零子序列问题 - 我的 C++ 解决方案有什么问题？

Question

Problem Statement: Given an array arr of n integers, count the number of non-empty subsequences of the given array such that their product of maximum element and minimum element is zero. Since this number can be huge, compute it modulo 10 ^ 9 + 7 A subsequence of an array is defined as the sequence obtained by deleting several elements from the array (possible none) without changing the order of the remaining elements.

Example Given n = 3, arr = [1, 0, – 2].

There are 7 subsequences of arr that are-

[1], minimum = 1, maximum =1, min * max = 1.

[1,0] minimum = 0, maximum=1, min * max=0

[ 1,0, – 2], minimum = – 2, maximum =1, min* max = -2.

[0], minimum = 0, maximum =0, min * max=0

[0,-2],minimum=-2,maximum=0, min* max=0,

[1, -2] minimum=-2, maximum=1,min* max=-2

[- 2] minimum =-2 maximum = – 2, min* max = 4.

There are 3 subsequences whose minimum * maximum = 0 that are

[1, 0], [0], [0, – 2]. Hence the answer is 3 .

I tried to come up with a solution, by counting the number of zeroes, positive numbers and negative numbers and then adding possible subsequences(2^n, per count) to an empty variable.

My answer is way off though, it's 10 when the expected answer is 3. Can someone please point out my mistake?

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

int zeroSubs(vector<int> arr){
    int x = 0, y = 0, z = 0, ans = 0;
    for(int i = 0; i < arr.size(); i++){
        if(arr[i] == 0) z++;
        else if(arr[i] < 0) x++;
        else y++;
    }
    ans += ((int)pow(2, z))*((int)pow(2, x));
    ans += ((int)pow(2, y))*((int)pow(2, z));
    ans += ((int)pow(2, z));

    return ans;
}

int32_t main()
{
//directly passed the sample test case as an array
    cout<<zeroSubs({1, 0, -2});
    return 0;
}

Answer 1

ans += ((1<<z)-1)*((1<<x)-1);
ans += ((1<<y)-1)*((1<<z)-1);
ans += ((1<<z)-1);

Made this slight change in the logic, thanks a lot to everyone for the valuable feedback. It works now.