Help me in solving AOCV203 problem

help
,
1

My issue

can anyone explain me this question, how they get output ??

My code

// Update the '_' in the code below to solve the problem

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

int main() 
{
	int t;
    cin >> t;
	
	while(t--)
	{
	    int N, K;
	    cin >> N >> K;
	    int A[N];
	    // Declare and initialise variables - pos, neg and divk
        // Note that we are reinitializing the variables to be 0 for each test case.
        int pos = 0;
        int neg = 0;
        int divk = 0;
        for(int i = 0; ____; i++)
        {
            // Input the elements to the array
            cin >> A[i];
        }
        for(int i = 0; i < N; i++)
        {
            // Count the negative elements of the array
            if(__ _ 0)
            {
                neg = neg + 1;
            }
            // Count the positive elements of the array
            else if(__ __ 0)
            {
                pos = pos + 1;
            }
            // Count if the given element is divisible by k
            if (__ __ 0)   
            {
                divk = divk + 1;
            }
        }
        cout << pos <<" "<< neg <<" "<< divk << endl;
	}
}

Learning course: Beginner DSA in C++
Problem Link: CodeChef: Practical coding for everyone

2

@jayambe7
the logic is just iterate through the loop and count for the numbers <0 and numbers >0 and the number %k==0 .
and then return the counts.

plzz refer the following solution for better understanding.

// Solution as follows

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

int main()
{
int t;
cin >> t;

while(t--)
{
    int N, K;
    cin >> N >> K;
    int A[N];
    // Declare and initialise variables - pos, neg and divk
    // Note that we are reinitializing the variables to be 0 for each test case.
    int pos = 0;
    int neg = 0;
    int divk = 0;
    for(int i = 0; i < N; i++)
    {
        // Input the elements to the array
        cin >> A[i];
    }
    for(int i = 0; i < N; i++)
    {
        // Count the negative elements of the array
        if(A[i] < 0)
        {
            neg = neg + 1;
        }
        // Count the positive elements of the array
        else if(A[i] > 0)
        {
            pos = pos + 1;
        }
        // Count if the given element is divisible by k
        if (A[i] % K == 0)   
        {
            divk = divk + 1;
        }
    }
    cout << pos <<" "<< neg <<" "<< divk << endl;
}

}

3

thank you sir