caioj1450:【快速傅里叶变换】大整数乘法

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
using namespace std;
const double PI=acos(-1.0);
struct Complex
{
    double r,i;
    Complex(){}
    Complex(double _r,double _i){r=_r;i=_i;}
    friend Complex operator + (const Complex &x,const Complex &y){return Complex(x.r+y.r,x.i+y.i);}
    friend Complex operator - (const Complex &x,const Complex &y){return Complex(x.r-y.r,x.i-y.i);}
    friend Complex operator * (const Complex &x,const Complex &y){return Complex(x.r*y.r-x.i*y.i,x.i*y.r+x.r*y.i);}
}a[410000],b[410000];
int n,m;
int R[410000];
void fft(Complex *y,int len,int on)
{
    for(int i=0;i<len;i++) if(i<R[i]) swap(y[i],y[R[i]]);
    for(int i=1;i<len;i<<=1)
    {
        Complex wn(cos(PI/i),sin(on*PI/i));
        for(int j=0;j<len;j+=(i<<1))
        {
            Complex w(1,0);
            for(int k=0;k<i;k++,w=w*wn)
            {
                Complex u=y[j+k];
                Complex v=w*y[j+k+i];
                y[j+k]=u+v;
                y[j+k+i]=u-v;
            }
        }
    }
    if(on==-1) for(int i=0;i<len;i++) y[i].r/=len;
}
void calc()
{
    m+=n;
    int L=0;
    for(n=1;n<=m;n<<=1) L++;
    for(int i=0;i<n;i++) R[i]=(R[i>>1]>>1)|(i&1)<<(L-1);
    fft(a,n,1);
    fft(b,n,1);
    for(int i=0;i<=n;i++) a[i]=a[i]*b[i];
    fft(a,n,-1);
}
char st[110000];
int d[410000];
int main()
{
    scanf("%s",st+1);
    n=strlen(st+1);n--;
    for(int i=0;i<=n;i++) a[i].r=double(st[i+1]-‘0‘);
    scanf("%s",st+1);
    m=strlen(st+1);m--;
    for(int i=0;i<=m;i++) b[i].r=double(st[i+1]-‘0‘);
    calc();
    for(int i=0;i<=m;i++) d[i]=int(a[m-i].r+0.5);
    for(int i=0;i<=m;i++)
    {
        d[i+1]+=d[i]/10;
        d[i]%=10;
    }
    int i=m;
    while(d[i+1]!=0)
    {
        i++;
        d[i+1]+=d[i]/10;
        d[i]%=10;
    }
    m=i;
    for(int i=m;i>=0;i--) printf("%d",d[i]);
    printf("\n");
    return 0;
}