A. Non-zero
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
Guy-Manuel and Thomas have an array a of n integers [a1,a2,…,an]. In one step they can add 1 to any element of the array. Formally, in one step they can choose any integer index i (1≤i≤n) and do ai:=ai+1.

If either the sum or the product of all elements in the array is equal to zero, Guy-Manuel and Thomas do not mind to do this operation one more time.

What is the minimum number of steps they need to do to make both the sum and the product of all elements in the array different from zero? Formally, find the minimum number of steps to make a1+a2+ … +an≠0 and a1⋅a2⋅ … ⋅an≠0.

Input
Each test contains multiple test cases.

The first line contains the number of test cases t (1≤t≤103). The description of the test cases follows.

The first line of each test case contains an integer n (1≤n≤100) — the size of the array.

The second line of each test case contains n integers a1,a2,…,an (−100≤ai≤100) — elements of the array .

Output
For each test case, output the minimum number of steps required to make both sum and product of all elements in the array different from zero.

Example
inputCopy
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
outputCopy
1
2
0
2
Note
In the first test case, the sum is 0. If we add 1 to the first element, the array will be [3,−1,−1], the sum will be equal to 1 and the product will be equal to 3.

In the second test case, both product and sum are 0. If we add 1 to the second and the third element, the array will be [−1,1,1,1], the sum will be equal to 2 and the product will be equal to −1. It can be shown that fewer steps can’t be enough.

In the third test case, both sum and product are non-zero, we don’t need to do anything.

In the fourth test case, after adding 1 twice to the first element the array will be [2,−2,1], the sum will be 1 and the product will be −4.

题意：给了n个数 你每次可以对一个数进行无限次加1，问你 让这堆数的和不为0并且乘积不为0，至少要操作多少次

思路:计算0的个数 和当前的总和 如果总和sum 加上0的个数 等于0 那么答案就是0的个数+1
否则答案就是0的个数

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define me(a,x) memset(a,x,sizeof a)
#define pb(a) push_back(a)
#define paii pair<int,int>
#define pali pair<ll,int>
#define pail pair<int,ll>
#define pall pair<ll,ll>
#define fi first
#define se second
int main()
{
    int t;cin>>t;
    while(t--){

        int n;cin>>n;
        vector<int> a(n+1);
        int sum=0;
        int flag=0;
        for(int i=1;i<=n;i++){

            cin>>a[i];sum+=a[i];
            if(!a[i]) flag++;
        }
        if(sum+flag==0) cout<<flag+1<<endl;
        else cout<<flag<<endl;


    }

   return 0;
}