two string are equal or not using (single hashing and double Hashing)

#include <bits/stdc++.h>

using namespace std;
#define ll long long int
const int mx = 1e5 + 5;
int p = 31, MOD = 1e9 + 7;
int pw[mx];

void pw_cal() {
    pw[0] = 1;
    for (int i = 1; i <= mx; i++) {
        pw[i] = (int) (1LL * pw[i - 1] * p) % MOD;
    }
}

int getHash(string s) {
    int res = 0;
    for (int i = 0; i < s.size(); i++) {
        res += 1LL * (s[i] - 'a' + 1) * pw[i] % MOD;
    }
    return res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    pw_cal();
    string s1, s2;
    cin >> s1 >> s2;
    cout << getHash(s1) << " " << getHash(s2) << endl;
    return 0;
}







#include <bits/stdc++.h>

using namespace std;
#define ll long long int
const int mx = 1e5 + 5;
int p1 = 31, p2 = 37, MOD = 1e9 + 9;
int pw1[mx + 5], pw2[mx + 5];

void pw_cal() {
    pw1[0] = 1;
    for (int i = 1; i <= mx; i++) {
        pw1[i] = (int) (1LL * pw1[i - 1] * p1) % MOD;
    }
    pw2[0] = 1;
    for (int i = 1; i <= mx; i++) {
        pw2[i] = (int) (1LL * pw2[i - 1] * p2) % MOD;
    }
}

pair<int, int> getHash(string s) {
    int res1 = 0;
    for (int i = 0; i < s.size(); i++) {
        res1 += 1LL * (s[i] - 'a' + 1) * pw1[i] % MOD;
        res1%=MOD;
    }
    int res2 = 0;
    for (int i = 0; i < s.size(); i++) {
        res2 += 1LL * (s[i] - 'a' + 1) * pw2[i] % MOD;
        res2%=MOD:
    }
    return {res1, res2};
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    pw_cal();
    string s1, s2;
    cin >> s1 >> s2;
    cout << (getHash(s1) == getHash(s2)) << endl;
    return 0;
}