CF1334G Substring Search

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字符串匹配要没$s_i-t_i=0 \ or p_{s_i}-t_i=0$

构造$\sum_{i=1}^{m} (s_i-t_{i+s-1})^2 (p_{s_i}-t_{s+i-1})^2$

瞎几把展开可以用这个网站

$v^2 x^2 - 2 v^2 x y - 2 v x^2 y + v^2 y^2 + 4 v x y^2 + x^2 y^2 - 2 v y^3 - 2 x y^3 + y^4$

然后卷积$NTT$即可.

  • 防止有傻逼卡$mod$，就让这个字母的值尽量不规律。
代码 
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#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pii pair<int, int>
#define mk make_pair
const int N = 8e5 + 10;
const int mod = 998244353;
const int P = 998244353, gi = 3;
const double pi = acos(-1.0);
int inc(int x, int y, int mo)
{
    if (x + y >= mo)
        return (x + y - mo);
    else
        return (x + y);
}

int del(int x, int y, int mo)
{
    if (x - y < 0)
        return (x - y + mo);
    else
        return (x - y);
}
int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while (c < '0' || c > '9')
    {
        if (c == '-')
            f = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        x = (x << 1) + (x << 3) + c - '0', c = getchar();
    return x * f;
}

int qpow(int a, int x, int mo)
{
    int res = 1;
    while (x)
    {
        if (x & 1)
            res = 1ll * res * a % mo;
        x >>= 1;
        a = 1ll * a * a % mo;
    }
    return res;
}
int rev[N];
void NTT(int *A, int n, int inv)
{
    for (int i = 0; i < n; i++)
        if (i < rev[i])
            swap(A[i], A[rev[i]]);
    for (int l = 1; l < n; l <<= 1)
    {
        int tt = qpow(gi, (P - 1) / (l << 1), P);
        int temp = (inv == 1 ? tt : qpow(tt, P - 2, P));
        for (int i = 0; i < n; i += (l << 1))
        {
            int omega = 1;
            for (int j = 0; j < l; j++, omega = 1ll * omega * temp % P)
            {
                int x = A[i + j], y = 1ll * omega * A[i + j + l] % P;
                A[i + j] = inc(x, y, P);
                A[i + j + l] = del(x, y, P);
            }
        }
    }
    int invv = qpow(n, P - 2, P);
    if (inv == -1)
        for (int i = 0; i < n; i++)
            A[i] = 1ll * A[i] * invv % P;
}

// void NTTX(int *a, int n, int *b, int m, int *ans)
// {
//     int ML = 1, bit = 0;
//     while (ML < n + m)
//         ML <<= 1, bit++;
//     for (int i = 0; i < ML; i++)
//         rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
//     NTT(a, ML, 1);
//     NTT(b, ML, 1);
//     for (int i = 0; i < ML; i++)
//         ans[i] = 1ll * a[i] * b[i] % P;
//     NTT(ans, ML, -1);
// }
//v^2 x^2 - 2 v^2 x y - 2 v x^2 y + v^2 y^2 + 4 v x y^2 + x^2 y^2 - 2 v y^3 - 2 x y^3 + y^4
int c[N], b[5][N], a[5][N];
int val[N], p[N], vis[N];
char s[N], t[N];
int main()
{
    //cout << 2333 << endl;
    for (int i = 1; i <= 26; i++)
        cin >> p[i];
    //cout << 2333 << endl;
    for (int i = 1; i <= 26; i++)
    {
        // int p = rand() % 100 + 1;
        // while (vis[p] == 1)
        // {
        //     p = rand() % 100 + 1;
        // }
        val[i] = i * i;
        //vis[p] = 1;
    }

    scanf("%s", t + 1);
    scanf("%s", s + 1);
    int n = strlen(s + 1);
    int m = strlen(t + 1);
    //cout << 2333 << endl;
    for (int i = 1; i <= m; i++)
    {
        int id = t[i] - 'a' + 1;
        int x = val[id], v = val[p[id]];
        a[0][i] = 1ll * v * v * x * x % mod;
        a[1][i] = (mod - 2 * v * v * x - 2 * v * x * x) % mod;
        a[2][i] = (v * v + 4 * v * x + x * x) % mod;
        a[3][i] = (mod - 2 * x - 2 * v) % mod;
        a[4][i] = 1;
    }
    for (int i = 1; i <= n; i++)
    {
        int id = s[i] - 'a' + 1;
        int y = val[id];
        b[0][i] = 1;
        b[1][i] = y;
        b[2][i] = y * y;
        b[3][i] = 1ll * y * y * y % mod;
        b[4][i] = 1ll * y * y * y * y % mod;
    }

    int ML = 1, bit = 0;
    while (ML < n + m)
        ML <<= 1, bit++;
    for (int i = 0; i < ML; i++)
        rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (bit - 1));
    for (int i = 0; i <= 4; i++)
    {
        reverse(b[i] + 1, b[i] + 1 + n);
        NTT(b[i], ML, 1);
        NTT(a[i], ML, 1);
        for (int j = 0; j < ML; j++)
            c[j] = (c[j] + 1ll * a[i][j] * b[i][j] % mod) % mod;
    }
    NTT(c, ML, -1);
    for (int s = 1; s <= n - m + 1; s++)
        if (c[n - s + 2] == 0)
            printf("1");
        else
            printf("0");
}