http://www.lydsy.com/JudgeOnline/problem.php?id=2194
相乘两项的下标 的 差相同
那么把某一个反过来就是卷积形式
fft优化
#include<cmath> #include<cstdio> #include<iostream> #include<algorithm> using namespace std; const int N=(1<<18)+2; const double pi=acos(-1); int r[N]; struct Complex { double a,b; Complex(double x_=0,double y_=0):a(x_),b(y_) {} Complex operator + (Complex p) { Complex c; c.a=a+p.a; c.b=b+p.b; return c; } Complex operator - (Complex p) { Complex c; c.a=a-p.a; c.b=b-p.b; return c; } Complex operator * (Complex p) { Complex c; c.a=a*p.a-b*p.b; c.b=a*p.b+b*p.a; return c; } }; typedef Complex E; E A[N],B[N],C[N]; int n; void read(int &x) { x=0; char c=getchar(); while(!isdigit(c)) c=getchar(); while(isdigit(c)) { x=x*10+c-‘0‘; c=getchar(); } } void fft(E *a,int f) { for(int i=0;i<n;++i) if(i<r[i]) swap(a[i],a[r[i]]); for(int i=1;i<n;i<<=1) { E wn(cos(pi/i),f*sin(pi/i)); for(int p=i<<1,j=0;j<n;j+=p) { E w(1,0); for(int k=0;k<i;++k,w=w*wn) { E x=a[j+k],y=w*a[j+k+i]; a[j+k]=x+y; a[j+k+i]=x-y; } } } } int main() { int cnt; read(cnt); int x,y; for(int i=0;i<cnt;++i) { read(x); read(y); A[cnt-1-i].a=x; B[i].a=y; } int m=cnt+cnt-2,l=0; for(n=1;n<=m;n<<=1) l++; for(int i=0;i<n;++i) r[i]=(r[i>>1]>>1)|((i&1)<<l-1); fft(A,1); fft(B,1); for(int i=0;i<n;++i) C[i]=A[i]*B[i]; fft(C,-1); for(int i=cnt-1;i>=0;--i) printf("%d\n",int(C[i].a/n+0.5)); }