【ZJOI2014】BZOJ3527力题解（FFT）

题目：BZOJ3527.
题目大意：给定一个长度为$n$的$q$数组，求：
$$E_j=\sum _{i<j}\frac{q_i}{(i-j{)}^2}-\sum _{i>j}\frac{q_i}{(i-j{)}^2}$$

其中$1\le n\le 10^5,1\le q_i<10^9$.

设$f_i=\frac{1}{i^2}$，开始大力推式子：
$$\begin{gathered} E_j=\sum _{i<j}{q}_i{f}_{j-i}-\sum _{i>j}{q}_i{f}_{i-j} \\ =\sum _{i=1}^{j-1}{q}_i{f}_{j-i}-\sum _{i=j+1}^n{q}_i{f}_{i-j} \end{gathered}$$

令$p_i=q_{n-i}$，那么：
$$\begin{gathered} E_j=\sum _{i=1}^{j-1}{q}_i{f}_{j-i}-\sum _{i=j+1}^n{p}_{n-i}{f}_{i-j} \\ =\sum _{i=1}^{j-1}{q}_i{f}_{j-i}-\sum _{i=1}^{n-j}{p}_{n-j-i}{f}_i \end{gathered}$$

发现前后两个都是卷积形式，用FFT优化即可.时间复杂度$O(n\log n)$.

代码如下：

#include<bits/stdc++.h>
  using namespace std;

#define Abigail inline void
typedef long long LL;

const int N=262144;
const double pi=acos(-1);

struct comp{
  double a,b;
  comp(double A=0,double B=0){a=A;b=B;}
  comp operator + (const comp &p){return comp(a+p.a,b+p.b);}
  comp operator - (const comp &p){return comp(a-p.a,b-p.b);}
  comp operator * (const comp &p){return comp(a*p.a-b*p.b,a*p.b+b*p.a);}
};

int n,m,len,rev[N+9];
comp pw[N+9],a[N+9],b[N+9],c[N+9];
double ans[N+9];

void Get_len(int n){
  int l=0;
  for (len=1;len<=n;len<<=1) ++l;
  for (int i=0;i<len;++i) rev[i]=rev[i>>1]>>1|(i&1)<<l-1; 
}

void FFT(comp *a,int n,int t){
  for (int i=0;i<n;++i)
    if (i<rev[i]) swap(a[i],a[rev[i]]);
  for (int i=1;1<<i<=n;++i){
  	int len=1<<i;
  	comp wn=comp(cos(2.0*pi/len),sin(2.0*pi/len)*(t?-1:1));
  	pw[0]=comp(1,0);
  	for (int j=1;j<len>>1;++j) pw[j]=pw[j-1]*wn;
  	for (int j=0;j<n;j+=len)
  	  for (int k=0;k<len>>1;++k){
  	  	comp x=a[j+k],y=pw[k]*a[j+k+(len>>1)];
  	  	a[j+k]=x+y;a[j+k+(len>>1)]=x-y;
  	  }
  }
}

Abigail into(){
  scanf("%d",&n);
  for (int i=1;i<=n;++i)
    scanf("%lf",&a[i].a);
}

Abigail work(){
  for (int i=0;i<=n;++i) b[i].a=a[n-i].a;
  for (int i=1;i<=n;++i) c[i].a=1.0/i/i;
  Get_len(n<<1);
  FFT(a,len,0);
  FFT(b,len,0);
  FFT(c,len,0);
  for (int i=0;i<len;++i) a[i]=a[i]*c[i],b[i]=b[i]*c[i];
  FFT(a,len,1);
  FFT(b,len,1);
  for (int i=1;i<=n;++i) ans[i]=(a[i].a-b[n-i].a)/len;
}

Abigail outo(){
  for (int i=1;i<=n;++i)
    printf("%.3lf\n",ans[i]);
  puts("");
}

int main(){
  into();
  work();
  outo();
  return 0;
}