BZOJ2179: FFT快速傅立叶 FFT实现高精度乘法

Code:

#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#define setIO(s) freopen(s".in","r",stdin)
#define maxn 200000
#define pi 3.1415926535898
using namespace std;
int len=1,l,r[maxn<<1];
long long ans[maxn]; 
char str1[maxn],str2[maxn]; 
struct Cpx{
    double x,y;
    Cpx(double t1=0,double t2=0){x=t1,y=t2;}
}A[maxn<<1],B[maxn<<1],C[maxn<<1];
Cpx operator+(Cpx a,Cpx b){ return Cpx(a.x+b.x,a.y+b.y);}
Cpx operator-(Cpx a,Cpx b){ return Cpx(a.x-b.x,a.y-b.y);}
Cpx operator*(Cpx a,Cpx b){ return Cpx(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
void FFT(Cpx *a,int n,int flag){
    for(int i=0;i<n;++i) if(i<r[i]) swap(a[i],a[r[i]]); 
    for(int mid=1;mid<n;mid<<=1){
        Cpx wn(cos(pi/mid),flag*sin(pi/mid)),x,y;
        for(int j=0;j<n;j+=(mid<<1)){
            Cpx w(1,0);
            for(int k=0;k<mid;++k){
                x=a[j+k],y=w*a[j+mid+k];
                a[j+k]=x+y,a[j+mid+k]=x-y;
                w=w*wn; 
            }
        }
    }
}
int main(){
    //setIO("input"); 
    int n; 
    scanf("%d",&n);
    scanf("%s%s",str1,str2);
    for(int i=0;i<n;++i) A[i].x=str1[n-i-1]-48;
    for(int i=0;i<n;++i) B[i].x=str2[n-i-1]-48; 
    while(len<n+n) len<<=1,++l; 
    for(int i=0;i<len;++i) 
        r[i]=(r[i>>1]>>1)|((i&1)<<(l-1));
    FFT(A,len,1),FFT(B,len,1);
    for(int i=0;i<len;++i) 
        C[i]=A[i]*B[i]; 
    FFT(C,len,-1); 
    for(int i=0;i<=len;++i) {
        ans[i]+=(int)(C[i].x/len+0.5); 
        if(ans[i]>=10) 
            ans[i+1]+=ans[i]/10,ans[i]%=10,
        len+=(len==i); 
    }
    while(!ans[len]&&len>=0) --len;  
    while(len>=0) printf("%d",(int)ans[len]),--len; 
    return 0; 
}