终于学会多项式开根了哈哈……
反正题解都烂大街了，我就不写了，直接贴代码算了……(犯懒ing)
1 /************************************************************** 2 Problem: 3625 3 User: _Angel_ 4 Language: C++ 5 Result: Accepted 6 Time:29048 ms 7 Memory:5944 kb 8 ****************************************************************/ 9 #include<cstdio> 10 #include<cstring> 11 #include<algorithm> 12 using namespace std; 13 const int maxn=262200,p=998244353,g=3,inv_2=499122177; 14 void NTT(int*,int,int); 15 void getsqrt(int*,int*,int); 16 void getinv(int*,int*,int); 17 int qpow(int,int,int); 18 int n,N=1,m,x,A[maxn],B[maxn]; 19 int main(){ 20 scanf("%d%d",&m,&n); 21 while(N<=n)N<<=1; 22 while(m--){ 23 scanf("%d",&x); 24 A[x]=(A[x]-4+p)%p; 25 } 26 A[0]=(A[0]+1)%p; 27 getsqrt(A,B,N); 28 B[0]=(B[0]+1)%p; 29 getinv(B,A,N); 30 for(int i=1;i<=n;i++)printf("%d ",(int)(A[i]*2ll%p)); 31 return 0; 32 } 33 void NTT(int *A,int n,int tp){ 34 for(int i=1,j=0,k;i<n-1;i++){ 35 k=n; 36 do j^=(k>>=1);while(j<k); 37 if(i<j)swap(A[i],A[j]); 38 } 39 for(int k=2;k<=n;k<<=1){ 40 int wn=qpow(g,(tp>0?(p-1)/k:(p-1)/k*(long long)(p-2)%(p-1)),p); 41 for(int i=0;i<n;i+=k){ 42 int w=1; 43 for(int j=0;j<(k>>1);j++,w=(long long)w*wn%p){ 44 int a=A[i+j],b=(long long)w*A[i+j+(k>>1)]%p; 45 A[i+j]=(a+b)%p; 46 A[i+j+(k>>1)]=(a-b+p)%p; 47 } 48 } 49 } 50 if(tp<0){ 51 int inv=qpow(n,p-2,p); 52 for(int i=0;i<n;i++)A[i]=(long long)A[i]*inv%p; 53 } 54 } 55 void getsqrt(int *A,int *C,int n){ 56 static int B[maxn],D[maxn]; 57 fill(C,C+(n<<1),0); 58 C[0]=1; 59 for(int k=2;k<=n;k<<=1){ 60 copy(A,A+k,B); 61 fill(B+k,B+(k<<1),0); 62 getinv(C,D,k); 63 NTT(B,k<<1,1); 64 NTT(D,k<<1,1); 65 for(int i=0;i<(k<<1);i++)B[i]=(long long)B[i]*D[i]%p; 66 NTT(B,k<<1,-1); 67 for(int i=0;i<k;i++)C[i]=(C[i]+B[i])%p*(long long)inv_2%p; 68 } 69 } 70 void getinv(int *A,int *C,int n){ 71 static int B[maxn]; 72 fill(C,C+(n<<1),0); 73 C[0]=qpow(A[0],p-2,p); 74 for(int k=2;k<=n;k<<=1){ 75 copy(A,A+k,B); 76 fill(B+k,B+(k<<1),0); 77 NTT(B,k<<1,1); 78 NTT(C,k<<1,1); 79 for(int i=0;i<(k<<1);i++)C[i]=C[i]*((2-(long long)B[i]*C[i]%p+p)%p)%p; 80 NTT(C,k<<1,-1); 81 fill(C+k,C+(k<<1),0); 82 } 83 } 84 int qpow(int a,int b,int p){ 85 int ans=1; 86 for(;b;b>>=1,a=(long long)a*a%p)if(b&1)ans=(long long)ans*a%p; 87 return ans; 88 }