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:
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #define maxn 262144 7 #define pi 3.14159265358979323846 8 using namespace std; 9 char ch; 10 int m,n; 11 bool ok; 12 long double ans[maxn]; 13 double q[maxn]; 14 void read(int &x){ 15 for (ok=0,ch=getchar();!isdigit(ch);ch=getchar()) if (ch=='-') ok=1; 16 for (x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar()); 17 if (ok) x=-x; 18 } 19 struct comp{ 20 long double rea,ima; 21 void clear(){rea=ima=0;} 22 comp operator +(const comp &x){return (comp){rea+x.rea,ima+x.ima};} 23 comp operator -(const comp &x){return (comp){rea-x.rea,ima-x.ima};} 24 comp operator *(const comp &x){return (comp){rea*x.rea-ima*x.ima,rea*x.ima+ima*x.rea};} 25 }a[maxn],b[maxn],c[maxn],tmp[maxn],w,wn; 26 void fft(comp *a,int st,int siz,int step,int op){ 27 if (siz==1) return; 28 fft(a,st,siz>>1,step<<1,op),fft(a,st+step,siz>>1,step<<1,op); 29 int x=st,x1=st,x2=st+step; 30 w=(comp){1,0},wn=(comp){cos(op*2*pi/siz),sin(op*2*pi/siz)}; 31 for (int i=0;i<(siz>>1);i++,x+=step,x1+=(step<<1),x2+=(step<<1),w=w*wn) 32 tmp[x]=a[x1]+(w*a[x2]),tmp[x+(siz>>1)*step]=a[x1]-(w*a[x2]); 33 for (int i=st;siz;i+=step,siz--) a[i]=tmp[i]; 34 } 35 int main(){ 36 read(m),n=1; 37 while (n<(m<<1)) n<<=1; 38 for (int i=0;i<m;i++) scanf("%lf",&q[i]); 39 40 for (int i=0;i<n;i++) a[i].clear(); 41 for (int i=0;i<n;i++) b[i].clear(); 42 for (int i=0;i<m;i++) a[i].rea=q[i]; 43 for (int i=1;i<m;i++) b[i].rea=1.0/i/i; 44 fft(a,0,n,1,1),fft(b,0,n,1,1); 45 for (int i=0;i<n;i++) c[i]=a[i]*b[i]; 46 fft(c,0,n,1,-1); 47 for (int i=0;i<m;i++) ans[i]=c[i].rea/n; 48 49 for (int i=0;i<n;i++) a[i].clear(); 50 for (int i=0;i<n;i++) b[i].clear(); 51 for (int i=0;i<m;i++) a[i].rea=q[m-i-1]; 52 for (int i=1;i<m;i++) b[i].rea=1.0/i/i; 53 fft(a,0,n,1,1),fft(b,0,n,1,1); 54 for (int i=0;i<n;i++) c[i]=a[i]*b[i]; 55 fft(c,0,n,1,-1); 56 57 for (int i=0;i<m;i++) printf("%.9lf ",(double)(ans[i]-c[m-i-1].rea/n)); 58 return 0; 59 }