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zc
2020-02-08 22:00:00
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假设我们要求的是A\cdot B \equiv 1 \pmod{x^n},

我们已经求出了B'满足A\cdot B' \equiv 1 \pmod{x^\frac n2}

\because A\cdot B \equiv 1 \pmod {x^n} \\ \because A\cdot B' \equiv 1 \pmod{x^\frac n2} \\ \therefore A \cdot (B-B') \equiv 0 \pmod{x^\frac n2} \\ \therefore B-B' \equiv 0 \pmod{x^\frac n2} \\ (B-B')^2 \equiv 0 \pmod{x^n} \\ B^2-2BB'+B'^2 \equiv 0 \pmod{x^n} \\ \therefore A(B^2-2BB'+B'^2) \equiv 0 \pmod{x^n} \\ AB^2-2ABB'+AB'^2 \equiv 0 \pmod{x^n} \\ \because A\cdot B \equiv 1 \pmod {x^n} \\ \therefore B-2B'+AB'^2\equiv 0 \pmod{x^n} \\ B\equiv 2B'-AB'^2 \pmod{x^n}

于是我们可以用倍增+NTT解决(复杂度\mathcal{O(n \log^2 n)})

#include<bits/stdc++.h>
namespace ZDY{
    #pragma GCC optimize(3)
    #define il __inline__ __attribute__ ((always_inline))
    #define register
    #define ll long long
    #define ull unsigned long long
    #define db double
    #define sht short
    #define MB template <class T>il
    #define Fur(i,x,y) for(int i(x);i<=y;++i)
    #define Fdr(i,x,y) for(int i(x);i>=y;--i)
    #define fl(i,x) for(int i(head[x]),to;to=e[i].to,i;i=e[i].nxt)
    #define clr(x,y) memset(x,y,sizeof(x))
    #define cpy(x,y) memcpy(x,y,sizeof(x))
    #define fin(s) freopen(s".in","r",stdin)
    #define fout(s) freopen(s".out","w",stdout)
    #define fcin ios::sync_with_stdio(false)
    #define l2(n) ((int)(log2(n)))
    #define inf 2122219134
    MB T ABS(T x){return x>0?x:-x;}
    MB T MAX(T x,T y){return x>y?x:y;}
    MB T MIN(T x,T y){return x<y?x:y;}
    MB T GCD(T x,T y){return y?GCD(y,x%y):x;}
    MB void SWAP(T &x,T &y){T t=x;x=y;y=t;}
}using namespace ZDY;using namespace std;
namespace IO{const int str=1<<20;static char in_buf[str],*in_s,*in_t;bool __=0;il char gc(){return (in_s==in_t)&&(in_t=(in_s=in_buf)+fread(in_buf,1,str,stdin)),in_s==in_t?__=1,EOF:*in_s++;}il void in(string &ch){ch.clear();if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}ch+=c;while((c=gc())!=EOF&&!isspace(c))ch+=c;if(c==EOF)__=1;}il void in(char &ch){if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF)__=1;else ch=c;}il void in(char *ch){*ch='\0';if(__)return;char c;while((c=gc())!=EOF&&isspace(c));if(c==EOF){__=1;return;}*ch=c;ch++;while((c=gc())!=EOF&&!isspace(c))*ch=c,ch++;if(c==EOF)__=1;*ch='\0';}template<typename T>il void in(T &x){if(__)return;char c=gc();bool f=0;while(c!=EOF&&(c<'0'||c>'9'))f^=(c=='-'),c=gc();if(c==EOF){__=1;return;}x=0;while(c!=EOF&&'0'<=c&&c<='9')x=x*10+c-48,c=gc();if(c==EOF)__=1;if(f)x=-x;}template<typename T,typename ... arr>il void in(T &x,arr & ... y){in(x),in(y...);}const char ln='\n';static char out_buf[str],*out_s=out_buf,*out_t=out_buf+str;il void flush(){fwrite(out_buf,1,out_s-out_buf,stdout);out_s=out_buf;}il void pt(char c){(out_s==out_t)?(fwrite(out_s=out_buf,1,str,stdout),*out_s++=c):(*out_s++=c);}il void out(const char* s){while(*s)pt(*s++);}il void out(char* s){while(*s)pt(*s++);}il void out(char c){pt(c);}il void out(string s){for(int i=0;s[i];i++)pt(s[i]);}template<typename T>il void out(T x){if(!x){pt('0');return;}if(x<0)pt('-'),x=-x;char a[50],t=0;while(x)a[t++]=x%10,x/= 10;while(t--)pt(a[t]+'0');}template<typename T,typename ... arr>il void out(T x,arr & ... y){out(x),out(y...);}}using namespace IO;
const int N=400011,P=998244353,G=3,Gi=332748118;
int n,l=0,limit=1,r[N],a[N],b[N],c[N];
il int pw(int x,int p){
    int ans=1;
    for(;p;x=1ll*x*x%P,p>>=1)if(p&1)ans=1ll*ans*x%P;
    return ans;
}
il void NTT(int *A,int typ){
    Fur(i,0,limit-1)
        if(i<r[i])SWAP(A[i],A[r[i]]);
    for(int len=1;len<limit;len<<=1){
        int Wn=pw(~typ?G:Gi,(P-1)/(len<<1));
        for(int i=0;i<limit;i+=(len<<1)){
            int w=1;
            Fur(k,0,len-1){
                int t=1ll*w*A[i+k+len]%P;
                A[i+k+len]=(A[i+k]-t+P)%P;
                A[i+k]=(A[i+k]+t)%P;
                w=1ll*w*Wn%P;
            }
        }
    }
    if(~typ)return;
    int inv=pw(limit,P-2);
    Fur(i,0,limit-1)A[i]=1ll*A[i]*inv%P;
}
void work(int p,int *a,int *b){
    if(p==1)return void(b[0]=pw(a[0],P-2));
    work(((p+1)>>1),a,b);
    limit=1,l=0;
    while(limit<(p<<1))limit<<=1,++l;
    Fur(i,0,limit-1)
        r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
    Fur(i,0,p-1)c[i]=a[i];
    Fur(i,p,limit-1)c[i]=0;
    NTT(c,1);NTT(b,1);
    Fur(i,0,limit-1)b[i]=1ll*b[i]*(2ll-1ll*c[i]*b[i]%P+P)%P;
    NTT(b,-1);
    Fur(i,p,limit-1)b[i]=0;
}
int main(){
    in(n);
    Fur(i,0,n-1)in(a[i]);
    work(n,a,b);
    Fur(i,0,n-1)out(b[i]," ");
    flush();
}
LG 4238 【模板】多项式乘法逆
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