BZOJ 3257 ZJOI2014快速傅裡葉變換
題目大意：給定n個點，第i個點和第j個點之間的庫侖力為(qi*qj)/(i-j)^2，定義左側為正方向，求每個點受的合力與電荷量的比值
#include
#include
#include
#include
#include
#define M 263000
#define PI 3.1415926535897932384626433832795028841971
using namespace std;
struct Complex{
long double a,b;
Complex() {}
Complex(long double _,long double __):a(_),b(__) {}
Complex operator + (const Complex &x) const
{
return Complex(a+x.a,b+x.b);
}
Complex operator - (const Complex &x) const
{
return Complex(a-x.a,b-x.b);
}
Complex operator * (const Complex &x) const
{
return Complex(a*x.a-b*x.b,a*x.b+b*x.a);
}
void operator *= (const Complex &x)
{
*this=(*this)*x;
}
}a[M],b[M],c[M];
int n;
long double q[M],ans[M];
void FFT(Complex x[],int n,int type)
{
static Complex temp[M];
if(n==1) return ;
int i;
for(i=0;i>1]=x[i],temp[i+n>>1]=x[i+1];
memcpy(x,temp,sizeof(Complex)*n);
Complex *l=x,*r=x+(n>>1);
FFT(l,n>>1,type);FFT(r,n>>1,type);
Complex root(cos(type*2*PI/n),sin(type*2*PI/n)),w(1,0);
for(i=0;i>1;i++,w*=root)
temp[i]=l[i]+w*r[i],temp[i+(n>>1)]=l[i]-w*r[i];
memcpy(x,temp,sizeof(Complex)*n);
}
int main()
{
int i,digit;
double x;
cin>>n;
for(digit=1;digit<=n<<1;digit<<=1);
for(i=0;i
