3527: [Zjoi2014]力

题目大意：给出n个数qi，定义 Fj为

      

令 Ei=Fi/qi，求Ei。

设A[i]=q[i],B[i]=1/(i^2)。

设C[i]=sigma(A[j]*B[i-j]),D[i]=sigma(A[n-j-1]*B[i-j])。

那么所求的E[i]=C[i]-D[i]。

不难发现C[i]已经是标准的卷积形式了，用FFT即可。

对于D[i],令A'[i]=A[n-i-1]，那么D[i]=sigma(A[j]*B[i-j])，于是也用FFT即可。

code:

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<cstring>
 #include<algorithm>
 #define maxn 262144
 #define pi 3.14159265358979323846
 using namespace std;
 char ch;
 int m,n;
 bool ok;
 long double ans[maxn];
 double q[maxn];
 void read(int &x){
     for (ok=,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=;
     for (x=;isdigit(ch);x=x*+ch-'',ch=getchar());
     if (ok) x=-x;
 }
 struct comp{
     long double rea,ima;
     void clear(){rea=ima=;}
     comp operator +(const comp &x){return (comp){rea+x.rea,ima+x.ima};}
     comp operator -(const comp &x){return (comp){rea-x.rea,ima-x.ima};}
     comp operator *(const comp &x){return (comp){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};}
 }a[maxn],b[maxn],c[maxn],tmp[maxn],w,wn;
 void fft(comp *a,int st,int siz,int step,int op){
     if (siz==) return;
     fft(a,st,siz>>,step<<,op),fft(a,st+step,siz>>,step<<,op);
     int x=st,x1=st,x2=st+step;
     w=(comp){,},wn=(comp){cos(op**pi/siz),sin(op**pi/siz)};
     for (int i=;i<(siz>>);i++,x+=step,x1+=(step<<),x2+=(step<<),w=w*wn)
         tmp[x]=a[x1]+(w*a[x2]),tmp[x+(siz>>)*step]=a[x1]-(w*a[x2]);
     for (int i=st;siz;i+=step,siz--) a[i]=tmp[i];
 }
 int main(){
     read(m),n=;
     while (n<(m<<)) n<<=;
     for (int i=;i<m;i++) scanf("%lf",&q[i]); 

     for (int i=;i<n;i++) a[i].clear();
     for (int i=;i<n;i++) b[i].clear();
     for (int i=;i<m;i++) a[i].rea=q[i];
     for (int i=;i<m;i++) b[i].rea=1.0/i/i;
     fft(a,,n,,),fft(b,,n,,);
     for (int i=;i<n;i++) c[i]=a[i]*b[i];
     fft(c,,n,,-);
     for (int i=;i<m;i++) ans[i]=c[i].rea/n; 

     for (int i=;i<n;i++) a[i].clear();
     for (int i=;i<n;i++) b[i].clear();
     for (int i=;i<m;i++) a[i].rea=q[m-i-];
     for (int i=;i<m;i++) b[i].rea=1.0/i/i;
     fft(a,,n,,),fft(b,,n,,);
     for (int i=;i<n;i++) c[i]=a[i]*b[i];
     fft(c,,n,,-); 

     for (int i=;i<m;i++) printf("%.9lf\n",(double)(ans[i]-c[m-i-].rea/n));
     return ;
 }