BZOJ 4259 FFT

思路:

为什么好多字符串的题都可以用FFT啊....

我们其实是要判断$\Sigma (a[i]-b[i])^2*a[i]*b[i]==0$

那就把a串翻转过来 把 上式展开

大力做几遍FFT就好啦~

//By SiriusRen
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define double long double
const double pi=acos(-1);
const int N=666666;
int nn,mm,n,L,R[N],all;
struct Complex{
    double x,y;Complex(){}
    Complex(double X,double Y){x=X,y=Y;}
}A[N],B[N],ans[N];
Complex operator+(Complex a,Complex b){return Complex(a.x+b.x,a.y+b.y);}
Complex operator-(Complex a,Complex b){return Complex(a.x-b.x,a.y-b.y);}
Complex operator*(Complex a,Complex b){return Complex(a.x*b.x-a.y*b.y,a.x*b.y+a.y*b.x);}
Complex operator/(Complex a,int b){return Complex(a.x/b,a.y/b);}
void FFT(Complex *E,int f){
    for(int i=0;i<n;i++)if(i<R[i])swap(E[i],E[R[i]]);
    for(int l=1;l<n;l<<=1){
        Complex wn=Complex(cos(pi/l),f*sin(pi/l));
        for(int j=0;j<n;j+=(l<<1)){
            Complex w=Complex(1,0);
            for(int k=0;k<l;k++,w=w*wn){
                Complex x=E[j+k],y=w*E[j+k+l];
                E[j+k]=x+y,E[j+k+l]=x-y;
            }
        }
    }
    if(!~f)for(int i=0;i<n;i++)E[i]=E[i]/n;
}
char a[N],b[N];
int main(){
    scanf("%d%d%s%s",&mm,&nn,a,b);
    for(n=1;n<=mm+nn;n<<=1)L++;
    for(int i=0;i<n;i++)R[i]=(R[i>>1]>>1)|((i&1)<<(L-1));
    reverse(a,a+mm);
    for(int i=0;i<mm;i++)if(a[i]=='*')a[i]=0;else a[i]=a[i]-'a'+1;
    for(int i=0;i<nn;i++)if(b[i]=='*')b[i]=0;else b[i]=b[i]-'a'+1;
    for(int i=0;i<mm;i++)A[i].x=a[i];
    for(int i=0;i<nn;i++)B[i].x=b[i]*b[i]*b[i];
    FFT(A,1),FFT(B,1);
    for(int i=0;i<n;i++)ans[i]=A[i]*B[i];
    for(int i=0;i<n;i++)A[i].x=A[i].y=B[i].x=B[i].y=0;
    for(int i=0;i<mm;i++)A[i].x=a[i]*a[i]*a[i];
    for(int i=0;i<nn;i++)B[i].x=b[i];
    FFT(A,1),FFT(B,1);
    for(int i=0;i<n;i++)ans[i]=ans[i]+A[i]*B[i];
    for(int i=0;i<n;i++)A[i].x=A[i].y=B[i].x=B[i].y=0;
    for(int i=0;i<mm;i++)A[i].x=a[i]*a[i];
    for(int i=0;i<nn;i++)B[i].x=b[i]*b[i];
    FFT(A,1),FFT(B,1);
    for(int i=0;i<n;i++)ans[i]=ans[i]-A[i]*B[i]-A[i]*B[i];
    FFT(ans,-1);
    for(int i=mm-1;i<nn;i++)if(fabs(ans[i].x)<1)all++;
    printf("%d\n",all);
    for(int i=mm-1;i<nn;i++)if(fabs(ans[i].x)<1)printf("%d ",i-mm+2);
}

 

posted @ 2017-03-10 19:19  SiriusRen  阅读(293)  评论(0编辑  收藏  举报