Codeforces Round #618 (Div. 2)-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

output

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.

这个题是说通过最小的修改次数，是数列和不能为0，乘积不能为0；
那么也即数列中不存在0，如果存在0的一定要改，存在0的只能变成1，那我们考虑变成1之后，的和是否等于0，如果等于，就在修改1个，即cnt+1。

#include<bits/stdc++.h>
using namespace std;
const int N=5e5;
#define read(a) scanf("%d",&a);
int  a[N];
int main()
{int t;read(t);while(t--){int n;read(n);long long sum=0;int cnt=0;for(int i=1;i<=n;i++){cin>>a[i];sum+=(long long)a[i];if(a[i]==0) cnt++;}if(cnt==0){if(sum!=0) cout<<0<<endl;else cout<<1<<endl;}else {if(cnt+sum==0) cout<<cnt+1<<endl;else cout<<cnt<<endl;}}
}

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