sol
这种字符串匹配的问题显然可以把一个串\(reverse\)过来然后用\(FFT\)做吧。
对每种字母分开考虑，设两个多项式\(A(x),B(x)\)，其中
\[A(x)=\sum_{i=0}^{n-1}[区间[i-k,i+k]内存在该种字符]x^i\]
\[B(x)=\sum_{i=0}^{n-1}[t[i]为该种字符]x^i\]
然后两个多项式卷一下就是这种字符在第\(i\)个位置上的匹配吧。
最后对每次的系数求和，若恰好等于\(|T|\)则说明匹配成功。
复杂度\(O(4n\log n)\)
code
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
const int _ = 8e5+5;
const double Pi = acos(-1);
struct Complex{
double rl,im;
Complex(){rl=im=0;}
Complex(double a,double b){rl=a,im=b;}
Complex operator + (Complex b)
{return Complex(rl+b.rl,im+b.im);}
Complex operator - (Complex b)
{return Complex(rl-b.rl,im-b.im);}
Complex operator * (Complex b)
{return Complex(rl*b.rl-im*b.im,rl*b.im+im*b.rl);}
}w[_],a[_],b[_];
int N,n,m,k,rev[_],l,tmp[_],ans[_],Ans;
char s[_],t[_],fg[4]={'A','G','C','T'};
void FFT(Complex *P,int opt)
{
for (int i=0;i<N;++i) if (i<rev[i]) swap(P[i],P[rev[i]]);
for (int i=1;i<N;i<<=1)
for (int p=i<<1,j=0;j<N;j+=p)
for (int k=0;k<i;++k)
{
Complex W=w[N/i*k];W.im*=opt;
Complex X=P[j+k],Y=W*P[j+k+i];
P[j+k]=X+Y;P[j+k+i]=X-Y;
}
if (opt==-1) for (int i=0;i<N;++i) P[i].rl/=N;
}
int main()
{
scanf("%d%d%d",&n,&m,&k);scanf("%s%s",s,t);reverse(t,t+m);
for (N=1;N<n+m;N<<=1) ++l;--l;
for (int i=0;i<N;++i) rev[i]=(rev[i>>1]>>1)|((i&1)<<l);
for (int i=0;i<N;++i) w[i]=Complex(cos(Pi*i/N),sin(Pi*i/N));
for (int zsy=0;zsy<4;++zsy)
{
for (int i=0;i<N;++i) a[i].rl=a[i].im=b[i].rl=b[i].im=0;
for (int i=0,pos=-1e9;i<n;++i)
{
if (s[i]==fg[zsy]) pos=i;
if (pos>=i-k) a[i].rl=1;
}
for (int i=n-1,pos=1e9;~i;--i)
{
if (s[i]==fg[zsy]) pos=i;
if (pos<=i+k) a[i].rl=1;
}
for (int i=0;i<m;++i) if (t[i]==fg[zsy]) b[i].rl=1;
FFT(a,1);FFT(b,1);
for (int i=0;i<N;++i) a[i]=a[i]*b[i];
FFT(a,-1);
for (int i=0;i<n;++i) ans[i]+=(int)(a[i].rl+0.5);
}
for (int i=0;i<n;++i) if (ans[i]==m) ++Ans;
printf("%d\n",Ans);return 0;
}