JavaScript Solution

let ans = 0
    if (n === 1) return "Bob"
    else ans++      // because n>0
    function isPrime(i) {
// optimal the loop condition
        let limit = Math.round(Math.sqrt(i)) 
        for (let mod = 2; mod <= limit; mod++) {
            if (i % mod === 0) {
                return false;
            }
        }
        return true
    }
// eleminate the even number except 2 to optimal the loop
    for (let i = 3; i <= n; i += 2) {   
        if (isPrime(i))
            ans++
    }
    return ans % 2 === 0 ? "Bob" : "Alice"

Here is Alice and Bob's Silly game problem solution - https://programs.programmingoneonone.com/2021/07/hackerrank-alice-and-bobs-silly-game-problem-solution.html

Linear Sieve+Prefix Sums

void sive(){
	memset(lp,0,sizeof lp);
	for (int i=2; i<=N; ++i) {
	if (lp[i] == 0) {
		lp[i] = i;
		pr.push_back (i);
	}
	for (int j=0; j<(int)pr.size() && pr[j]<=lp[i] && i*pr[j]<=N; ++j){
		lp[i * pr[j]] = pr[j];
	}
    }
    a[0]=a[1]=0;
    for(int i=2;i<mxn;i++){
        if(lp[i]==i){
            a[i]=a[i-1]+1;
            continue;
        }
        a[i]=a[i-1];
    }
}
void test_case(){
	int n,cnt=0;
	read(n);
	if(n<2){
		cout<<"Bob"<<endl;
		return;
	}
	cnt=a[n];
	cout<<(cnt&1?"Alice":"Bob")<<endl;
}
int main(){
	int tt;
    sync();
	read(tt);
	sive();
	while(tt--){
		test_case();
	}
}

Here is my simple algorithm to solve the problem. I won't tell the code as so many people here have already done without explaining anything. I created a function which takes int n input and returns total count of prime numbers form 1 to n including. Now, if n is even, answer is "Bob" else it is "Alice". Now this method passes three test cases but time limit exceeds in two other test case. What I observed is that, on single test case, multiple inputs are being taken, so calculating everytime the count of prime numbers is redundant. So, I created some global arrays to store the count and when it repeats, I supply it from there. Reply if you need more explanation or code.

In C#: public static string sillyGame(int num) { if (num < 3) return num % 2 == 0 ? "Alice" : "Bob"; bool[] primes = new bool[num + 1]; primes[0] = primes[1] = true; for (int idx = 4; idx < num + 1; idx += 2) primes[idx] = true; // exclude int cnt = 1; // {2} for (int i = 3; i <= num; i += 2) { if (!primes[i]) { cnt++; for (int idx = 2 * i; idx < num + 1; idx += i) primes[idx] = true; } } return primes.Where(p => !p).Count() % 2 == 1 ? "Alice" : "Bob"; }

Looks similar to ojava