题面

题解

生成函数这厮到底还有什么是办不到的……

首先对于一个数\(i\)，如果存在的话可以取无限多次，那么它的生成函数为\[\sum_{j=0}^{\infty}x^{ij}={1\over 1-x^i}\]

然后我们设\(a_i\in [0,1]\)表示这个数是否存在这个集合里，那么给出了\(F\)，满足

\[F(x)=\prod_{i=1}^n\left({1\over 1-x^i}\right)^{a_i}\]

然后我们现在就是要求出\(a_i\)

首先我们要知道一个东西\[\ln(1-x^i)=-\sum_{j=1}^{\infty}\frac{x^{ij}}{j}\]

证明抄\(Cyhlnj\)的

\[\ln F(x)=G(x)\\\frac{F'(x)}{F(x)}=G'(x)\\\frac{-ix^{i-1}}{1-x^i}=G'(x)\\-\sum_{j=0}^{\infty} ix^{i-1+ij}=G'(x)\\-\sum_{j=0}^{\infty}\frac{ix^{i+ij}}{i+ij}=G(x)\\-\sum_{j=1}^{\infty}\frac{x^{ij}}{j}=G(x)\]

我们先对原来的式子两边取\(\ln\)，再用上面的式子代入

\[ \begin{align} -\ln F(x) &=\sum_{i=1}^n a_i\ln (1-x^i)\\ &=-\sum_{i=1}^n a_i\sum_{j=1}^{\infty}\frac{x^{ij}}{j}\\ \end{align} \]

然后我们枚举\(d=ij\)，有

\[\ln F(x)=\sum_{d=1}^{\infty} x^d\sum_{i\mid d}a_i{i\over d}\]

然后我们现在就求出了\(\sum_{i\mid d}a_i{i\over d}\)，这个我们直接枚举倍数，然后让每一个数的倍数减去它的贡献就行了

最后一个问题是，这里的模数不一定有原根，所以得拆系数

//minamoto#include
    
     #define R register#define ll long long#define fp(i,a,b) for(R int i=a,I=b+1;i
     
      I;--i)#define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)using namespace std;char buf[1<<21],*p1=buf,*p2=buf;inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}int read(){    R int res,f=1;R char ch;    while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);    for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');    return res*f;}char sr[1<<21],z[20];int K=-1,Z=0;inline void Ot(){fwrite(sr,1,K+1,stdout),K=-1;}void print(R int x){    if(K>1<<20)Ot();if(x<0)sr[++K]='-',x=-x;    while(z[++Z]=x%10+48,x/=10);    while(sr[++K]=z[Z],--Z);sr[++K]=' ';}const int N=6e5+5;const double Pi=acos(-1.0);int P;inline int add(R int x,R int y){return x+y>=P?x+y-P:x+y;}inline int dec(R int x,R int y){return x-y<0?x-y+P:x-y;}inline int mul(R int x,R int y){return 1ll*x*y-1ll*x*y/P*P;}int ksm(R int x,R int y){    R int res=1;    for(;y;y>>=1,x=mul(x,x))if(y&1)res=mul(res,x);    return res;}struct cp{    double x,y;    cp(R double xx=0,R double yy=0){x=xx,y=yy;}    inline cp operator +(const cp &b)const{return cp(x+b.x,y+b.y);}    inline cp operator -(const cp &b)const{return cp(x-b.x,y-b.y);}    inline cp operator *(const cp &b)const{return cp(x*b.x-y*b.y,x*b.y+y*b.x);}    inline cp operator *(const double &b)const{return cp(x*b,y*b);}}O[N];int r[N],c[N],d[N],e[N],f[N],g[N],h[N],n,len,ans;void FFT(cp *A,int ty,int lim){    fp(i,0,lim-1)if(i
      
       >1]>>1)|((i&1)<<(l-1));    for(R int i=1;i
       
        <<=1)fp(k,0,i-1)O[i+k]=cp(cos(Pi*k/i),sin(Pi*k/i));    fp(i,0,len-1){        A[i].x=a[i]>>15,B[i].x=a[i]&32767;        C[i].x=b[i]>>15,D[i].x=b[i]&32767;        A[i].y=B[i].y=C[i].y=D[i].y=0;    }fp(i,len,lim-1)A[i]=B[i]=C[i]=D[i]=0;    FFT(A,1,lim),FFT(B,1,lim),FFT(C,1,lim),FFT(D,1,lim);    fp(i,0,lim-1)        F[i]=A[i]*C[i],G[i]=A[i]*D[i]+B[i]*C[i],H[i]=B[i]*D[i];    FFT(F,-1,lim),FFT(G,-1,lim),FFT(H,-1,lim);    fp(i,0,len-1)c[i]=(((ll)(F[i].x+0.5)%P<<30)+((ll)(G[i].x+0.5)<<15)+((ll)(H[i].x+0.5)))%P;    fp(i,len,lim-1)c[i]=0;}void Inv(int *a,int *b,int len){    if(len==1)return b[0]=ksm(a[0],P-2),void();    Inv(a,b,len>>1);    Mul(a,b,c,len),Mul(c,b,d,len);    fp(i,0,len-1)b[i]=dec(add(b[i],b[i]),d[i]);}void Ln(int *a,int *b,int len){    fp(i,1,len-1)e[i-1]=mul(a[i],i);    Inv(a,f,len),Mul(e,f,b,len);    fd(i,len-1,1)b[i]=mul(b[i-1],ksm(i,P-2));}int main(){//  freopen("testdata.in","r",stdin);    n=read(),P=read();    int len=1;while(len<=n)len<<=1;    fp(i,1,n)g[i]=read();    g[0]=1,Ln(g,h,len);    fp(i,1,n)h[i]=mul(i,h[i]);    fp(i,1,n)for(R int j=i+i;j<=n;j+=i)h[j]=dec(h[j],h[i]);    fp(i,1,n)if(h[i])++ans;    print(ans),sr[K]='\n';    fp(i,1,n)if(h[i])print(i);    return Ot(),0;}