标签:print cst style void algo 字符 ret efi else
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); }
标签:print cst style void algo 字符 ret efi else
原文地址:http://www.cnblogs.com/SiriusRen/p/6532747.html