BZOJ 2179 [快速傅里叶变换 高精度乘法]

#include <iostream>
#include <cstdio>
#include <string>
#include <algorithm>
#include <cmath>
using namespace std;
const int N=3e5+5;
const double PI=acos(-1);
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1;c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘;c=getchar();}
    return x*f;
}
struct Vector{
    double x,y;
    Vector(double a=0,double b=0):x(a),y(b){}
};
typedef Vector CD;
Vector operator +(Vector a,Vector b){return Vector(a.x+b.x,a.y+b.y);}
Vector operator -(Vector a,Vector b){return Vector(a.x-b.x,a.y-b.y);}
Vector operator *(Vector a,Vector b){return Vector(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
Vector operator /(Vector a,int b){return Vector(a.x/b,a.y/b);}
Vector conj(Vector a){return Vector(a.x,-a.y);}

struct FastFourierTransform{
    CD omega[N],omegaInv[N];
    void ini(int n){
        for(int k=0;k<n;k++) 
            omega[k]=CD(cos(2*PI/n*k),sin(2*PI/n*k)),
            omegaInv[k]=conj(omega[k]);
    }
    void transform(CD *a,int n,CD *omega){
        int k=0;
        while((1<<k)<n) k++;
        for(int i=0;i<n;i++){
            int t=0;
            for(int j=0;j<k;j++) if(i&(1<<j)) t|=(1<<(k-j-1));
            if(i>t) swap(a[t],a[i]);
        }
        
        for(int l=2;l<=n;l<<=1){
            int m=l>>1;
            for(CD *p=a;p!=a+n;p+=l)
                for(int k=0;k<m;k++){
                    CD t=omega[n/l*k]*p[k+m];
                    p[k+m]=p[k]-t;
                    p[k]=p[k]+t;
                }
        }
    }
    void DFT(CD *a,int n){transform(a,n,omega);}
    void IDFT(CD *a,int n){
        transform(a,n,omegaInv);
        for(int i=0;i<n;i++) a[i]=a[i]/n;
    }
    void FFT(CD *a,CD *b,int m){
        int n=1;
        while(n<m) n<<=1;
        ini(n);
        DFT(a,n);DFT(b,n);
        for(int i=0;i<n;i++) a[i]=a[i]*b[i];
        IDFT(a,n);
    }
}fft;
CD A[N],B[N];
int n,m,c[N];
char s1[N],s2[N];
int main(){
    freopen("in","r",stdin);
    n=read();m=n+n-1;
    scanf("%s%s",s1,s2);
    for(int i=0;i<n;i++) A[i].x=s1[n-i-1]-‘0‘,B[i].x=s2[n-i-1]-‘0‘;
    fft.FFT(A,B,m);
    for(int i=0;i<m;i++) c[i]=int(A[i].x+0.5);//printf("c %d\n",c[i]);
    for(int i=0;i<m;i++) c[i+1]+=c[i]/10,c[i]%=10;
    while(c[m]) m++;
    for(int i=m-1;i>=0;i--) printf("%d",c[i]);

    return 0;
}