HackerEarth Sherlock and XOR problem solution

#include <cstdlib>
#include <stdio.h>
#include <cstring>
#include <complex>
#include <vector>
#include <cmath>
#include <ctime>
#include <iostream>
#include <numeric>
#include <algorithm>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <iomanip>
#include <utility>
#include <locale>
#include <sstream>
#include <string>
#include <istream>
#include <ostream>
#include<assert.h>

using namespace std;

#define FOR(i,n) for(i=0;i<n;i++)
#define FORI(i,a,n) for(i=a;i<n;i++)
#define FORC(it,C) for(it=C.begin();it!=C.end();it++)
#define scanI(x) scanf("%d",&x)
#define scanD(x) scanf("%lf",&x)
#define print(x) printf("%dn",x)
#define ll  long long
#define number32 4294967296ull
#define MAX 100001
#define len(s) s.length()
#define ff first
#define ss second
#define IN(A, B, C)  (B) <= (A) && (A) <= (C)
#define AIN(A, B, C) assert(IN(A, B, C))
using namespace std;

typedef map<int,int> Mii;
typedef map<char,int> Mci;
typedef pair<int,int> Pii;
typedef vector<int> vi;
typedef vector<int> vc;
typedef vector<ll> vl;

typedef map<int,int>::iterator Miii;
typedef map<char,int>::iterator Mcii;

ll A[MAX+1];
int main()
{
    int t,i;
    int n;
    cin>>t;
    AIN(t,1,10);
    ll ceven,codd;
    ll ans;
    while(t--)
    {
        ceven=0;
        codd=0;
        cin>>n;
        AIN(n,1,100000);
        FOR(i,n)
        {
            cin>>A[i];
            AIN(A[i],0,1000000000);
            if(A[i]%2==0)
            {
                ceven++;
            }
            else codd++;
            
        }
        ans=ceven*codd;
        cout<<ans<<endl;
    }
    return 0;
}