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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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import sys
MOD=998244353

def mult(a, b, K):
    res = [0] * K
    for i in range(K):
        ai = a[i]
        if ai:
            for j in range(K):
                res[(i + j) % K] = (res[(i + j) % K] + ai * b[j]) % MOD
    return res

def power(a, e, K):
    res = [0] * K
    res[0] = 1
    while e:
        if e & 1:
            res = mult(res, a, K)
        a = mult(a, a, K)
        e >>= 1
    return res

def main():
    input = sys.stdin.read
    data = list(map(int, input().split()))
   
    ptr = 0
    N = data[ptr]
    P = data[ptr + 1]
    K = data[ptr + 2]
    ptr += 3
   
    M = data[ptr]
    ptr += 1
   
    L = data[ptr:ptr + M]
    ptr += M
   
    C = [P // K + (1 if j > 0 and j <= P % K else 0) for j in range(K)]
    LC = [0] * K
   
    for x in L:
        LC[x % K] += 1
   
    Cn = [(C[j] - LC[j]) % MOD for j in range(K)]
   
    T = power(C, N, K)[0]
    B = power(Cn, N, K)[0]
   
    print((T - B) % MOD)

main()


Lucky Charm
Rubrik โœ…
nt FUNC(vector<int> &v){
    unordered_set<int> st(v.begin(),v.end());

    for(int i=0;i<31;i++){
        if(st.find(1<<i)==st.end()){
            return 1<<i;
        }
    }

    return -1;

}

Aptiv โœ…
#define ll long long
#define N 110
#define MOD 1000000007
int solution(string p, string q, string r) {
    int n = p.size();
    int m = q.size();
    int k = r.size();

    // 1-indexing
    p = "*" + p;
    q = "*" + q;
    r = "*" + r;

    ll dp[N][N][N] = {};  // Initialize the dp array to zero
    dp[0][0][0] = 1;      // Base case

    // Fill the dp table
    for (int i = 0; i <= n; i++) {
        for (int j = 0; j <= m; j++) {
            for (int len = 0; len < k; len++) {
                if (dp[i][j][len]) {
                    // Check the next characters from string p
                    for (int ni = i + 1; ni <= n; ni++) {
                        if (p[ni] == r[len + 1]) {
                            dp[ni][j][len + 1] = (dp[ni][j][len + 1] + dp[i][j][len]) % MOD;
                        }
                    }
                    // Check the next characters from string q
                    for (int nj = j + 1; nj <= m; nj++) {
                        if (q[nj] == r[len + 1]) {
                            dp[i][nj][len + 1] = (dp[i][nj][len + 1] + dp[i][j][len]) % MOD;
                        }
                    }
                }
            }
        }
    }

    // Calculate the final answer
    ll ans = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            ans = (ans + dp[i][j][k]) % MOD;
        }
    }

    return ans;  // Return the calculated result
}

Uber โœ…
๐Ÿ‘1
def ss(wordlist):
    a = []
    for word in wordlist:
        if len(word) % 2 == 0:
            a.append(word[::-1])
        else:
            a.append(word.upper())
    return a


Uber โœ…
๐Ÿ‘1
from collections import Counter
def digit_root(n):
    if n == 0:
        return 0
    return 1 + (n - 1) % 9
def ss(readings):
    reduced_readings = [digit_root(num) for num in readings]
    freq = Counter(reduced_readings)
    v = max(freq.keys(), key=lambda x: (freq[x], x))
    return v

Uber โœ