块
r=min(n/(t1=n/l),m/(t2=m/l)),
t=1ll*s[r]*inv[l-1]%P,
ans=1ll*ans*power(t,1ll*t1*t2%(P-1))%P;
printf("%d\n",ans);
}
return 0;
}
P5518 [MtOI2019]幽灵乐团 / 莫比乌斯反演基础练习题
这个题要麻烦得多
全程大力推式子 虽然非常不基础 但确实是针对性很强的练习题
等于
\[\large\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\left(\dfrac{ij}{\gcd(i,j)\gcd(i,k)}\right)^{f(type)} \]可以转化为以下两个子问题：
\[\large M(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^Ci^{f(type)} \]\[\large N(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\gcd(i,j)^{f(type)} \]答案为
\[\large \dfrac{M(A,B,C)\times M(B,A,C)}{N(A,B,C)\times N(A,C,B)} \]下面按 \(type\) 的取值分为三种情况讨论
\(\Large \textbf{1. type=0,f(type)=1}\)
\[\large M(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^Ci=\prod_{i=1}^Ai^{BC}=(A!)^{BC} \]预处理阶乘，快速幂回答询问，复杂度为 \(O(n+T\log n)\) （ \(n\) 与 \(A,B,C\) 同级， \(T\) 为询问组数，下同）
\[\large N(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\gcd(i,j)=\left(\prod_{i=1}^A\prod_{j=1}^B\gcd(i,j)\right)^C \]括号内的式子和上题基本一致 过程不详细写了
设 \(D=\min(A,B)\) ，则
依然是预处理括号内的式子 然后整除分块
时间复杂度 \(O(n\log n+T\sqrt n\log n)\)
预处理也可以做到 \(O(n\log \log n)\) ，但对本题而言基本没啥优化效果
\(\Large \textbf{2. type=1,f(type)=i×j×k}\)
\[\large M(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^Ci^{i\times j\times k}=\prod_{i=1}^Ai^{i(\sum_{j=1}^Bj)(\sum_{k=1}^Ck)} \]令 \(S(n)=\sum\limits_{i=1}^ni=\dfrac{n(n+1)}2\)
\[\large M(A,B,C)=\left(\prod_{i=1}^Ai^i\right)^{S(B)S(C)} \]括号内的式子可以 \(O(n\log n)\) 预处理，然后快速幂回答询问
\[\large N(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\gcd(i,j)^{i\times j\times k}=\left(\prod_{i=1}^A\prod_{j=1}^B\gcd(i,j)^{i\times j}\right)^{S(C)} \]括号里的式子拿出来推一下
\[\large \begin{aligned}\prod_{i=1}^A\prod_{j=1}^B\gcd(i,j)^{i\times j} &=\prod_{d=1}^D\prod_{i=1}^A\prod_{j=1}^Bd^{i\times j\times [(i,j)=d]} =\prod_{d=1}^D\prod_{i=1}^{A/d}\prod_{j=1}^{B/d}d^{id\times jd\times [(i,j)=1]} =\prod_{d=1}^Dd^{d^2\sum_{i=1}^{A/d}\sum_{j=1}^{B/d}ij[(i,j)=1]}\\ &=\prod_{d=1}^Dd^{d^2\sum_{i=1}^{A/d}\sum_{j=1}^{B/d}ij\sum_{t\mid (i,j)}~\mu(t)} =\prod_{d=1}^Dd^{d^2\sum_{t=1}^{D/d}\mu(t)\sum_{i=1}^{A/dt}it\sum_{j=1}^{B/dt}jt} =\prod_{d=1}^Dd^{d^2\sum_{t=1}^{D/d}\mu(t)t^2S(A/dt)S(B/dt)}\\ &=\prod_{d=1}^D\prod_{t=1}^{D/d}d^{\mu(t)(dt)^2S(A/dt)S(B/dt)} =\prod_{t'=1}^D\prod_{d\mid t'}d^{\mu(t'/d)t'^2S(A/t')S(B/t')} =\prod_{t'=1}^D\left(\prod_{d\mid t'}d^{\mu(t'/d)}\right)^{t'^2S(A/t')S(B/t')} \end{aligned}\]推到这里就可以了，代回 \(N(A,B,C)\) 的表达式
\[\large N(A,B,C)=\prod_{t=1}^D\left(\prod_{d\mid t}d^{\mu(t/d)}\right)^{t^2S(A/t)S(B/t)S(C)} \]括号里的式子预处理过了 然后还是整除分块
时间复杂度 \(O(n\log n+T\sqrt n\log n)\)
\(\Large\textbf{3. type=2,f(type)=gcd(i,j,k)}\)
\[\large M(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C i^{\gcd(i,j,k)}=\prod_{i=1}^Ai^{\sum_{j=1}^B\sum_{k=1}^C\gcd(i,j,k)} \]指数可以考虑欧拉反演
\[\large \sum_{j=1}^B\sum_{k=1}^C\gcd(i,j,k) =\sum_{j=1}^B\sum_{k=1}^C\sum_{d\mid \gcd(i,j,k)}\varphi(d) =\sum_{d\mid i}\varphi(d)\sum_{j=1}^B\sum_{k=1}^C[d|j][d|k] =\sum_{d\mid i}\varphi(d)(B/d)(C/d)\]设 \(E=\min(A,B,C)\)
\[\large M(A,B,C)=\prod_{i=1}^Ai^{\sum_{d\mid i}\varphi(d)(B/d)(C/d)} =\prod_{d=1}^E\left(\prod_{i'=1}^{A/d}i'd\right)^{\varphi(d)(B/d)(C/d)} =\prod_{d=1}^E\left((A/d)!\times d^{A/d}\right)^{\varphi(d)(B/d)(C/d)}\]\[\large M(A,B,C)=\prod_{d=1}^E(A/d)!^{\varphi(d)(B/d)(C/d)}\times \prod_{d=1}^Ed^{\varphi(d)(A/d)(B/d)(C/d)} \]两部分都可以整除分块做
接下来是本题最难处理的式子了
\[\large N(A,B,C)=\prod_{i=1}^A\prod_{j=1}^B\prod_{k=1}^C\gcd(i,j)^{\gcd(i,j,k)} \]改为枚举 \(\gcd(i,j)\)
\[\large N(A,B,C)=\prod_{d=1}^Dd^{\sum_{i=1}^A\sum_{j=1}^B[(i,j)=d]\sum_{k=1}^C\gcd(d,k)} \]指数还是常规反演
\[\large \sum_{i=1}^A\sum_{j=1}^B[(i,j)=d]=\sum_{i=1}^{A/d}\sum_{j=1}^{B/d}\sum_{t\mid (i,j)}\mu(t)=\sum_{t=1}^{D/d}\mu(t)(A/dt)(B/dt) \]\[\large \sum_{k=1}^C\gcd(d,k)=\sum_{k=1}^C\sum_{t'\mid (d,k)}\varphi(t')=\sum_{t'\mid d}\varphi(t')(C/t') \]\[\large N(A,B,C)=\prod_{d=1}^Dd^{\sum_{t=1}^{D/d}\mu(t)(A/dt)(B/dt)\sum_{t'\mid d}\varphi(t')(C/t')} \]枚举因数看起来不对劲，换一下枚举方式
\[\large N(A,B,C)=\prod_{t'=1}^E\prod_{d'=1}^{D/t'}(d't')^{\varphi(t')(C/t')\sum_{t=1}^{D/d't'}\mu(t)(A/d't't)(B/d't't)} \]式子有点丑 所以替换一次字母（
