Help me in solving COUNTN problem

help
,
1

My issue

how to reduce time limit

My code

def is_prime_recursive(num, divisor=None):
    if divisor is None:
        divisor = num - 1
    if num <= 1:
        return False
    if divisor == 1:
        return True
    if num % divisor == 0:
        return False
    return is_prime_recursive(num, divisor - 1)
for i in range(int(input())):
    k = int(input())
    if k%2==1:
        p = 0
        for j in range(2, k+1):
            if is_prime_recursive(j):
                p+=k*j
        print(p)

    else:
         print(k*2)

Problem Link: Sum of N Practice Coding Problem

2

you can first find all the prime no. in 1 to 1e6
then calculate prefix sum of prime no.
then for each testcase find the smallest prime factor possible in O(sqrt(k))
then cout the sum for that prime no. * k

3

nhi ho raha samajh nhi aa raha pura solution likhdo pls

4

int main() {
std::ios_base::sync_with_stdio(false);
cin.tie(NULL);
int t;
cin >> t;
vector < ll > prime;
vector < int > a(1000005, 0);
prime.push_back(2);
for (ll i = 3; i < 1000005; i += 2) {
if (a[i] == 1) continue;
a[i] = 1;
prime.push_back(i);
for (ll j = i * i; j < 1000005; j += i) {
a[j] = 1;
}
}
unordered_map < ll, ll > mp;
ll sum = 0;
for (int i = 0; i < prime.size(); i++) {
sum += prime[i];
mp[prime[i]] = sum;
}
while (t–) {
ll n;
cin >> n;
ll ans = 0;
if (n%2==0) ans = 2;
else {
ans=n;
for (long long i = 3; i <= sqrt(n); i += 2) {
if (n % i == 0) {
ans= i;
break;
}
}

    }
    ans=mp[ans];
    ans *= n;
    cout << ans << endl;
}
cerr << "Time : " << 1000 * ((double) clock()) / CLOCKS_PER_SEC << "ms" << endl;
return 0;

}