Two Sets II ( CSES )

#include <iostream>
#include <vector>
#include <algorithm>
#include <numeric>
#include <cmath>
#include <queue>
using namespace std;
const int N = 500*501/4, mod = 1e9 + 7;
long long n, dp[N+1], m;
void read()
{
   cin >> n;
}
void solve()
{
   if (n * (n+1) % 4 != 0)
   {
       cout << 0; return ;
   }
   fill(dp + 1, dp + N + 1, 0);
   m = n * (n + 1)/4;
   dp[0] = 1;
   for (int i = 1; i <= n; ++ i)
   {
       for (int j = m; j >= i; -- j)
       {
           dp[j] = (dp[j] + dp[j - i]) % mod;
       }
   }
   cout << (dp[m] * (mod + 1)/2) % mod; // Nghich dao modulo
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    read();
    solve();
}