BZOJ 2194 快速傅立叶之二

最新推荐文章于 2020-11-24 12:47:06 发布

Helloat123

最新推荐文章于 2020-11-24 12:47:06 发布

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分类专栏：数学文章标签： FFT

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本文探讨了卷积的基本概念，并通过一个具体的C++实现案例介绍了如何使用快速傅里叶变换（FFT）来高效地计算两个多项式的卷积。此方法广泛应用于信号处理和计算机科学中。

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还没搞清楚卷积到底是啥...大概就是多项式乘法?

//By Richard
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <ctime>
#define rep(x,y,z) for (int x=(y);(x)<=(z);(x)++)
#define per(x,y,z) for (int x=(y);(x)>=(z);(x)--)
#define log2(x) (31-__builtin_clz(x))
#define mod (int)(1e9+7)
#define inf 0x3f3f3f3f
#define cls(x) memset(x,0,sizeof(x))
#ifdef DEBUG
#define debugdo(X) X
#define debugndo(X)
#define debugout(X) cout<<(#X)<<"="<<(X)<<endl
#else
#define debugdo(X)
#define debugndo(X) X
#define debugout(X)
#endif // debug
#ifdef ONLINE_JUDGE
#define debugdo(X)
#define debugndo(X)
#define debugout(X)
#endif
#define putarray(x,n) rep(iiii,1,n) printf("%d ",x[iiii])
#define mp make_pair
using namespace std;
typedef pair<int,int> pairs;
typedef long long LL;
/////////////////////read3.0////////////////////////////////////
template <typename T>
inline void read(T &x){char ch;x=0;bool flag=false;ch=getchar();while (ch>'9'||ch<'0') {ch=getchar();if (ch=='-') flag=true;}while ((ch<='9'&&ch>='0')){x=x*10+ch-'0';ch=getchar();}if (flag) x*=-1;}
template <typename T>
inline void read(T &x,T &y){read(x);read(y);}
/////////////////variables&functions////////////////////
const double pi=3.1415926535897932384626433832795028841971693993751058209749446;
const int maxn=1024768;
struct cp
{
	double x,y;
	cp(double a=0,double b=0):x(a),y(b){}
	cp operator+(cp a){return cp(x+a.x,y+a.y);}
	cp operator-(cp a){return cp(x-a.x,y-a.y);}
	cp operator*(cp a){return cp(x*a.x-y*a.y,x*a.y+y*a.x);}
	cp operator/(int a){return cp(x/a,y/a);}
}A[maxn],B[maxn];
int l,n,r[maxn],nn,m;
inline void swap(int &x,int &y){int t=x;x=y;y=t;}
void FFT(cp *a,int n,int op)
{
	rep(i,0,n-1) if (i<r[i]) swap(a[i],a[r[i]]);
	for (int i=1;i<n;i<<=1)
	{
		cp wn(cos(pi/i),sin(pi/i)*op);
		for (int j=0;j<n;j+=i<<1)
		{
			cp w(1,0);
			for (int k=0;k<i;k++,w=w*wn)
			{
				cp x=a[j+k],y=w*a[i+j+k];
				a[j+k]=x+y;
				a[i+j+k]=x-y;
			}
		}
	}
}
void mul(cp *a,cp *b)
{
	m+=n-2;
	for (l=0,n=1;n<=m;n<<=1,l++);
	rep(i,0,n-1) r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
	FFT(a,n,1);FFT(b,n,1);
	// rep(i,0,n-1) printf("%d %d\n",(int)A[i].x,(int)B[i].x);
	rep(i,0,n-1) a[i]=a[i]*b[i];
	FFT(a,n,-1);
	rep(i,0,n-1) a[i].x=a[i].x/n;
}
int main()
{
	read(n);
	nn=n;
	rep(i,0,n-1) read(A[i].x,B[n-i-1].x);
	m=n;
	mul(A,B);
	// rep(i,0,n-1) printf("%d %d\n",(int)A[i].x,(int)B[i].x);
	rep(i,m-nn+1,m) printf("%d\n",(int)(A[i].x+0.5));
	return 0;
}