FFT高精度乘法

Posted 2021-08-16 爷灬傲奈我何123

#include<iostream>
#include<cmath>
#include<vector>
using namespace std;
const int N=400010;
struct Complex{
    double x;
    double y;
    Complex()
    {
        
    }
    Complex (double _x,double _y)
    {
        x=_x,y=_y;
    }
    Complex operator+(const Complex&tt)const
    {
        return {x+tt.x,y+tt.y};
    }
    Complex operator-(const Complex&tt)const
    {
        return {x-tt.x,y-tt.y};
    }
    Complex operator*(const Complex&tt)const
    {
        return {x*tt.x-y*tt.y,x*tt.y+y*tt.x};
    }
}a[N],b[N];
string s1,s2;
int rev[N];
int ans[N];
int tot;
void fft(Complex a[],int inv)
{
    for(int i=0;i<tot;i++)
        if(i<rev[i])
        swap(a[i],a[rev[i]]);
    for(int mid=1;mid<tot;mid*=2)
    {
        auto w1=Complex(cos(acos(-1)/mid),inv*sin(acos(-1)/mid));
        for(int i=0;i<tot;i+=mid*2)
        {
            auto wk=Complex(1,0);
            for(int j=0;j<mid;j++,wk=wk*w1)
            {
                auto x=a[i+j];
                auto y=a[i+j+mid]*wk;
                a[i+j]=x+y,a[i+j+mid]=x-y;
            }
        }
    }
}
signed main()
{
    cin>>s1>>s2;
    int n=s1.size();
    int m=s2.size();
    n--;
    m--;
    for(int i=0;i<=n;i++)
        a[i].x=s1[n-i]-'0';
    for(int i=0;i<=m;i++)
        b[i].x=s2[m-i]-'0';
    int bit=0;
    while((1<<bit)<n+m+1)   bit++;
    tot=(1<<bit);
    for(int i=1;i<tot;i++)
        rev[i]=(rev[i>>1]>>1)|((i&1)<<(bit-1));
    fft(a,1);
    fft(b,1);
    for(int i=0;i<tot;i++)
    a[i]=a[i]*b[i];
    fft(a,-1);
    for(int i=0;i<tot;i++)
    a[i].x/=tot;
    int tt=0;
    vector<int>v;
    for(int i=0;i<n+m+1||tt;i++)
    {
        tt=tt+(int)(a[i].x+0.5);
        v.push_back(tt%10);
        tt/=10;
    }
    while(v.size()&&v.back()==0)    v.pop_back();
    for(int i=v.size()-1;i>=0;i--)
        cout<<v[i];
    return 0;
}