bzoj3527: [Zjoi2014]力

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int maxn=400005;
const double PI=acos(-1);
struct node{
    double real,imag;
    void clear(){real=imag=0;}
    node operator +(const node &x){return (node){real+x.real,imag+x.imag};}
    node operator -(const node &x){return (node){real-x.real,imag-x.imag};}
    node operator *(const node &x){return (node){real*x.real-imag*x.imag,real*x.imag+imag*x.real};}
}q[maxn],p[maxn],A[maxn],t1,t2,w,wn;
int m,n,len,rev[maxn];
int Rev(int x){
    int temp=0;
    for (int i=1;i<=len;i++){temp<<=1,temp+=(x&1),x>>=1;}
    return temp;
}
void FFT(node *a,int op){
    for (int i=0;i<n;i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
    for (int s=2;s<=n;s<<=1){
        wn=(node){cos(2.0*op*PI/s),sin(2.0*op*PI/s)};
        for (int i=0;i<n;i+=s){
            w=(node){1,0};
            for (int j=i;j<i+s/2;j++,w=w*wn){
                t1=a[j],t2=w*a[j+s/2];
                a[j]=t1+t2,a[j+s/2]=t1-t2;
            }
        }
    }
}
int main(){
    scanf("%d",&m); n=1,len=0;
    while (n<(m<<1)) n<<=1,len++;
    for (int i=0;i<n;i++) rev[i]=Rev(i);
    for (int i=0;i<n;i++) p[i].clear(),q[i].clear();
    for (int i=1;i<=m;i++) scanf("%lf",&q[i].real);
    for (int i=0;i<m;i++) p[i].real=-1.0/(i-m)/(i-m);
    p[m].real=0; for (int i=m+1;i<n;i++) p[i].real=1.0/(i-m)/(i-m);
    FFT(q,1),FFT(p,1);
    for (int i=0;i<n;i++) A[i]=q[i]*p[i];
    FFT(A,-1);
    for (int i=0;i<n;i++) A[i].real=1.0*A[i].real/n;
    for (int i=1;i<=m;i++) printf("%.3lf
",A[m+i].real);
    return 0;
}

题目大意；题意上网找吧。

做法：我们令A[i+n]=E[n],然后修改一个数组的定义，就是裸的卷积了，直接FFT，详见16年国家集训队论文。