FFT的常数优化

卡得一手好常数。。学习了。。(似乎只对FFT有效)

JZOJ 4349

 #include <bits/stdc++.h>
 #define LL long long
 #define DB long double
 using namespace std;
 const int mo=;
 const DB pi=acos(-);
 int K,T,p[],q[],f[],n,m,k,a[],rev[],ans;
 DB COS[],SIN[],Cos[],Sin[];
 struct cp{
     DB x,y;
     cp (DB p=,DB q=){x=p,y=q;}
 }w[],N[];
 cp operator+(cp a,cp b){
     return cp(a.x+b.x,a.y+b.y);
 }
 cp operator-(cp a,cp b){
     return cp(a.x-b.x,a.y-b.y);
 }
 cp operator*(cp a,cp b){
     return cp(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);
 }
 LL po(LL x,int y){
     LL z=;
     for (;y;y>>=,x=x*x%mo)
     if (y&) z=z*x%mo;
     return z;
 }
 LL C(int n,int m){
     return n<m?:(n<mo?1ll*p[n]*q[m]%mo*q[n-m]%mo:C(n%mo,m%mo)*C(n/mo,m/mo)%mo);
 }
 void fft(cp a[],int n,int d=){
     for (int i=;i<n;++i) if (i<rev[i]) swap(a[i],a[rev[i]]);
     for (int m=,k=;m<=n;k=m,m<<=){
         cp wn;
         wn=d>?cp(COS[m],SIN[m]):cp(Cos[m],Sin[m]);
         N[]=cp(,);
         for (int i=;i<k;++i) N[i]=N[i-]*wn;
         for (int i=;i<n;i+=m)
         for (int j=i;j<i+k;++j){
             cp u=a[j],v=a[j+k]*N[j-i];
             a[j]=u+v; a[j+k]=u-v;
         }
     }
     if (d==-) for (int i=;i<=n;++i) a[i].x/=n;
 }
 void preF(){
     f[]=; f[]=; f[]=;
     int t=,X=,Y=<<t;
     for (int i=;i<=;++i){
         int x=,y=;
         for (int j=;j<=;++j,x*=)
             (f[i]+=(C(,j)*x*(Y-X+y)%mo))%=mo,(y+=C(t,i-j-))%=mo;
         f[i]=(f[i]%mo+mo)%mo;
         for (int j=;j<=;++j)
             X=(X*-C(t,i-))%mo,++t;
         X=(X+C(t,i))%mo; Y=Y*%mo;
     }
 }
 int main(){
     for (int i=;i<=;i<<=)
         COS[i]=cos(pi*/i),Cos[i]=cos(pi*-/i),
         SIN[i]=sin(pi*/i),Sin[i]=sin(pi*-/i);
     scanf("%d",&T);
     p[]=q[]=;
     for (int i=,x;i<mo;++i){
         for (x=i;x%mo==;x/=mo);
         p[i]=1ll*p[i-]*x%mo;
     }
     q[mo-]=po(p[mo-],mo-);
     for (int i=mo-,x;i;--i){
         for (x=i+;x%mo==;x/=mo);
         q[i]=1ll*q[i+]*x%mo;
     }
     preF();
     while (T--){
         scanf("%d%d",&n,&K);
         for (int i=;i<=n;++i) a[i]=f[i];
         for (m=;m<n;m<<=);
         for (int i=n+;i<=m;++i) a[i]=; n=m;
         for (m=,k=;m<=n;k=m,m<<=){
             for (int i=;i<m;++i) rev[i]=rev[i>>]>>|(i&)*(m>>);
             for (int i=;i<=n;i+=m){
                 w[]=cp(,);
                 for (int j=i;j<i+k;++j) w[j-i+]=cp(a[j]+a[j+k],a[j]-a[j+k]);
                 for (int j=k+;j<m;++j) w[j]=cp(,);
                 fft(w,m);
                 for (int j=;j<m;++j) w[j]=w[j]*w[j];
                 fft(w,m,-);
                 for (int j=i;j<i+m-;++j) a[j]=(LL)round(w[j-i+].x/)%mo;
                 a[i+m-]=(LL)round(w[].x/-)%mo;
             }
         }
         ans=;
         for (int i=K;i<=n;++i) (ans+=a[i])%=mo;
         printf("%d\n",(ans%mo+mo)%mo);
     }
     return ;
 }

DAISHI