bzoj2179 FFT快速傅立葉

bzoj2179 FFT快速傅立葉

Description

Input

Output

Sample Input

Sample Output

HINT

Source

#include
#define F(i,j,n) for(int i=j;i<=n;i++)
#define D(i,j,n) for(int i=j;i>=n;i--)
#define ll long long
#define maxn 200000
using namespace std;
const double pi=acos(-1.0);
int n,m,len,rev[maxn],ans[maxn];
char s[maxn];
struct cp
{
	double x,y;
	inline cp operator +(cp a){return (cp){x+a.x,y+a.y};}
	inline cp operator -(cp a){return (cp){x-a.x,y-a.y};}
	inline cp operator *(cp a){return (cp){x*a.x-y*a.y,x*a.y+y*a.x};}
}a[maxn],b[maxn],c[maxn];
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
inline void fft(cp *x,int n,int flag)
{
	F(i,0,n-1) if (rev[i]>i) swap(x[rev[i]],x[i]);
	for(int m=2;m<=n;m<<=1)
	{
		cp wn=(cp){cos(2.0*pi/m*flag),sin(2.0*pi/m*flag)};
		for(int i=0;i>1;
			F(j,0,mid-1)
			{
				cp u=x[i+j],v=x[i+j+mid]*w;
				x[i+j]=u+v;x[i+j+mid]=u-v;
				w=w*wn;
			}
		}
	}
}
int main()
{
	n=read();
	scanf("%s",s);
	F(i,0,n-1) a[i].x=s[n-1-i]-'0';
	scanf("%s",s);
	F(i,0,n-1) b[i].x=s[n-1-i]-'0';
	m=1;n=2*n-1;
	while (m<=n) m<<=1,len++;n=m;
	F(i,0,n-1)
	{
		int t=i,ret=0;
		F(j,1,len) ret<<=1,ret|=t&1,t>>=1;
		rev[i]=ret;
	}
	fft(a,n,1);fft(b,n,1);
	F(i,0,n-1) c[i]=a[i]*b[i];
	fft(c,n,-1);
	F(i,0,n-1) ans[i]=(c[i].x/n)+0.5;
	F(i,0,n-1) ans[i+1]+=ans[i]/10,ans[i]%=10;
	n++;
	while (!ans[n]&&n) n--;
	D(i,n,0) putchar(ans[i]+'0');
	return 0;
}