大数A*B(洛谷的,常数小一些)

大数A*B(FFT)

题目链接

递归实现

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
const int N = 4e6+10;
const double pi = acos(-1.0);
using namespace std;
typedef long long ll;
struct complex{
    double x,y;
    complex(double xx=0,double yy=0){x=xx,y=yy;}
    complex operator +(const complex &a){return complex(x+a.x,y+a.y);}
    complex operator -(const complex &a){return complex(x-a.x,y-a.y);}
    complex operator *(const complex &a){return complex(x*a.x-y*a.y,x*a.y+y*a.x);}
};

void fft(complex* a,int n,int type){
    if(n==1) return;
    complex a1[n+1],a2[n+1];
    for(int i=0;i<=n;i+=2)
        a1[i>>1]=a[i],a2[i>>1]=a[i+1];
    fft(a1,n>>1,type);
    fft(a2,n>>1,type);
    complex Wn=complex(cos(2.0*pi/n),type*sin(2.0*pi/n)),w = complex(1,0);
    for(int i=0;i<(n>>1);i++,w=w*Wn){
        a[i]=a1[i]+w*a2[i];
        a[i+(n>>1)]=a1[i]-w*a2[i];
    }
}
complex a[N],b[N];
int main(){
    int n,m,x;
    scanf("%d",&n);
    scanf("%d",&m);
    for(int i=0;i<=n;i++) scanf("%d",&x),a[i].x = x,a[i].y=0;
    for(int i=0;i<=m;i++) scanf("%d",&x),b[i].x = x,b[i].y=0;
    int len = 1;
    while(len <= (n+m)) len *= 2;
    fft(a,len,1);
    fft(b,len,1);
    for(int i=0;i<=len;i++){
        a[i] = a[i] * b[i];
    }
    fft(a,len,-1);
    for(int i=0;i<=(n+m);i++) printf("%0.lf ",fabs(a[i].x/(len)));
    return 0;
}
//8 64 90 50 113 160 105 64 61 65 20
//8 64 90 50 113 160 105 64 61 65 20

迭代实现

#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <iostream>
const int N = 4e6+10;
const double pi = acos(-1.0);
using namespace std;
typedef long long ll;
struct complex{
    double x,y;
    complex(double xx=0,double yy=0){x=xx,y=yy;}
    complex operator +(const complex &a){return complex(x+a.x,y+a.y);}
    complex operator -(const complex &a){return complex(x-a.x,y-a.y);}
    complex operator *(const complex &a){return complex(x*a.x-y*a.y,x*a.y+y*a.x);}
};
int r[N],ans[N];
char s1[N],s2[N];
complex a[N],b[N];
void fft(complex* a,int n,int type){
    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){
        complex Wn=complex(cos(pi/mid),type*sin(pi/mid));
        for(int R=mid<<1,j=0;j<n;j+=R){
            complex w(1,0);
            for(int k=0;k<mid;k++,w=w*Wn){
                complex x=a[j+k],y=w*a[j+mid+k];
                a[j+k]=x+y;
                a[j+mid+k]=x-y;
            }
        }
    }
}
int main(){
    int n,m;
    scanf("%s%s",s1,s2);
    n = strlen(s1);
    m = strlen(s2);
    for(int i=n-1;i>=0;i--){
        a[n-i-1].x = s1[i] - '0';
        a[n-i-1].y = 0;
    }
    for(int i=m-1;i>=0;i--){
        b[m-i-1].x = s2[i] - '0';
        b[m-i-1].y = 0;
    }
    int len = 1,l=0;
    while(len < (n+m)) len *= 2,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++){
        a[i] = a[i] * b[i];
    }
    fft(a,len,-1);
    for(int i=0;i<=len;i++){
        ans[i] += (int)(a[i].x/len+0.5);
        ans[i+1]+=ans[i]/10;
        ans[i]%=10;
    }
    while(!ans[len]&&len>0)len--;
    for(int i=len;i>=0;i--)printf("%d",ans[i]);

    return 0;
}