BZOJ4451 : [Cerc2015]Frightful Formula

$(i,1)$对答案的贡献为$l_iC(2n-i-2,n-i)a^{n-1}b^{n-i}$。

$(1,i)$对答案的贡献为$t_iC(2n-i-2,n-i)*a^{n-i}b^{n-1}$。

$(i,j)$的$c$对答案的贡献为$cC(2n-i-j,n-i)a^{n-j}b^{n-i}$。

$c$总的贡献为：

\[\begin{eqnarray*}
&&c\sum_{i=2}^n\sum_{j=2}^nC(2n-i-j,n-i)a^{n-j}b^{n-i}\\
&=&c\sum_{i=2}^n\sum_{j=2}^n(2n-i-j)!\times\frac{a^{n-j}}{(n-j)!}\times\frac{b^{n-i}}{(n-i)!}\\
&=&c\sum_{i=2}^n\sum_{j=2}^n(2n-i-j)!A_jB_i
\end{eqnarray*}\]

设

\[\begin{eqnarray*}
A_i=\frac{a^{n-i}}{(n-i)!}\\
B_i=\frac{b^{n-i}}{(n-i)!}
\end{eqnarray*}\]

则

\[\begin{eqnarray*}
ans+=c\sum_{i=4}^{2n}(2n-i)!\sum_{j=0}^i A_{i-j}B{j}
\end{eqnarray*}\]

FFT mod any prime即可。

时间复杂度$O(n\log n)$。

 

#include<cstdio>
#include<cmath>
#include<algorithm>
using namespace std;
const int N=524300,P=1000003,M=1000;
int n,a,b,c,i,j,k,pos[N],ans;
int pa[N],pb[N],fac[N],inv[N],A[N],B[N],C[N];
inline void read(int&a){char c;while(!(((c=getchar())>='0')&&(c<='9')));a=c-'0';while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';}
namespace FFT{
struct comp{
  long double r,i;comp(long double _r=0,long double _i=0){r=_r;i=_i;}
  comp operator+(const comp x){return comp(r+x.r,i+x.i);}
  comp operator-(const comp x){return comp(r-x.r,i-x.i);}
  comp operator*(const comp x){return comp(r*x.r-i*x.i,r*x.i+i*x.r);}
  comp conj(){return comp(r,-i);}
}A[N],B[N];
int a0[N],b0[N],a1[N],b1[N];
const long double pi=acos(-1.0);
void FFT(comp a[],int n,int t){
  for(int i=1;i<n;i++)if(i<pos[i])swap(a[i],a[pos[i]]);
  for(int d=0;(1<<d)<n;d++){
    int m=1<<d,m2=m<<1;
    long double o=pi*2/m2*t;comp _w(cos(o),sin(o));
    for(int i=0;i<n;i+=m2){
      comp w(1,0);
      for(int j=0;j<m;j++){
        comp&A=a[i+j+m],&B=a[i+j],t=w*A;
        A=B-t;B=B+t;w=w*_w;
      }
    }
  }
  if(t==-1)for(int i=0;i<n;i++)a[i].r/=n;
}
void mul(int*a,int*b,int*c){//c=a*b
  int i,j;
  for(i=0;i<k;i++)A[i]=comp(a[i],b[i]);
  FFT(A,k,1);
  for(i=0;i<k;i++){
    j=(k-i)&(k-1);
    B[i]=(A[i]*A[i]-(A[j]*A[j]).conj())*comp(0,-0.25);
  }
  FFT(B,k,-1);
  for(i=0;i<k;i++)c[i]=((long long)(B[i].r+0.5))%P;
}
//输入两个多项式，求a*b mod P，保存在c中，c不能为a或b
void mulmod(int*a,int*b,int*c){
  int i;
  for(i=0;i<k;i++)a0[i]=a[i]/M,b0[i]=b[i]/M;
  for(mul(a0,b0,a0),i=0;i<k;i++){
    c[i]=1LL*a0[i]*M*M%P;
    a1[i]=a[i]%M,b1[i]=b[i]%M;
  }
  for(mul(a1,b1,a1),i=0;i<k;i++){
    c[i]=(a1[i]+c[i])%P,a0[i]=(a0[i]+a1[i])%P;
    a1[i]=a[i]/M+a[i]%M,b1[i]=b[i]/M+b[i]%M;
  }
  for(mul(a1,b1,a1),i=0;i<k;i++)c[i]=(1LL*M*(a1[i]-a0[i]+P)+c[i])%P;
}
}
int main(){
  read(n),read(a),read(b),read(c);
  for(pa[0]=i=1;i<=n;i++)pa[i]=1LL*pa[i-1]*a%P;
  for(pb[0]=i=1;i<=n;i++)pb[i]=1LL*pb[i-1]*b%P;
  for(fac[0]=i=1;i<=n+n;i++)fac[i]=1LL*fac[i-1]*i%P;
  for(inv[0]=inv[1]=1,i=2;i<=n;i++)inv[i]=1LL*(P-inv[P%i])*(P/i)%P;
  for(i=1;i<=n;i++)inv[i]=1LL*inv[i]*inv[i-1]%P;
  for(i=1;i<=n;i++){
    read(j);
    if(i>1)ans=(1LL*fac[n+n-i-2]*inv[n-i]%P*pa[n-1]%P*pb[n-i]%P*j+ans)%P;
  }
  for(i=1;i<=n;i++){
    read(j);
    if(i>1)ans=(1LL*fac[n+n-i-2]*inv[n-i]%P*pa[n-i]%P*pb[n-1]%P*j+ans)%P;
  }
  ans=1LL*ans*inv[n-2]%P;
  for(k=1;k<=n;k<<=1);k<<=1;
  j=__builtin_ctz(k)-1;
  for(i=0;i<k;i++)pos[i]=pos[i>>1]>>1|((i&1)<<j);
  for(i=2;i<=n;i++)A[i]=1LL*pa[n-i]*inv[n-i]%P;
  for(i=2;i<=n;i++)B[i]=1LL*pb[n-i]*inv[n-i]%P;
  FFT::mulmod(A,B,C);
  for(i=4;i<=n+n;i++)ans=(1LL*C[i]*fac[n+n-i]%P*c+ans)%P;
  return printf("%d",ans),0;
}