bzoj3527: [Zjoi2014]力

Posted 2020-08-09 oyzx~

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 using namespace std;
 7 const int maxn=400005;
 8 const double PI=acos(-1);
 9 struct node{
10     double real,imag;
11     void clear(){real=imag=0;}
12     node operator +(const node &x){return (node){real+x.real,imag+x.imag};}
13     node operator -(const node &x){return (node){real-x.real,imag-x.imag};}
14     node operator *(const node &x){return (node){real*x.real-imag*x.imag,real*x.imag+imag*x.real};}
15 }q[maxn],p[maxn],A[maxn],t1,t2,w,wn;
16 int m,n,len,rev[maxn];
17 int Rev(int x){
18     int temp=0;
19     for (int i=1;i<=len;i++){temp<<=1,temp+=(x&1),x>>=1;}
20     return temp;
21 }
22 void FFT(node *a,int op){
23     for (int i=0;i<n;i++) if (i<rev[i]) swap(a[i],a[rev[i]]);
24     for (int s=2;s<=n;s<<=1){
25         wn=(node){cos(2.0*op*PI/s),sin(2.0*op*PI/s)};
26         for (int i=0;i<n;i+=s){
27             w=(node){1,0};
28             for (int j=i;j<i+s/2;j++,w=w*wn){
29                 t1=a[j],t2=w*a[j+s/2];
30                 a[j]=t1+t2,a[j+s/2]=t1-t2;
31             }
32         }
33     }
34 }
35 int main(){
36     scanf("%d",&m); n=1,len=0;
37     while (n<(m<<1)) n<<=1,len++;
38     for (int i=0;i<n;i++) rev[i]=Rev(i);
39     for (int i=0;i<n;i++) p[i].clear(),q[i].clear();
40     for (int i=1;i<=m;i++) scanf("%lf",&q[i].real);
41     for (int i=0;i<m;i++) p[i].real=-1.0/(i-m)/(i-m);
42     p[m].real=0; for (int i=m+1;i<n;i++) p[i].real=1.0/(i-m)/(i-m);
43     FFT(q,1),FFT(p,1);
44     for (int i=0;i<n;i++) A[i]=q[i]*p[i];
45     FFT(A,-1);
46     for (int i=0;i<n;i++) A[i].real=1.0*A[i].real/n;
47     for (int i=1;i<=m;i++) printf("%.3lf\n",A[m+i].real);
48     return 0;
49 }
50 

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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。