\[\large N(A,B,C)=\prod_{i=1}^E\prod_{j=1}^{D/i}(ij)^{\varphi(i)(C/i)\sum_{k=1}^{D/ij}\mu(k)(A/ijk)(B/ijk)} \]似乎很难继续推了
但通过看官方题解我们了解到一个技巧，把 \(i,j\) 拆开分别算贡献
对于第一部分，枚举 \(t=jk\)
\[\large\prod_{i=1}^Ei^{\varphi(i)(C/i)\sum_{t=1}^{D/i}~(A/it)(B/it)\sum_{k\mid t}\mu(k)} \]\[\large\prod_{i=1}^Ei^{\varphi(i)(C/i)\sum_{t=1}^{D/i}~(A/it)(B/it)[t=1]} \]发现只有 \(t=1\) 的情况能对答案产生贡献
\[\large\prod_{i=1}^Ei^{\varphi(i)(A/i)(B/i)(C/i)} \]而我们在 \(M(A,B,C)\) 中也算出过相同的式子：
\[\large M(A,B,C)=\prod_{d=1}^E(A/d)!^{\varphi(d)(B/d)(C/d)}\times \prod_{d=1}^Ed^{\varphi(d)(A/d)(B/d)(C/d)} \]所以直接约掉就好了
最后来处理第二部分
\[\large \prod_{i=1}^E\prod_{j=1}^{D/i}j^{\varphi(i)(C/i)\sum_{k=1}^{D/ij}\mu(k)(A/ijk)(B/ijk)} \]\[\large \prod_{i=1}^E\left(\prod_{j=1}^{D/i}j^{\sum_{k=1}^{D/ij}\mu(k)(A/ijk)(B/ijk)}\right)^{\varphi(i)(C/i)} \]同样枚举 \(t=jk\)
\[\large \prod_{i=1}^E\left(\prod_{t=1}^{D/i}\prod_{j\mid t}j^{\mu(t/j)(A/it)(B/it)}\right)^{\varphi(i)(C/i)} \]简单整理一下
\[\large \prod_{i=1}^E\left(\prod_{t=1}^{D/i}\left(\prod_{d\mid t}d^{\mu(t/d)}\right)^{(A/it)(B/it)}\right)^{\varphi(i)(C/i)} \]最内层括号中的式子预处理过了
可以两层整除分块做，时间复杂度 \(O(n^{3/4}\log n)\)
还有个 \(O(n^{2/3}\log n)\) 的做法，虽然我没有写，而且也证不来复杂度，但还是提一下
之前我们算过
\[\large\prod_{i=1}^A\prod_{j=1}^B\gcd(i,j)=\prod_{t=1}^D\left(\prod_{d\mid T}d^{\mu(t/d)}\right)^{(A/t)(B/t)} \]带入到现在的式子里就是
\[\large\prod_{i=1}^E\left(\prod_{j=1}^{A/i}\prod_{k=1}^{B/i}\gcd(j,k)\right)^{\varphi(i)(C/i)} \]预处理 \(A,B\le n^{2/3}\) 的所有 \(\prod\limits_{i=1}^A\prod\limits_{j=1}^B\gcd(i,j)\)
\(i\) 需要一层整除分块
对于括号内的部分，当 \(i\ge n^{1/3}\) 时直接用预处理的值即可，当 \(i<n^{1/3}\) 时需要再套一层整除分块
时间复杂度证明可类比杜教筛 我都不会证
至此这道题的推式子部分终于结束了
写 \(O(n\log n+Tn^{3/4}\log n)\) 或者 \(O(n^{4/3}+Tn^{2/3}\log n)\) 的做法都能通过
细节很多 需要耐心调试
但很小的错误都能导致答案产生巨大误差 所以能过样例就基本就没问题了 调试难度不大
象征性地放一下代码（虽然丑到只有编译器看得懂
#include<stdio.h>
#define M0(A,B) power(fac[A],1ll*B*C%P2)
#define M1(A,B) power(pow[A],1ll*S1(B)*S1(C)%P2)
const int N=100001; int A,B,C,D,E,T,P,P2,t,cnt,ans,p[10000];
int s[N],inv[N],S[N],INV[N],fac[N],pow[N],phi[N],v[N];
inline int min(int x,int y) { return x<y?x:y; }
inline int S1(int x) { return (1ll*x*(x+1)>>1)%P2; }
