• 洛谷 P4245 【模板】任意模数NTT

• 贴个板子，4次DFT。

Code

#include<cstdio> #include<algorithm> #include<cmath> #include<cctype> using namespace std; typedef long long LL; const int N=1e5+5; const double Pi=acos(-1); struct comp {double r,i;inline comp(){}inline comp(double rr,double ii){r=rr,i=ii;}friend inline comp operator +(comp x,comp y){return comp(x.r+y.r,x.i+y.i);}friend inline comp operator -(comp x,comp y){return comp(x.r-y.r,x.i-y.i);}friend inline comp operator *(comp x,comp y){return comp(x.r*y.r-x.i*y.i,x.r*y.i+x.i*y.r);}friend inline comp operator /(comp x,double y){return comp(x.r/y,x.i/y);}friend inline comp conj(comp x){return comp(x.r,-x.i);} }; int rev[N<<2],a[N],b[N],ans[N<<1]; comp a1[N<<2],b1[N<<2],a2[N<<2],b2[N<<2],wn[N<<2]; inline int read() {int X=0,w=0; char ch=0;while(!isdigit(ch)) w|=ch=='-',ch=getchar();while(isdigit(ch)) X=(X<<3)+(X<<1)+(ch^48),ch=getchar();return w?-X:X; } void write(int x) {if(x>9) write(x/10);putchar(x%10+'0'); } inline void FFT(comp *y,int len,int ff) {for(int i=0;i<len;i++)if(i<rev[i]) swap(y[i],y[rev[i]]);for(int h=2,d=len>>1;h<=len;h<<=1,d>>=1)for(int i=0,e=h>>1;i<len;i+=h)for(int j=0,cnt=0;j<e;j++,cnt+=d){comp u=y[i+j],v=wn[cnt]*y[i+j+e];y[i+j]=u+v;y[i+j+e]=u-v;}if(ff==-1){reverse(y+1,y+len);for(int i=0;i<len;i++) y[i]=y[i]/len;} } int main() {int n=read()+1,m=read()+1,p=read();for(int i=0;i<n;i++) a[i]=read();for(int i=0;i<m;i++) b[i]=read();int len=1,ll=0;while(len<n+m) len<<=1,ll++;for(int i=0;i<len;i++) rev[i]=rev[i>>1]>>1|(i&1)<<ll-1;for(int i=0;i<=len;i++) wn[i]=comp(cos(2*Pi*i/len),sin(2*Pi*i/len));for(int i=0;i<n;i++) a1[i]=comp(a[i]>>15,a[i]&32767);for(int i=0;i<m;i++) b1[i]=comp(b[i]>>15,b[i]&32767);FFT(a1,len,1),FFT(b1,len,1);for(int i=0;i<len;i++){int j=len-i&len-1;comp ea1=(a1[i]+conj(a1[j]))*comp(0.5,0);comp ea0=(a1[i]-conj(a1[j]))*comp(0,-0.5);comp eb1=(b1[i]+conj(b1[j]))*comp(0.5,0);comp eb0=(b1[i]-conj(b1[j]))*comp(0,-0.5);a2[i]=ea1*eb1+ea1*eb0*comp(0,1);b2[i]=ea0*eb1+ea0*eb0*comp(0,1);}FFT(a2,len,-1),FFT(b2,len,-1);for(int i=0;i<n+m-1;i++){int e1=(LL)round(a2[i].r)%p;int e2=(LL)round(a2[i].i)%p;int e3=(LL)round(b2[i].r)%p;int e4=(LL)round(b2[i].i)%p;ans[i]=((((LL)e1<<30)+((LL)e2+e3<<15)+e4)%p+p)%p;}for(int i=0;i<n+m-1;i++) write(ans[i]),putchar(' ');return 0; }

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