题目链接：洛谷、BZOJ2179

//将乘数拆成 a0*10^n + a1*10^(n-1) + ... + a_n-1的形式

//可以发现多项式乘法就模拟了竖式乘法 所以用FFT即可 注意处理进位

//n位*n位最多就只有2n位了

//putchar的速度。。还是快的

#include <cmath>

#include <cstdio>

#include <cctype>

#include <algorithm>

#define gc() getchar()

const int N=6e4+5;

const double PI=acos(-1);

int n,ans[N<<1];

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);}

}A[N*3],B[N*3];

inline void read(Complex *a)

{

	int cnt=0;register char c=gc();

	for(;!isdigit(c);c=gc());

	for(;isdigit(c);a[cnt++].x=(double)(c-'0'),c=gc());

}

void Fast_Fourier_Transform(Complex *a,int lim,int opt)

{

	for(int j=0,i=0; i<lim; ++i)

	{

		if(i>j) std::swap(a[i],a[j]);

		for(int l=lim>>1; (j^=l)<l; l>>=1);

	}

	for(int i=2; i<=lim; i<<=1)

	{

		int mid=i>>1;

		Complex Wn(cos(2.0*PI/i),opt*sin(2.0*PI/i)),t;

		for(int j=0; j<lim; j+=i)

		{

			Complex w(1,0);

			for(int k=0; k<mid; ++k,w=w*Wn)

				a[j+mid+k]=a[j+k]-(t=w*a[j+mid+k]),

				a[j+k]=a[j+k]+t;

		}

	}

}

int main()

{

	scanf("%d",&n);

	read(A), read(B);

	std::reverse(A,A+n), std::reverse(B,B+n);

	int lim=1;

	while(lim <= n<<1) lim<<=1;

	Fast_Fourier_Transform(A,lim,1);

	Fast_Fourier_Transform(B,lim,1);

	for(int i=0; i<=lim; ++i) A[i]=A[i]*B[i];

	Fast_Fourier_Transform(A,lim,-1);

	int t=n<<1;

	for(int i=0; i<=t; ++i) ans[i]=(int)(A[i].x/lim+0.5);

	for(int i=0; i<t; ++i) ans[i+1]+=ans[i]/10, ans[i]%=10;

	while(!ans[t] && t) --t;

	while(~t) putchar(ans[t--]+'0');

	return 0;

}