I have a doubt about the implementation of the Rabin-Karp Algorithm given on CP Algorithms.com

Problem: Given two strings — a pattern s  and a text t, determine if the pattern appears in the text and if it does, enumerate all its occurrences in O(|s| + |t|) time.

Solution:

vector<int> rabin_karp(string const& s, string const& t) {
    const int p = 31; 
    const int m = 1e9 + 9;
    int S = s.size(), T = t.size();

    vector<long long> p_pow(max(S, T)); 
    p_pow[0] = 1; 
    for (int i = 1; i < (int)p_pow.size(); i++) 
        p_pow[i] = (p_pow[i-1] * p) % m;

    vector<long long> h(T + 1, 0); 
    for (int i = 0; i < T; i++)
        h[i+1] = (h[i] + (t[i] - 'a' + 1) * p_pow[i]) % m; 
    long long h_s = 0; 
    for (int i = 0; i < S; i++) 
        h_s = (h_s + (s[i] - 'a' + 1) * p_pow[i]) % m; 

    vector<int> occurences;
    for (int i = 0; i + S - 1 < T; i++) { 
        long long cur_h = (h[i+S] + m - h[i]) % m; 
        if (cur_h == h_s * p_pow[i] % m)
            occurences.push_back(i);
    }
    return occurences;
}

In the last comparison:

if (cur_h == h_s * p_pow[i] % m)
            occurences.push_back(i);

Why do we need to multiply h_s by p_pow[i], shouldn't it be just if(cur_h == h_s)

Please help.

Problem Link

Why can't I flip all the 1's in the string, that way number of '01' substrings == number of '10' substrings == 0 ?

What rating problems should I practice to be able to solve Div2 C problem.

I have followed a pattern for practicing problems. When I started I used to practice 800 rating problems, then when I was able to solve A problem in 2-3 Div2 contests, I moved on to solve 1000-1100 rating problems and when I was able to solve B problem in 2-3 Div2 contests I moved on 1200-1300 rating problems.

But, I have noticed that most of Div2 C problems are of 1600-1700 rating (I may be wrong). So, should I start solving 1600-1700 rating problems? Also, will jumping from 1300 to 1700 be actually beneficial for me?

Info which might give you insights to my level

Highest rating: 1201

Current rating: 1159

Number of 1200-1300 problems solve: 70(1200), 15(1300)

Problem Link

My Approach:

Only make the first element of vector A less than vector B. So if B[0] < A[0], iterate over A and find the first element smaller than B[0] by this we will also know the number of swaps required to change the first element of A. Then iterate over B and find the first element greater than A[0] by this we will know the number of swaps required to change the first element of B. Print the minimum number of swaps.

solution Link ,546th test case fails.