话说FFT该不会真的只能用来做这种板子吧……
我们把两个数字的每一位都看作多项式的系数
然后这就是一个多项式乘法
上FFT就好了
然后去掉前导零
(然而连FFT的板子都背不来orz,而且空间又开小了……)
//minamoto
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
char sr[<<],z[];int C=-,Z;
inline void Ot(){fwrite(sr,,C+,stdout),C=-;}
inline void print(int x){
if(C><<)Ot();if(x<)sr[++C]=,x=-x;
while(z[++Z]=x%+,x/=);
while(sr[++C]=z[Z],--Z);
}
const int N=2e5+;const double Pi=acos(-1.0);
int r[N],l=,limit=,c[N],n;char sa[N],sb[N];
struct complex{
double x,y;
complex(double xx=,double yy=){x=xx,y=yy;}
inline complex operator +(complex b){return complex(x+b.x,y+b.y);}
inline complex operator -(complex b){return complex(x-b.x,y-b.y);}
inline complex operator *(complex b){return complex(x*b.x-y*b.y,x*b.y+y*b.x);}
}a[N],b[N];
void FFT(complex *a,int type){
for(int i=;i<limit;++i)
if(i<r[i]) swap(a[i],a[r[i]]);
for(int mid=;mid<limit;mid<<=){
complex Wn(cos(Pi/mid),type*sin(Pi/mid));
for(int R=mid<<,j=;j<limit;j+=R){
complex w(,);
for(int k=;k<mid;++k,w=w*Wn){
complex x=a[j+k],y=w*a[j+k+mid];
a[j+k]=x+y,a[j+k+mid]=x-y;
}
}
}
}
int main(){
// freopen("testdata.in","r",stdin);
scanf("%d",&n),--n;
scanf("%s%s",sa,sb);
for(int i=;i<=n;++i) a[i].x=sa[n-i]-'',b[i].x=sb[n-i]-'';
while(limit<=n*) limit<<=,++l;
for(int i=;i<=limit;++i) r[i]=(r[i>>]>>)|((i&)<<(l-));
FFT(a,),FFT(b,);
for(int i=;i<=limit;++i) a[i]=a[i]*b[i];
FFT(a,-);
for(int i=;i<=limit;++i) c[i]=(int)(a[i].x/limit+0.5);
for(int i=;i<=limit;++i)
if(c[i]>){
c[i+]+=c[i]/,c[i]%=;
if(i+>limit) ++limit;
}
for(int i=limit;i>=;--i)
if(c[i]==) --limit;
else break;
for(int i=limit;i>=;--i) print(c[i]);
Ot();
return ;
}