FFT

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感觉也不是很会FFT，先把板子敲会再说吧= =

两条模板题_(:зゝ∠)_

UOJ #34

#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>using namespace std;#define f(i,n) for (int i=0;i<n;i++)const int N=4*1e5+5;const double pi=acos(-1);int n,m,k,ans[N]; struct CP{double x,y;CP operator +(const CP &A)const{return (CP){x+A.x,y+A.y};}CP operator -(const CP &A)const{return (CP){x-A.x,y-A.y};} CP operator *(const CP &A)const{return (CP){x*A.x-y*A.y,x*A.y+y*A.x};}}a[N],b[N];void FFT(CP a[],int flag){for (int i=1,j=0;i<k-1;i++){for (int s=k;(~j)&s;j^=(s>>=1));if (i<j) swap(a[i],a[j]);}for (int i=2;i<=k;i<<=1){CP wn=(CP){cos(2*pi/i),flag*sin(2*pi/i)};for (int j=0;j<k;j+=i){CP w=(CP){1,0};for (int t=j;t<j+i/2;t++){CP x=a[t],y=a[t+i/2]*w;a[t]=x+y;a[t+i/2]=x-y;w=w*wn;}}}if (flag==1) return;f(i,k) a[i].x/=k; }int main(){ //freopen("34.in","r",stdin);scanf("%d%d",&n,&m);n++;m++;int x; f(i,n) {scanf("%d",&x);a[i].x=x;}f(i,m) {scanf("%d",&x);b[i].x=x;}for (k=1;k<(n+m);k<<=1);FFT(a,1);FFT(b,1);f(i,k) a[i]=a[i]*b[i];FFT(a,-1);f(i,k) ans[i]=(int)(a[i].x+0.5);f(i,n+m-1)printf("%d%c",ans[i],i==n+m-2?'\n':' ');return 0; }

BZOJ 2179

#include <cstdio>#include <algorithm>#include <cstring>#include <cstdlib>#include <cmath>using namespace std;#define f(i,x) for (int i=0;i<x;i++)const int N=3*1e5;const double pi=acos(-1);int n,m,t,ans[N],len;struct CP{double x,y;CP operator +(const CP &A)const{return (CP){x+A.x,y+A.y};}CP operator -(const CP &A)const{return (CP){x-A.x,y-A.y};}CP operator *(const CP &A)const{return (CP){x*A.x-y*A.y,x*A.y+y*A.x};}}A[N],B[N];void fft(CP a[],int flag){for (int i=1,j=0;i<t-1;i++){for (int s=t;(~j)&s;j^=(s>>=1));if (i<j) swap(a[i],a[j]);}for (int i=2;i<=t;i<<=1){CP wn=(CP){cos(2*pi/i),flag*sin(2*pi/i)};for (int j=0;j<t;j+=i){CP w=(CP){1,0};for (int k=j;k<j+i/2;k++){CP x=a[k],y=a[k+i/2]*w;a[k]=x+y;a[k+i/2]=x-y;w=w*wn;} } }if (flag==1) return ;f(i,t) a[i].x/=t;}int main(){scanf("%d\n",&n);for (t=1;t<=n+n;t<<=1);f(i,n){int x=getchar()-'0';A[n-i-1].x=x;}getchar();f(i,n){int x=getchar()-'0';B[n-i-1].x=x;}fft(A,1);fft(B,1);f(i,t) A[i]=A[i]*B[i];fft(A,-1);f(i,t) ans[i]+=(int)(A[i].x+0.5);f(i,n+n) if (ans[i]){len=i;ans[i+1]+=ans[i]/10;ans[i]%=10; }for (int i=len;i>=0;i--) printf("%d",ans[i]);return 0;}

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