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Luogu P4173 残缺的字符串

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题目描述

https://www.luogu.com.cn/problem/P4173

Solution

这是一个很经典的做法

我们考虑构造一个多项式，使得如果 $S_i=T_i$，那么这个式子是 $0$，其它情况下均为一个大于 $0$ 的数字，另外还要能处理通配符

那么我们能够得到这么一个式子 $(S_i-T_i)^2\times S_i\times T_i$

然后我们考虑怎么对于 $S$ 的每个 $pre_i$ 都按照这个式子求一下

我们考虑首先将 $a$ 到 $z$ 映射到 $1$ 到 $26$，通配符为 $0$

然后把 $T$ 翻转一下，然后我们直接将 $S$ 和 $T$ 卷积起来，卷积得到的 $A_i$ 就是 $pre_i$ 的值

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#include <iostream>
#include <vector>
#include <algorithm>
#define maxn 1050000
#define ll long long
using namespace std;

const int p = 998244353;
const int G = 3;
const int Gi = 332748118;

ll pow_mod(ll x, ll n) {
    ll s = 1;
    for (; n; n >>= 1) {
        if (n & 1) s = s * x % p;
        x = x * x % p;
    }
    return s;
}

int n, m, s[maxn], t[maxn], f[maxn], A[maxn], B[maxn];

char S[maxn], T[maxn];

int a[maxn], b[maxn];
inline int add(int x, int y) { return (x += y) >= p ? x - p : x; }

int P[maxn];
void init_P(int n) {
    int l = 0; while ((1 << l) < n) ++l;
    for (int i = 0; i < n; ++i) P[i] = (P[i >> 1] >> 1) | ((i & 1) << l - 1);
}

void NTT(int *a, int n, int type) {
    static int w[maxn];
    for (int i = 0; i < n; ++i) if (i < P[i]) swap(a[i], a[P[i]]);
    for (int i = 2, m = 1; i <= n; m = i, i *= 2) {
        ll wn = pow_mod(type > 0 ? G : Gi, (p - 1) / i);
        w[0] = 1; for (int j = 1; j < m; ++j) w[j] = wn * w[j - 1] % p;
        for (int j = 0; j < n; j += i) 
            for (int k = 0; k < m; ++k) {
                ll t1 = a[j + k], t2 = 1ll * a[j + k + m] * w[k] % p; 
                a[j + k] = add(t1, t2);
                a[j + k + m] = add(t1, p - t2);
            }
    }
    if (type < 0) {
        ll inv = pow_mod(n, p - 2);
        for (int i = 0; i < n; ++i) a[i] = inv * a[i] % p; 
    }
}

void Mul(int *a, int *B, int n1, int n2) {
    int n = 1; while (n < n1 + n2 - 1) n <<= 1; init_P(n);
    for (int i = 0; i < n2; ++i) b[i] = B[i];
    fill(a + n1, a + n, 0); fill(b + n2, b + n, 0);
    NTT(a, n, 1); NTT(b, n, 1);
    for (int i = 0; i < n; ++i) a[i] = 1ll * a[i] * b[i] % p; 
    NTT(a, n, -1); 
}

void solve() {
    for (int i = 0; i < n; ++i) A[i] = s[i] * s[i] * s[i];
    for (int i = 0; i < m; ++i) B[i] = t[i];
    Mul(A, B, n, m);
    for (int i = 0; i < n; ++i) f[i] = A[i];

    for (int i = 0; i < n; ++i) A[i] = s[i];
    for (int i = 0; i < m; ++i) B[i] = t[i] * t[i] * t[i];
    Mul(A, B, n, m);
    for (int i = 0; i < n; ++i) f[i] = (f[i] + A[i]) % p;

    for (int i = 0; i < n; ++i) A[i] = s[i] * s[i];
    for (int i = 0; i < m; ++i) B[i] = t[i] * t[i];
    Mul(A, B, n, m);
    for (int i = 0; i < n; ++i) f[i] = (f[i] - 2ll * A[i]) % p;

    vector<int> ans;
    for (int i = m - 1; i < n; ++i)
        if ((f[i] + p) % p == 0) ans.push_back(i - m + 2);
    cout << ans.size() << "\n";
    for (int t : ans) cout << t << " ";
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> m >> n >> T >> S;
    for (int i = 0; i < m; ++i)
        if (T[i] == '*') t[i] = 0;
        else t[i] = T[i] - 'a' + 1;
    for (int i = 0; i < n; ++i)
        if (S[i] == '*') s[i] = 0;
        else s[i] = S[i] - 'a' + 1;
    reverse(t, t + m); solve();
    return 0;
}