bzoj3527: [Zjoi2014]力
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
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。
