Codeforces #250E the child and Binary fast Fourier transform __codeforces

The topic: Given a set S, for I=1...M i=1...m How many binary trees satisfy each node's weights in the set S S and the weights and the I-I
constructs the answer polynomial f (x) f (x) and set S S, then the generating function C (x) c (x), then root node is a binary tree, and the right subtree is a two-pronged tree, the weight of itself must be in set S, in addition to the case of an empty tree
There are f (x) =f2 (x) c (x) +1 f (x) =f^2 (x) c (x) +1
Solution f (x) =1±1−4c (x) √2c (x) =21±1−4c (x) √f (x) =\frac{1\pm\sqrt{1-4c (x)}}{2c (x)}=\frac2{1\pm\sqrt{1-4c (x)}}
If the denominator is not reversible if a minus sign is taken below the equation, go to
Get F (x) =21+1−4C (x) √f (x) =\frac2{1+\sqrt{1-4c (x)}}
about polynomial negation and polynomial open root see picks blog
CF on each point 7s time can cross BZ I'm really not in the mood. The card constant
[cover face Bear] I'm all fast Fourier transform the

#include <cstdio> #include <cstring> #include <iostream> #include <algorithm> #define M 525000 #d

Efine P 998244353 #define G 3 #define INV2 499122177 using namespace std;

int n,m,l;

int a[m],b[m];
    Long Long quick_power (long long X,long long y) {long long re=1;
        while (y) {if (y&1) (re*=x)%=p; (x*=x)%=p;
    y>>=1;
return to re;
    } void FFT (int a[],int n,int type) {static int temp[m];
    int i;
    if (n==1) return;
    for (i=0;i<n;i+=2) temp[i>>1]=a[i],temp[i+n>>1]=a[i+1];
    memcpy (A,temp,sizeof (a[0]) *n);
    int *l=a,*r=a+ (n>>1);
    FFT (L,n>>1,type);
    FFT (R,n>>1,type);
    Long Long W=quick_power (G, Long Long) (P-1)/n*type% (P-1), wn=1;
    For (i=0;i<n>>1;i++, (wn*=w)%=p) temp[i]= (l[i]+wn*r[i))%p,temp[i+ (n>>1)]= (l[i]-wn*r[i]%p+p)%P;
memcpy (A,temp,sizeof (a[0]) *n);
    } void Get_inv (int a[],int b[],int n) {static int temp[m];
    int i; IF (n==1) {b[0]=quick_power (a[0],p-2);
    return;
    } GET_INV (a,b,n>>1);
    memcpy (Temp,a,sizeof (a[0]) *n);
    memset (Temp+n,0,sizeof (a[0]) *n);
    FFT (temp,n<<1,1);
    FFT (b,n<<1,1);
    for (i=0;i<n<<1;i++) temp[i]= (Long Long) b[i]* (long Long) temp[i]*b[i]%p+p)%P;
    FFT (temp,n<<1,p-2);
    Long Long Inv=quick_power (n<<1,p-2);
    for (i=0;i<n;i++) b[i]=temp[i]*inv%p;
memset (B+n,0,sizeof (a[0]) *n);
    } void Get_root (int a[m],int b[m],int N)//0-ori 1-inv {static int temp[m],b_inv[m];
    int i;
        if (n==1) {b[0]=1;
    return;
    } get_root (a,b,n>>1);
    memset (B_inv,0,sizeof (a[0]) *n);
    GET_INV (B,b_inv,n);
    memcpy (Temp,a,sizeof (a[0]) *n);
    memset (Temp+n,0,sizeof (a[0]) *n);
    FFT (temp,n<<1,1);
    FFT (b,n<<1,1);
    FFT (b_inv,n<<1,1);
  for (i=0;i<n<<1;i++) temp[i]= (Long Long) inv2* (b[i]+ (Long Long) temp[i]*b_inv[i]%p)%P;  FFT (temp,n<<1,p-2);
    Long Long Inv=quick_power (n<<1,p-2);
    for (i=0;i<n;i++) b[i]=temp[i]*inv%p;
memset (B+n,0,sizeof (a[0]) *n);
    int main () {int i,x;
    cin>>n>>m;
    for (l=1;l<=m;l<<=1); for (i=1;i<=n;i++) {s