inline int power(int x,int y) {
int ans=1;
while (y) (y&1)&&(ans=1ll*ans*x%P),x=1ll*x*x%P,y>>=1;
return ans;
}
void prework() {
phi[1]=fac[0]=pow[0]=S[0]=INV[0]=1;
s[0]=s[1]=inv[0]=inv[1]=1;
for (int i=2; i<N; ++i) {
v[i]||(p[++cnt]=i,phi[i]=i-1);
for (int j=1; j<=cnt&&(t=p[j]*i)<N; ++j) {
v[t]=1,phi[t]=phi[i]*(p[j]-1);
if (i%p[j]==0) { phi[t]+=phi[i]; break; }
}
(phi[i]+=phi[i-1])>=P2&&(phi[i]-=P2);
}
for (int i=2; i<N; ++i) s[i]=i,inv[i]=1ll*(P-P/i)*inv[P%i]%P;
for (int i=1; i<=cnt; ++i)
for (int j=(N-1)/p[i],t=j*p[i]; j; --j,t-=p[i])
s[t]=1ll*s[t]*inv[j]%P,inv[t]=1ll*inv[t]*s[j]%P;
for (int i=1; i<N; ++i)
S[i]=1ll*S[i-1]*power(s[i],1ll*i*i%P2)%P,
INV[i]=1ll*INV[i-1]*power(inv[i],1ll*i*i%P2)%P,
s[i]=1ll*s[i-1]*s[i]%P,inv[i]=1ll*inv[i-1]*inv[i]%P,
fac[i]=1ll*fac[i-1]*i%P,pow[i]=1ll*pow[i-1]*power(i,i)%P;
}
int N0(int A,int B,int C) {
int D=min(A,B),ans=1;
for (int l=1,r,t1,t2; l<=D; l=r+1)
r=min(A/(t1=A/l),B/(t2=B/l)),
ans=1ll*ans*power(1ll*s[r]*inv[l-1]%P,1ll*t1*t2%P2)%P;
return power(ans,P2-C);
}
int N1(int B,int C) {
D=min(A,B),ans=1;
for (int l=1,r,t,t1,t2; l<=D; l=r+1)
r=min(A/(t1=A/l),B/(t2=B/l)),
t=1ll*S1(t1)*S1(t2)%P2,
ans=1ll*ans*power(1ll*S[r]*INV[l-1]%P,t)%P;
return power(ans,P2-S1(C));
}
int M2(int A,int B) {
E=min(min(A,B),C),ans=1;
for (int l=1,r,t,t1,t2,t3; l<=E; l=r+1)
r=min(min(A/(t1=A/l),B/(t2=B/l)),C/(t3=C/l)),
t=1ll*t2*t3%P2*(phi[r]-phi[l-1]+P2)%P2,
ans=1ll*ans*power(fac[t1],t)%P;
return ans;
}
int N2(int B,int C) {
E=min(D=min(A,B),C),ans=1;
for (int l=1,r,t1,t2,t3; l<=E; l=r+1)
r=min(min(A/(t1=A/l),B/(t2=B/l)),C/(t3=C/l)),
ans=1ll*ans*N0(t1,t2,1ll*t3*(phi[r]-phi[l-1]+P2)%P2)%P;
return ans;
}
int main() {
scanf("%d%d",&T,&P),P2=P-1,prework();
while (T--)
scanf("%d%d%d",&A,&B,&C),t=1ll*M0(A,B)*M0(B,A)%P,
printf("%d ",1ll*t*N0(A,B,C)%P*N0(A,C,B)%P),
printf("%d ",1ll*M1(A,B)*M1(B,A)%P*N1(B,C)%P*N1(C,B)%P),
printf("%d\n",1ll*M2(A,B)*M2(B,A)%P*N2(B,C)%P*N2(C,B)%P);
return 0;
}
标签：large,sum,mid,varphi,MtOI2019,乐团,mu,SDOI2017,prod 来源： https://www.cnblogs.com/REKonib/p/15019534.html