Luo Gu [P4173] incomplete string

Title effect: Given a text string and a template string, a string containing the wildcard, seeking the number of text strings may exactly match the position of the text string.

Solution: use convolution to solve the string matching problem.
Numeric string matching the wildcard represented as \ [\ sum \ limits_ {i = 0} ^ {m - 1} (a [i] - b [i + k]) ^ 2 a [i] b [i + k] = 0 \] . Three convolution can be calculated directly after the start.
Note that: not all of the a [i] b [i + k] are circular convolution multiplication depends on the need to form a ring.

code show as below

#include <bits/stdc++.h>
using namespace std;
typedef complex<double> cp;
const double eps = 1e-8;
const double pi = acos(-1);

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> m >> n;
    string s, t;
    cin >> t >> s;
    int tot = 1, bit = 0;
    while (tot <= n + m) {
        tot <<= 1;
        ++bit;
    }
    vector<int> rev(tot);
    for (int i = 0; i < tot; i++) {
        rev[i] = rev[i >> 1] >> 1 | (i & 1) << bit - 1;
    }
    vector<cp> a1(tot), a2(tot), a3(tot), b1(tot), b2(tot), b3(tot);
    auto work = [](char ch) {
        if (ch == '*') {
            return 0;
        } else {
            return ch - 'a' + 1;
        }
    };
    for (int i = 0; i < m; i++) {
        int x = work(t[i]);
        a1[m - 1 - i] = x;
        a2[m - 1 - i] = x * x;
        a3[m - 1 - i] = x * x * x;
    }
    for (int i = 0; i < n; i++) {
        int x = work(s[i]);
        b1[i] = x;
        b2[i] = x * x;
        b3[i] = x * x * x;
    }
    auto fft = [=](vector<cp> &v, int opt) {
        for (int i = 0; i < tot; i++) {
            if (i < rev[i]) {
                swap(v[i], v[rev[i]]);
            }
        }
        for (int mid = 1; mid < tot; mid <<= 1) {
            cp wn(cos(pi / mid), opt * sin(pi / mid));
            for (int j = 0; j < tot; j += mid << 1) {
                cp w(1, 0);
                for (int k = 0; k < mid; k++) {
                    cp x = v[j + k], y = w * v[j + mid + k];
                    v[j + k] = x + y, v[j + mid + k] = x - y;
                    w *= wn;
                }
            }
        }
        if (opt == -1) {
            for (int i = 0; i < tot; i++) {
                v[i].real(round(v[i].real() / tot));
            }
        }
    };
    auto calc = [=](vector<cp> &a, vector<cp> &b) {
        fft(a, 1), fft(b, 1);
        for (int i = 0; i < tot; i++) {
            a[i] *= b[i];
        }
        fft(a, -1);
    };
    calc(a3, b1), calc(a2, b2), calc(a1, b3);
    int ans = 0;
    vector<int> pos;
    for (int k = 0; k < n - m + 1; k++) {
        double ret = a3[m + k - 1].real() - 2 * a2[m + k - 1].real() + a1[m + k - 1].real();
        if (fabs(ret) < eps) {
            ++ans;
            pos.push_back(k + 1);
        }
    }
    cout << ans << endl;
    for (auto p : pos) {
        cout << p << " ";
    }
    return 0;
}