UOJ62 怎样跑的更快

$$b_i=\sum _{j=1}^n(i,j{)}^{c-d}\ast i^d{j}^d{z}_j$$
$$\frac{b_i}{i^d}=\sum _{j=1}^n{z}_j{j}^d(i,j{)}^{c-d}$$记$(i,j)=x,B_i=\frac{b_i}{i^d},Z_i=z_i{i}^d,f_i(d)={\sum }_{(i,j)=d}{Z}_j,F_i(d)={\sum }_{d|(i,j)}{Z}_j$
$$\frac{b_i}{i^d}=\sum _{x|i}^{n}f(x)x^{c-d}$$$$\frac{b_i}{i^d}=\sum _{x|i}^n{x}^{c-d}\sum _{k}F(kx)\mu (k)$$$$\frac{b_i}{i^d}=\sum _{kx|i}^n{x}^{c-d}F(kx)\mu (k)$$$$B(i)=\sum _{T|i}^{n}F(T)\sum _{x|T}\mu (T/x)x^{c-d}$$$$令g(T)=\mu \ast i{d}^{c-d}(T)$$得到$$B(i)=\sum _{T|i}F(T)g(T)$$那么$$F(T)g(T)=\sum _{ij=T}\mu (j)B(i)$$让后一路反推回去得到$z$就行了

#include
#include
#include
#define N 100010
#define M 998244353
#define LL long long
using namespace std;
inline LL pow(LL x,LL k,int& s){
	while(k<0) k+=M-1;
	for(s=1;k;x=x*x%M,k>>=1) k&1?s=s*x%M:0;
}
int mu[N],c,d,n,q,inv[N];
int f[N],g[N],b[N],B[N],z[N],Z[N],F[N],xd[N],xcd[N],G[N];
int main(){
	scanf("%d%d%d%d",&n,&c,&d,&q);
	mu[1]=1;
	for(int i=1;i<=n;++i){
		pow(i,d,xd[i]);
		pow(i,c-d,xcd[i]);
		pow(xd[i],M-2,inv[i]);
		for(int j=i+i;j<=n;j+=i) mu[j]-=mu[i];
	}
	for(int i=1;i<=n;++i) if(mu[i])
		for(int j=1,k=i;k<=n;++j,k+=i)
			g[k]=(g[k]+mu[i]*xcd[j])%M;
	while(q--){
		for(int i=1;i<=n;++i){
			scanf("%d",b+i);
			B[i]=(LL)b[i]*inv[i]%M;
		}
		memset(G,0,n+1<<2);
		for(int i=1;i<=n;++i) if(mu[i])
			for(int j=1,k=i;k<=n;++j,k+=i)
				G[k]=(G[k]+mu[i]*B[j])%M;
		for(int i=1;i<=n;++i)
			if(!g[i]&&G[i]){
				puts("-1");
				goto end;
			} else{
				pow(g[i],M-2,F[i]); 
				F[i]=(LL)F[i]*G[i]%M;
			}
		memset(Z,0,n+1<<2);
		for(int i=1;i<=n;++i) if(mu[i])
			for(int j=1,k=i;k<=n;++j,k+=i)
				Z[j]=(Z[j]+mu[i]*F[k]%M)%M;
		for(int i=1;i<=n;++i){
			z[i]=(LL)Z[i]*inv[i]%M;
			printf("%d ",(z[i]+M)%M);
		}
		puts(""); end:;
	}
}