【題目連結】

【前置技能】

• FFT/NTT

【題解】

• 字串中出現了萬用字元，一般的字串演算法就失去效果了。
• 先忽略萬用字元的問題。令每個位置Ak=∑i=0LenT−1(Si+k−LenT+1−Ti)$A_k={\displaystyle \sum _{i=0}^{LenT-1}(S_{i+k-LenT+1}-T_i)}$，若Ak=0$A_k=0$，則說明T$T$與S[k−LenT+1,k]$S[k-LenT+1,k]$匹配。但發現這可能會出錯，相減項有可能會正負抵消，如字串ab和ba就會被認為是匹配的。得到一種解決方法：將求和中的減法變成相減之後平方就可以避免抵消的問題了。
• 把模式串翻轉一下，把平方展開，發現是卷積的形式。那麼萬用字元的問題也很好想明白了，在式子上再乘上
  Ti∗Si$T_i\ast S_i$，其中萬用字元的值為零即可。
• 最後的式子：Ak=∑i=0k(Tk−i∗S3i)−2∗∑i=0k(T2k−i∗S2i)+∑i=0k(T3k−i∗Si)$A_k={\displaystyle \sum _{i=0}^k(T_{k-i}\ast S_i^3)-2\ast {\displaystyle \sum _{i=0}^k(T_{k-i}^2\ast S_i^2)+{\displaystyle \sum _{i=0}^k(T_{k-i}^3\ast S_i)}}}$。
• 如果FFT精度處理得很不好的話，有可能會過不去。
• 時間複雜度O(NlogN)$O(NlogN)$

【程式碼】

#include<bits/stdc++.h>
#define INF 0x3f3f3f3f
#define LL  long long
#define MAXS    300010
#define MAXN    1048577
 

#define eps 1
using namespace std;
const double pi = acos(-1);
int n, m, s[MAXN], t[MAXN], cnt; 
char S[MAXS], T[MAXS];
int N, LOG, rev[MAXN];
struct dot{double x, y;}A[MAXN], B[MAXN], C[MAXN], tmp[MAXN], tnp[MAXN];
dot operator + (dot a, dot b) {return (dot){a.x + b.x, a.y + b.y};}
dot operator - (dot a, dot b) {return
 
 (dot){a.x - b.x, a.y - b.y};}
dot operator * (dot a, dot b) {return (dot){a.x * b.x - a.y * b.y, a.x * b.y + a.y * b.x};}
dot operator / (dot a, double k) {return (dot){a.x / k, a.y / k};}

template <typename T> void chkmin(T &x, T y){x = min(x, y);}
template <typename T> void chkmax(T &x, T y){x = max(x, y);}
template <typename T> void read(T &x){
    x = 0; int f = 1; char ch = getchar();
    while (!isdigit(ch)) {if (ch == '-') f = -1; ch = getchar();}
    while (isdigit(ch)) {x = x * 10 + ch - '0'; ch = getchar();}
    x *= f;
}

void fft_init(){
    N = 1, LOG = 0;
    while (N < n + m) N <<= 1, ++LOG;
    for (int i = 1; i < N; ++i)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (LOG - 1));
}

void fft(dot *a, int opt){
    for (int i = 0; i < N; ++i)
        if (rev[i] < i) swap(a[rev[i]], a[i]);
    for (int len = 2; len <= N; len <<= 1){
        dot delta = (dot){cos(pi * 2 / len), sin(pi * 2 / len * opt)};
        for (int i = 0; i < N; i += len){
            dot cur = (dot){1, 0};
            for (int j = i, t = i + len / 2; t < i + len; ++j, ++t){
                dot tmp = a[j], tnp = a[t] * cur;
                a[j] = tmp + tnp, a[t] = tmp - tnp;
                cur = cur * delta;
            }
        }
    }
    if (opt == -1){
        for (int i = 0; i < N; ++i)
            a[i] = a[i] / (1.0 * N);
    }
}

int id(char ch){
    if (ch == '*') return 0;
    else return (ch - 'a' + 1);
}

bool sam(double a, double b){
    return (a >= b - eps && a <= b + eps);
}

int main(){
    read(n), read(m);
    scanf("%s", S);
    scanf("%s", T);
    for (int i = 0; i < n; ++i)
        s[i] = id(S[n - i - 1]);
    for (int i = 0; i < m; ++i)
        t[i] = id(T[i]);
    fft_init();
    for (int i = 0; i < N; ++i)
        tmp[i].x = s[i] * s[i] * s[i], tnp[i].x = t[i], tmp[i].y = 0, tnp[i].y = 0;
    fft(tmp, 1), fft(tnp, 1);
    for (int i = 0; i < N; ++i)
        A[i] = tmp[i] * tnp[i];
    for (int i = 0; i < N; ++i)
        tmp[i].x = s[i] * s[i], tnp[i].x = t[i] * t[i], tmp[i].y = 0, tnp[i].y = 0;
    fft(tmp, 1), fft(tnp, 1);
    for (int i = 0; i < N; ++i)
        B[i] = tmp[i] * tnp[i];
    for (int i = 0; i < N; ++i)
        tmp[i].x = s[i], tnp[i].x = t[i] * t[i] * t[i], tmp[i].y = 0, tnp[i].y = 0;
    fft(tmp, 1), fft(tnp, 1);
    for (int i = 0; i < N; ++i)
        C[i] = tmp[i] * tnp[i];
    fft(A, -1), fft(B, -1), fft(C, -1);
    for (int i = 0; i < N; ++i)
        A[i] = A[i] - (B[i] / 0.5) + C[i];
    for (int i = n - 1; i < m; ++i)
        if (sam(A[i].x, 0)) ++cnt;
    printf("%d\n", cnt);
    int fla = 0;
    for (int i = n - 1; i < m; ++i)
        if (sam(A[i].x, 0)) {
            if (fla) printf(" ");
            if (!fla) fla = 1;
            printf("%d", i - (n - 2));
        }
    if (cnt != 0) printf("\n");
    return 0;
}