bzoj2179: FFT快速傅立叶

一句话题意:给出两个n位10进制整数x和y,你需要计算x*y。n<=60000

思路:FFT就是神奇的公式多。。。感觉还没有完全理解

#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long
const int maxn=200010;
const double pi=M_PI;
using namespace std;
struct plex{
	double r,i;
	void clear(){r=i=0;}
}tmp[maxn];
plex operator +(plex a,plex b){return (plex){a.r+b.r,a.i+b.i};}
plex operator -(plex a,plex b){return (plex){a.r-b.r,a.i-b.i};}
plex operator *(plex a,plex b){return (plex){a.r*b.r-a.i*b.i,a.r*b.i+a.i*b.r};}
char s[maxn];int nn,ans[maxn];

struct DFT{
	plex a[maxn];
	void read(){
		scanf("%s",s);
		int n=strlen(s),m;
		for (int i=n-1;i>=0;i--) a[n-i-1].r=s[i]-'0';
		for (m=1;m<=n;m<<=1);m<<=1;
		nn=max(nn,m);
	}
	void fft(int bg,int step,int size,int op){
		if (size==1) return;
		fft(bg,step*2,size/2,op),fft(bg+step,step*2,size/2,op);
		plex w=(plex){1,0},t=(plex){cos(2*pi/size),sin(2*op*pi/size)};//w是n次单位根,乘一次t就能得到下一个w
		int p=bg,p0=bg,p1=bg+step;
		for (int i=0;i<size/2;i++){
			tmp[p]=a[p0]+w*a[p1];
			tmp[p+size/2*step]=a[p0]-w*a[p1];
			p+=step,p0+=step*2,p1+=step*2,w=w*t;
		}
		for (int i=bg;size;size--,i+=step) a[i]=tmp[i];
	}
}a,b,c;

int main(){
	scanf("%d",&nn);
	a.read(),b.read();
   	a.fft(0,1,nn,1),b.fft(0,1,nn,1);
	for (int i=0;i<nn;i++) c.a[i]=a.a[i]*b.a[i];//点值表示时直接乘
	c.fft(0,1,nn,-1);ll x=0;
	for (int i=0;i<nn;i++){
		c.a[i].r/=nn;//逆变换时要多除一个n
		x+=1ll*round(c.a[i].r);
		ans[i]=x%10,x/=10;
	}
	for (;nn&&!ans[nn];nn--);//去前导零
	printf("%d",ans[nn--]);
	for (int i=nn;i>=0;i--) printf("%01d",ans[i]);puts("");
	return 0;
}

直接上FFT

posted @ 2015-07-15 17:44  orzpps  阅读(69)  评论(0编辑  收藏  举报