H - Non-zero CodeForces - 1300A

H - Non-zero

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
Input
4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1
Output
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：给你一个数组，然后你进行一个步骤：随机的将里面一个数加1，然后保证：a1+a2+ … +an≠0 and a1⋅a2⋅ … ⋅an≠0.我们将所用步骤打印下来即可。
解题思路：输入每个数，判断是不是0，是0我们即将这个数加1,步骤加一；然后得出总和，如果总和为0，步骤再加一即可。

代码：

#include
#define ull unsigned long long
#define pi acos(-1)
using namespace std;
typedef long long ll;
const int maxn = 1e7+5;
int a[150];
int main(){
	int t;
	cin>>t;
	while(t--){
		int n,sum=0,step=0;
		cin>>n;
		for(int i=1;i<=n;i++){
			cin>>a[i];
			if(a[i]==0){
				a[i]+=1;
				step+=1;
			}
			sum+=a[i];
		}
		if(sum==0) step+=1;
		cout<<step<<'\n';
	} 
	return 0;
}