洛谷4173（fft带通配符字符串匹配）

题目链接：https://www.luogu.org/problemnew/show/P4173

思路：就是FFT的匹配，具体可以去洛谷看题解（PS：第二个大佬的题解写的挺好的emmm）

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <bitset>
#include <cmath>
#include <cctype>
#include <unordered_map>
#include <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const ll inff = 0x3f3f3f3f3f3f3f3f;
#define FOR(i,a,b) for(int i(a);i<=(b);++i)
#define FOL(i,a,b) for(int i(a);i>=(b);--i)
#define REW(a,b) memset(a,b,sizeof(a))
#define inf int(0x3f3f3f3f)
#define si(a) scanf("%d",&a)
#define sl(a) scanf("%I64d",&a)
#define sd(a) scanf("%lf",&a)
#define ss(a) scanf("%s",a)
#define mod ll(998244353)
#define pb push_back
#define eps 1e-7
#define lc d<<1
#define rc d<<1|1
#define Pll pair<ll,ll>
#define P pair<int,int>
#define pi acos(-1)
int n,m,e[2000008],f[2000008];
char c[2000008],d[2000008];
struct CP{
    double x,y;
    CP(){} CP(double a,double b):x(a),y(b){}
    CP operator+(const CP&r) const{return CP(x+r.x,y+r.y);}
    CP operator-(const CP&r) const{return CP(x-r.x,y-r.y);}
    CP operator*(const CP&r) const{return CP(x*r.x-y*r.y,x*r.y+y*r.x);}
}a[2000008],b[2000008],t,ans[2000008];
inline void Swap(CP&a,CP&b) {t=a;a=b;b=t;}
inline void fft(CP*a,int f)
{
    int i,j,k;
    for(i=j=0;i<n;i++)
    {
        if(i>j) Swap(a[i],a[j]);
        for(k=n>>1;(j^=k)<k;k>>=1);
    }
    for(int i=1;i<n;i<<=1)
    {
        CP wn(cos(pi/i),f*sin(pi/i));
        for(int j=0;j<n;j+=i<<1)
        {
            CP w(1,0);
            for(int k=0;k<i;k++,w=w*wn)
            {
                CP x=a[j+k],y=w*a[i+j+k];
                a[j+k]=x+y;a[i+j+k]=x-y;
            }
        }
    }
    if(f==-1) FOR(i,0,n) a[i].x/=n;
}
int main()
{
    cin.tie(0);
    cout.tie(0);
    cin>>m>>n;
    ss(d),ss(c);
    int qw=n,zz=m,k=0;
    reverse(d,d+m);
    FOR(i,0,n-1) e[i]=(c[i]=='*')?0:(c[i]-'a'+1);
    FOR(i,0,m-1) f[i]=(d[i]=='*')?0:(d[i]-'a'+1);
    for(m=n+m,n=1;n<=m;n<<=1);
    FOR(i,0,n-1) a[i]=CP(e[i]*e[i]*e[i],0),b[i]=CP(f[i],0);
    fft(a,1),fft(b,1);
    FOR(i,0,n-1) ans[i]=ans[i]+a[i]*b[i];
    FOR(i,0,n-1) a[i]=CP(f[i]*f[i]*f[i],0),b[i]=CP(e[i],0);
    fft(a,1),fft(b,1);
    FOR(i,0,n-1) ans[i]=ans[i]+a[i]*b[i];
    FOR(i,0,n-1) a[i]=CP(e[i]*e[i],0),b[i]=CP(f[i]*f[i],0);
    fft(a,1),fft(b,1);
    FOR(i,0,n-1) ans[i]=ans[i]-CP(2,0)*a[i]*b[i];
    fft(ans,-1);
    for(int i=zz-1;i<qw;i++) if(fabs(ans[i].x)<=0.1) k++;
    cout<<k<<endl;
    for(int i=zz-1;i<qw;i++) if(fabs(ans[i].x)<=0.1) printf("%d ",i-zz+2);
    return 0;
}