FFT
做的第二道用到FFT的……好吧其实还是模板题-_-b
百度上说好像分治也能做……不过像FFT这种敲模板的还是省事=。=
/**************************************************************
Problem: 2179
User: Tunix
Language: C++
Result: Accepted
Time:1236 ms
Memory:9184 kb
****************************************************************/ //BZOJ 2179
#include<cmath>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#define rep(i,n) for(int i=0;i<n;++i)
#define F(i,j,n) for(int i=j;i<=n;++i)
#define D(i,j,n) for(int i=j;i>=n;--i)
using namespace std;
void read(int &v){
v=; int sign=; char ch=getchar();
while(ch<''||ch>''){ if (ch=='-') sign=-; ch=getchar();}
while(ch>=''&&ch<=''){ v=v*+ch-''; ch=getchar();}
v*=sign;
}
/******************tamplate*********************/
#define debug
const int N=;
const double pi=acos(-1.0);
struct comp{
double r,i;
comp(double _r=0.0,double _i=0.0):r(_r),i(_i){}
// comp(){}
comp operator+(const comp &b)const{return comp(r+b.r,i+b.i);}
comp operator-(const comp &b)const{return comp(r-b.r,i-b.i);}
comp operator*(const comp &b)const{return comp(r*b.r-i*b.i,r*b.i+i*b.r);}
}a[N],b[N],c[N]; void FFT(comp *a,int n,int type){
for(int i=,j=;i<n-;++i){//只需改变1～n-2,0和n-1两个位置不变
for(int s=n;j^=s>>=,~j&s;);
if (i<j) swap(a[i],a[j]);
}
for(int m=;m<n;m<<=){
double u=pi/m*type; comp wm(cos(u),sin(u));
for(int i=;i<n;i+=(m<<)){
comp w(,);
rep(j,m){
comp &A=a[i+j+m],&B=a[i+j],t=w*A;
A=B-t; B=B+t; w=w*wm;
}
}
}
if (type==-) rep(i,n) a[i].r/=n;
}
char s1[N],s2[N];
int ans[N];
int main(){
int n,k;
read(n);
scanf("%s%s",s1,s2);
rep(i,n){
a[i].r=s1[n-i-]-'';
b[i].r=s2[n-i-]-'';
}
for(k=;k<=n*;k<<=);
FFT(a,k,); FFT(b,k,);
F(i,,k) c[i]=a[i]*b[i];
FFT(c,k,-);
F(i,,n*)
ans[i]=c[i].r+0.4;
int temp=;
F(i,,n*){
if (ans[i]) temp=i;
ans[i+]+=ans[i]/;
ans[i]%=;
}
D(i,temp,) printf("%d",ans[i]);
return ;
}