字符串哈希/双哈希模板

struct Hash {
    using u64 = unsigned long long;
    u64 base = 13331;
    vector<u64> pow, hash;
    Hash(string &s) {
        s = " " + s;
        int N = s.size();
        pow.resize(N + 1), hash.resize(N + 1);
        pow[0] = 1, hash[0] = 0;
        for (int i = 1; i < s.size(); i ++) {
            pow[i] = pow[i - 1] * base;
            hash[i] = hash[i - 1] * base + s[i];
        }
    }

    u64 get(int l, int r) {
        return hash[r] - hash[l - 1] * pow[r - l + 1];
    }

    //拼接两个子串
    u64 link(int l1, int r1, int l2, int r2) {
        return get(l1, r1) * pow[r2 - l2 + 1] + get(l2, r2);
    }

    bool same(int l1, int r1, int l2, int r2) {
        return get(l1, r1) == get(l2, r2);
    }

};

struct DoubleHash {
    const int n;
    const i64 Base1 = 29, MOD1 = 1e9 + 7;
    const i64 Base2 = 131, MOD2 = 1e9 + 9;
    vector<i64> ha1, ha2, pow1, pow2;
    vector<i64> rha1, rha2;
    DoubleHash(string &s, int n1) : n(n1), ha1(n + 1), ha2(n + 1), pow1(n + 1), pow2(n + 1), rha1(n + 1), rha2(n + 1) {
        pow1[0] = pow2[0] = 1;
        for (int i = 1; i <= n; i++) {
            pow1[i] = pow1[i - 1] * Base1 % MOD1;
            pow2[i] = pow2[i - 1] * Base2 % MOD2;
        }
        for (int i = 1; i <= n; i++) {
            ha1[i] = (ha1[i - 1] * Base1 + s[i]) % MOD1;
            ha2[i] = (ha2[i - 1] * Base2 + s[i]) % MOD2;
            rha1[i] = (rha1[i - 1] * Base1 + s[n - i + 1]) % MOD1;
            rha2[i] = (rha2[i - 1] * Base2 + s[n - i + 1]) % MOD2;
        }
    }
    pair<i64, i64> get(int l, int r) {
        i64 res1 = ((ha1[r] - ha1[l - 1] * pow1[r - l + 1]) % MOD1 + MOD1) % MOD1;
        i64 res2 = ((ha2[r] - ha2[l - 1] * pow2[r - l + 1]) % MOD2 + MOD2) % MOD2;
        return {res1, res2};
    }
    //反哈希
    pair<i64, i64> get_rhash(int l, int r) {
        i64 res1 = ((rha1[n - l + 1] - rha1[n - r] * pow1[r - l + 1]) % MOD1 + MOD1) % MOD1;
        i64 res2 = ((rha2[n - l + 1] - rha2[n - r] * pow2[r - l + 1]) % MOD2 + MOD2) % MOD2;
        return {res1, res2};
    }
    //判断s[l, r]是否为回文串
    bool is_palindrome(int l, int r) {
        return get(l, r) == get_rhash(l, r);
    }
    pair<i64, i64> add(pair<i64, i64> aa, pair<i64, i64> bb) {
        i64 res1 = (aa.first + bb.first) % MOD1;
        i64 res2 = (aa.second + bb.second) % MOD2;
        return {res1, res2};
    }
    //aa *= Base的k次方
    pair<i64, i64> mul(pair<i64, i64> aa, i64 kk) {
        i64 res1 = aa.first * pow1[kk] % MOD1;
        i64 res2 = aa.second * pow2[kk] % MOD2;
        return {res1, res2};
    }
    //拼接字符串 r1 < l2  s = s1 + s2
    pair<i64, i64> link(int l1, int r1, int l2, int r2) {
        return add(mul(get(l2, r2), r1 - l1 + 1), get(l1, r1));
    }
};