I understood the tutorial O(n.sqrt(n)) solution, but can't seem to understand why my solution with the same complexity is exceeding the time limit. Any help would be greatly appreciated!

#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define err() cout << "==================" << endl;
#define err1(a) cout << #a << " " << a << endl;
#define err2(a, b) cout << #a << " " << a << " " << #b << " " << b << endl;
#define err3(a, b, c) cout << #a << " " << a << " " << #b << " " << b << " " << #c << " " << c << " " << endl;
const ll mod = 998244353;

void solve(int testCase) {
    ll i = 1, j, n, k;
    cin >> n >> k;
    vector<ll> dp1(n + 1, 0), dp2(n + 1, 0), ans(n + 1, 0);
    dp1[0] = 1;
    ans[0] = 1;
    for (i = 1; k * i + ((i * (i - 1)) / 2) <= n; i++) {
        for (j = 0; j <= n; j++) {
            if (j - (k + i - 1) >= 0) {
                dp1[j] = (dp1[j] + dp1[j - (k + i - 1)]) % mod;
                dp2[j] = dp1[j - (k + i - 1)] % mod;
            }
        }
        for (j = 0; j <= n; j++) {
            ans[j] = (ans[j] + dp2[j]) % mod;
            dp1[j] = dp2[j] % mod;
            dp2[j] = 0;
        }
    }
    for (i = 1; i <= n; i++) cout << ans[i] % mod << " ";
    cout << endl;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll t = 1;
    // cin >> t;
    for (int testCase = 1; testCase <= t; testCase++)
        solve(testCase);
    return 0;
}

167477187

I noticed that many people were confused by the mathematics used in the soltuion of the editorial of this Div2.C question and decided to share with you my easy method for solving it. I use the property of distributivity of the lcm fucnction over gcd to simplify the solution. For three integers $$$ a, b, c$$$ we have

• $$$ \gcd(lcm(a, b), lcm(a, c)) = lcm(a, \gcd(b, c))$$$
• $$$ lcm(\gcd(a, b), \gcd(a, c)) = \gcd(a, lcm(b, c)) $$$

Proof: GCD and LCM Distribute Over Each Other

Hence for an array of integers,

• $$$ \gcd(lcm(a_0, a_1), lcm(a_0, a_2) \dots lcm(a_0, a_n)) = lcm(a_0, gcd(a_1, a_2 \dots a_n)) $$$
• $$$ \gcd(lcm(a_1, a_2), lcm(a_1, a_3) \dots lcm(a_1, a_n)) = lcm(a_1, gcd(a_2, a_3 \dots a_n)) $$$
• $$$\dots$$$ $$$and$$$ $$$so$$$ $$$on$$$

Therefore, for every element of the array we can precalculate the $$$\gcd$$$ of its next elements. Then we can take the lcm of that precalculated value with the element and store it in a new array. This way all possible pairs will be covered. The answer of the problem will be the $$$\gcd$$$ of these elements. Time complexity of the $$$\gcd(m, n)$$$ function is $$$\log_2v$$$ where $$$v=\max(m, n)$$$. Hence, time complexity of the solution will be $$$O(n\cdot\log_2maxval)$$$ where $$$maxval$$$ is the maximum of the array. Here is the solution:

#include <bits/stdc++.h>
using namespace std;
 
long long lcm(long long a, long long b) { return (a*b)/__gcd(a, b); }
 
int main()
{
	long long n, ans=0;
	cin>>n;
	vector<long long> a(n), pregcd(n);
	
	for(int i=0; i<n; i++)
		cin>>a[i];
	
	pregcd[n-2]=a[n-1];	
	for(int i=n-3; i>=0; i--)
		pregcd[i]=__gcd(pregcd[i+1], a[i+1]);
	
	for(int i=0; i<n-1; i++)
		pregcd[i]=lcm(pregcd[i], a[i]);
	
	for(int i=0; i<n-1; i++)
		ans=__gcd(ans, pregcd[i]);
	cout<<ans<<endl;	  
}

This is my first time writing a blog so suggestions are welcome! Edit: Changed the data type of variables from int to long long.