題目大意：給定n個(gè)點(diǎn)，第i個(gè)點(diǎn)和第j個(gè)點(diǎn)之間的庫(kù)侖力為(qi*qj)/(i-j)^2，定義左側(cè)為正方向，求每個(gè)點(diǎn)受的合力與電荷量的比值

#include<cmath>#include<cstdio>#include<cstring>#include<iostream>#include #define M 263000#define PI 3.1415926535897932384626433832795028841971using 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<n;i+=2) 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<n>>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<n;i++) -="c[n-i-1].a;" .a="1.0/i/i;" i="1;i<n;i++)" lf="" pre="" return=""></p></n;i++)></n></n;i+=2)></iostream></cstring></cstdio></cmath>