题目大意：给定n个点，第i个点和第j个点之间的库仑力为(qi*qj)/(i-j)^2，定义左侧为正方向，求每个点受的合力与电荷量的比值
#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>