【BZOJ】【2179】FFT快速傅里叶

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 ;
 }