bzoj2179: FFT快速傅立叶

来源：互联网发布：win10声音优化软件编辑：程序博客网时间：2024/04/30 19:51

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 longconst 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wint 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nx+=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

0 0