多项式求逆:http://blog.miskcoo.com/2015/05/polynomial-inverse
注意:直接在点值表达下做$B(x) \equiv 2B'(x) - A(x)B'^2(x) \pmod {x^n}$是可以的,但是一定要注意,这一步中有一个长度为n的和两个长度为(n/2)的多项式卷积,因此要在DFT前就扩展FFT点值表达的“长度”到2n,否则会出错(调了1.5个小时)
1 #prag\ 2 ma GCC optimize(2) 3 #include<cstdio> 4 #include<algorithm> 5 #include<cstring> 6 #include<vector> 7 #include<cmath> 8 using namespace std; 9 #define fi first 10 #define se second 11 #define mp make_pair 12 #define pb push_back 13 typedef long long ll; 14 typedef unsigned long long ull; 15 const int md=998244353; 16 const int N=2097152; 17 int rev[N]; 18 void init(int len) 19 { 20 int bit=0,i; 21 while((1<<(bit+1))<=len) ++bit; 22 for(i=0;i<len;++i) 23 rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1)); 24 } 25 ll poww(ll a,ll b) 26 { 27 ll base=a,ans=1; 28 for(;b;b>>=1,base=base*base%md) 29 if(b&1) 30 ans=ans*base%md; 31 return ans; 32 } 33 void dft(int *a,int len,int idx)//要求len为2的幂 34 { 35 int i,j,k,t1,t2;ll wn,wnk; 36 for(i=0;i<len;++i) 37 if(i<rev[i]) 38 swap(a[i],a[rev[i]]); 39 for(i=1;i<len;i<<=1) 40 { 41 wn=poww(idx==1?3:332748118,(md-1)/(i<<1)); 42 for(j=0;j<len;j+=(i<<1)) 43 { 44 wnk=1; 45 for(k=j;k<j+i;++k,wnk=wnk*wn%md) 46 { 47 t1=a[k];t2=a[k+i]*wnk%md; 48 a[k]+=t2; 49 (a[k]>=md) && (a[k]-=md); 50 a[k+i]=t1-t2; 51 (a[k+i]<0) && (a[k+i]+=md); 52 } 53 } 54 } 55 if(idx==-1) 56 { 57 ll ilen=poww(len,md-2); 58 for(i=0;i<len;++i) 59 a[i]=a[i]*ilen%md; 60 } 61 } 62 int f[N],g[N],t1[N]; 63 int n,n1; 64 void p_inv(int *f,int *g,int len)//g=f^(-1);f,g长度不小于2^(ceil(log2(len))+1) 65 { 66 g[0]=poww(f[0],md-2); 67 for(int i=2,j;i<(len<<1);i<<=1) 68 { 69 init(i<<1); 70 memcpy(t1,f,sizeof(int)*i); 71 memset(t1+i,0,sizeof(int)*i); 72 memset(g+(i>>1),0,sizeof(int)*(i+(i>>1))); 73 dft(t1,i<<1,1);dft(g,i<<1,1); 74 for(j=0;j<(i<<1);++j) 75 g[j]=ll(g[j])*(2+ll(md-g[j])*t1[j]%md)%md; 76 dft(g,i<<1,-1); 77 } 78 } 79 int main() 80 { 81 int i,t; 82 scanf("%d",&n);n1=n; 83 for(i=0;i<n;++i) 84 scanf("%d",g+i); 85 for(t=1;t<n;t<<=1); 86 n=t; 87 p_inv(g,f,n); 88 for(i=0;i<n1;++i) 89 printf("%d ",f[i]); 90 return 0; 91 }