BZOJ4259 残缺的字符串 FFT

首先是csy博客的题解，但csy的并不十分优秀。

事实上，我们可以求出f[i] = sigma(A[j] * inv[B[i - j]])，当f[i] <= n时，匹配，否则不匹配。这样就只需要做一次FFT或者NTT即可。

#include<cstdio>
#include<cmath>
#include<algorithm>
#define N 1048576
using namespace std;
char sa[N],sb[N];int n,m,i,j,k,a[N],b[N],ans,q[N];
struct comp{
  double r,i;comp(double _r=0,double _i=0){r=_r;i=_i;}
  comp operator+(const comp&x){return comp(r+x.r,i+x.i);}
  comp operator-(const comp&x){return comp(r-x.r,i-x.i);}
  comp operator*(const comp&x){return comp(r*x.r-i*x.i,r*x.i+i*x.r);}
}A[N],B[N],C[N];
const double pi=acos(-1.0);
void FFT(comp*a,int n,int t){
  for(int i=1,j=0;i<n-1;i++){
    for(int s=n;j^=s>>=1,~j&s;);
    if(i<j)swap(a[i],a[j]);
  }
  for(int d=0;(1<<d)<n;d++){
    int m=1<<d,m2=m<<1;
    double o=pi/m*t;comp _w(cos(o),sin(o));
    for(int i=0;i<n;i+=m2){
      comp w(1,0);
      for(int j=0;j<m;j++){
        comp &A=a[i+j+m],&B=a[i+j],t=w*A;
        A=B-t;B=B+t;w=w*_w;
      }
    }
  }
  if(t==-1)for(int i=0;i<n;i++)a[i].r/=n;
}
int main(){
  scanf("%d%d%s%s",&m,&n,sa,sb);
  for(i=0,j=m-1;i<j;i++,j--)swap(sa[i],sa[j]);
  for(i=0;i<m;i++)if(sa[i]!='*')a[i]=sa[i]-'a'+1;
  for(i=0;i<n;i++)if(sb[i]!='*')b[i]=sb[i]-'a'+1;
  for(k=1;k<n+m;k<<=1);
  for(i=0;i<k;i++)A[i]=comp(a[i]*a[i]*a[i],0),B[i]=comp(b[i],0);
  for(FFT(A,k,1),FFT(B,k,1),i=0;i<k;i++)C[i]=C[i]+A[i]*B[i];
  for(i=0;i<k;i++)A[i]=comp(a[i],0),B[i]=comp(b[i]*b[i]*b[i],0);
  for(FFT(A,k,1),FFT(B,k,1),i=0;i<k;i++)C[i]=C[i]+A[i]*B[i];
  for(i=0;i<k;i++)A[i]=comp(a[i]*a[i],0),B[i]=comp(b[i]*b[i],0);
  for(FFT(A,k,1),FFT(B,k,1),i=0;i<k;i++)C[i]=C[i]-A[i]*B[i]*comp(2,0);
  FFT(C,k,-1);
  for(i=m-1;i<n;i++)if(C[i].r<0.5)q[++ans]=i-m+2;
  for(printf("%d
",ans),i=1;i<ans;i++)printf("%d ",q[i]);
  if(ans)printf("%d",q[ans]);
  return 0;
}

 NTT版本代码

#include<bits/stdc++.h>
using namespace std;

const int MOD = 998244353;
const int N = 1048576;
const int M = 998244353;

int WM[N + 2], IWM[N + 2];
vector<int> IW[N + 2], W[N + 2];

int Pw(int a, int b, int p)
{
    int v = 1;
    for(; b; b >>= 1, a = 1ll * a * a % p)
        if (b & 1)
            v = 1ll * v * a % p;
    return v;
}

void ycl()
{
    for (int m = 2; m <= N; m <<= 1)
    {
        WM[m] = Pw(3, (M - 1) / m, M), IWM[m] = Pw(3, (M - 1) / m * (m - 1), M);
        int o = 1;
        W[m].push_back(o);
        for (int i = 1; i < m; ++i)
            o = 1ll * o * WM[m] % M, W[m].push_back(o);
        o = 1;
        IW[m].push_back(o);
        for (int i = 1; i < m; ++i)
            o = 1ll * o * IWM[m] % M, IW[m].push_back(o);
    }
}

void NTT(int *a, int n, int f = 1)
{
    int i, j, k, m, w, u, v;
    for (i = n >> 1, j = 1; j < n - 1; ++j)
    {
        if (i > j)
            swap(a[i], a[j]);
        for (k = n >> 1; k <= i; k >>= 1)
            i ^= k;
        i ^= k;
    }
    for (m = 2; m <= n; m <<= 1)
        for (i = 0; i < n; i += m)
            for (j = i; j < i + (m >> 1); ++j)
                if (a[j] || a[j + (m >> 1)])
                {
                    u = a[j];
                    v = 1ll * (f == 1 ? W[m][j - i] : IW[m][j - i]) * a[j + (m >> 1)] % M;
                    if ((a[j] = u + v) >= M)
                        a[j] -= M;
                    if ((a[j + (m >> 1)] = u - v) < 0)
                        a[j + (m >> 1)] += M;
                }
    if (f == -1)
        for (w = Pw(n, M - 2, M), i = 0; i < n; ++i)
            a[i] = 1ll * a[i] * w % M;
}

char sa[N], sb[N], st[N];
int a[5][N], b[5][N], c[N], q[N];

int main()
{
    ycl();
    int m, n;
    scanf("%d%d", &m, &n);
    scanf("%s%s", sa, sb);
    for (int i = 0, j = m - 1; i < j; ++i, --j)
        swap(sa[i],sa[j]);
    for (int i = 0; i < m; ++i)
        a[1][i] = sa[i] - 'a' + 1;
    for (int i = 0; i < n; ++i)
        b[1][i] = sb[i] - 'a' + 1;
    for (int i = 0; i < m; ++i)
        if(sa[i] == '*')
            a[1][i] = 0;
    for (int i = 0; i < n; ++i)
        if(sb[i] == '*')
            b[1][i] = 0;
    int K;
    for (K = 1; K < n + m; K <<= 1);
    for (int i = 2; i <= 3; ++i)
    {
        for (int j = 0; j < m; ++j)
            a[i][j] = a[i - 1][j] * a[1][j];
        for (int j = 0; j < n; ++j)
            b[i][j] = b[i - 1][j] * b[1][j];
    }
    for (int i = 1; i < 4; ++ i)
    {
        NTT(a[i], K);
        NTT(b[i], K);
    }
    for (int i = 0; i < K; ++ i)
    {
        c[i] = (c[i] + 1LL * a[3][i] * b[1][i] % MOD) % MOD;
        c[i] = (c[i] + MOD - 2LL * a[2][i] * b[2][i] % MOD) % MOD;
        c[i] = (c[i] + 1LL * a[1][i] * b[3][i] % MOD) % MOD;
    }
    NTT(c, K, -1);
    
    int ans(0);
    for (int i = m - 1; i < n; ++i)
        if (!c[i])
            q[++ans] = i - m + 2;
    printf("%d
", ans);
    for (int i = 1; i < ans; ++i)
        printf("%d ", q[i]);
    if (ans)
        printf("%d
", q[ans]);

    return 0;
}