题意:给你两个字符串,问你第一个在第二个中出现过多少次,并输出位置,匹配时是模糊匹配*可和任意一个字符匹配

题解:fft加速字符串匹配;

假设上面的串是s,s长度为m,下面的串是p,p长度为n,先考虑没有*的情况那么\(\sum_{j=1}^m(s_{i+j}-p_j)^2=0\)就表示能够从i开始匹配,现在考虑有*的情况,我们只需要让有*的和任意字符匹配即可,那么把公式变成\(\sum_{j=1}^m(s_{i+j}-p_j)^2*s_{i+j}*p_j)=0\),但是fft正向匹配太慢了,我们把方向变一下,\(\sum_{j=1}^m(s_{i+j}-p_{n-j})^2*s_{i+j}*p_{n-j}=0\),把式子分解一下得\(\sum_{j=1}^m(s_{i+j}^3*p_{n-j}+s_{i+j}*p_{n-j}^3-2*s_{i+j}^2*p_{n-j}^2)=0\),这样把这6个多项式跑一边fft,就得到了答案,

//#pragma comment(linker, "/stack:200000000")

//#pragma GCC optimize("Ofast,no-stack-protector")

//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

//#pragma GCC optimize("unroll-loops")

#include<bits/stdc++.h>

#define fi first

#define se second

#define db double

#define mp make_pair

#define pb push_back

#define pi acos(-1.0)

#define ll long long

#define vi vector<int>

#define mod 1000000007

#define ld long double

#define C 0.5772156649

#define ls l,m,rt<<1

#define rs m+1,r,rt<<1|1

#define pll pair<ll,ll>

#define pil pair<int,ll>

#define pli pair<ll,int>

#define pii pair<int,int>

//#define cd complex<double>

#define ull unsigned long long

#define base 1000000000000000000

#define Max(a,b) ((a)>(b)?(a):(b))

#define Min(a,b) ((a)<(b)?(a):(b))

#define fio ios::sync_with_stdio(false);cin.tie(0)

template<typename T>

inline T const& MAX(T const &a,T const &b){return a>b?a:b;}

template<typename T>

inline T const& MIN(T const &a,T const &b){return a<b?a:b;}

inline void add(ll &a,ll b){a+=b;if(a>=mod)a-=mod;}

inline void sub(ll &a,ll b){a-=b;if(a<0)a+=mod;}

inline ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}

inline ll qp(ll a,ll b){ll ans=1;while(b){if(b&1)ans=ans*a%mod;a=a*a%mod,b>>=1;}return ans;}

inline ll qp(ll a,ll b,ll c){ll ans=1;while(b){if(b&1)ans=ans*a%c;a=a*a%c,b>>=1;}return ans;}

using namespace std;

const double eps=1e-8;

const ll INF=0x3f3f3f3f3f3f3f3f;

const int N=300000+10,maxn=50000+10,inf=0x3f3f3f3f;

struct cd{

    db x,y;

    cd(db _x=0.0,db _y=0.0):x(_x),y(_y){}

    cd operator +(const cd &b)const{

        return cd(x+b.x,y+b.y);

    }

    cd operator -(const cd &b)const{

        return cd(x-b.x,y-b.y);

    }

    cd operator *(const cd &b)const{

        return cd(x*b.x - y*b.y,x*b.y + y*b.x);

    }

    cd operator /(const db &b)const{

        return cd(x/b,y/b);

    }

}a[N<<3],b[N<<3],c[N<<3],d[N<<3],e[N<<3],f[N<<3];

int rev[N<<3];

void getrev(int bit)

{

    for(int i=0;i<(1<<bit);i++)

        rev[i]=(rev[i>>1]>>1) | ((i&1)<<(bit-1));

}

void fft(cd *a,int n,int dft)

{

    for(int i=0;i<n;i++)

        if(i<rev[i])

            swap(a[i],a[rev[i]]);

    for(int step=1;step<n;step<<=1)

    {

        cd wn(cos(dft*pi/step),sin(dft*pi/step));

        for(int j=0;j<n;j+=step<<1)

        {

            cd wnk(1,0);

            for(int k=j;k<j+step;k++)

            {

                cd x=a[k];

                cd y=wnk*a[k+step];

                a[k]=x+y;a[k+step]=x-y;

                wnk=wnk*wn;

            }

        }

    }

    if(dft==-1)for(int i=0;i<n;i++)a[i]=a[i]/n;

}

char s[N],p[N];

int main()

{

    int n,m;

    scanf("%d%d",&n,&m);

    scanf("%s%s",s+1,p+1);

    int sz=0;

    while((1<<sz)<m)sz++;

    sz++,getrev(sz);

    for(int i=0;i<=(1<<sz);i++)

        a[i]=b[i]=c[i]=d[i]=e[i]=f[i]=0;

    for(int i=1,te;i<=n;i++)

    {

        if(s[i]=='*')te=0;

        else te=s[i]-'a'+1;

        a[m-i]=te*te*te,b[m-i]=-2*te*te,c[m-i]=te;

    }

    for(int i=1,te;i<=m;i++)

    {

        if(p[i]=='*')te=0;

        else te=p[i]-'a'+1;

        d[i]=te*te*te,e[i]=te*te,f[i]=te;

    }

    fft(a,(1<<sz),1),fft(b,(1<<sz),1),fft(c,(1<<sz),1);

    fft(d,(1<<sz),1),fft(e,(1<<sz),1),fft(f,(1<<sz),1);

    for(int i=0;i<=(1<<sz);i++)

        d[i]=a[i]*f[i]+b[i]*e[i]+c[i]*d[i];

    fft(d,(1<<sz),-1);

//    for(int i=0;i<=m-n;i++)

//        printf("%d ",(int)((d[m+i].x+0.5)/(1<<sz)));

//    puts("");

    vi ans;

    for(int i=0;i<=m-n;i++)

        if((int)(d[m+i].x+0.5/(1<<sz))==0)

            ans.pb(i+1);

    printf("%d\n",ans.size());

    for(int i=0;i<ans.size();i++)

        printf("%d ",ans[i]);

    puts("");

    return 0;

}

/********************

********************/