Guy-Manuel and Thomas have an array aa of nn integers [a1,a2,…,ana1,a2,…,an]. In one step they can add 11 to any element of the array. Formally, in one step they can choose any integer index ii (1≤i≤n1≤i≤n) and do ai:=ai+1ai:=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+a1+a2+ …… +an≠0+an≠0 and a1⋅a2⋅a1⋅a2⋅ …… ⋅an≠0⋅an≠0.
Input
Each test contains multiple test cases.
The first line contains the number of test cases tt (1≤t≤1031≤t≤103). The description of the test cases follows.
The first line of each test case contains an integer nn (1≤n≤1001≤n≤100) — the size of the array.
The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (−100≤ai≤100−100≤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 00. If we add 11 to the first element, the array will be [3,−1,−1], the sum will be equal to 11 and the product will be equal to 33.
In the second test case, both product and sum are 00. If we add 11 to the second and the third element, the array will be [−1,1,1,1], the sum will be equal to 22 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 11 twice to the first element the array will be [2,−2,1], the sum will be 11 and the product will be −4.
题目大意：给出一组数，通过给个别数+1的方式，确保这组数无论全部相加，还是全部相乘，都不会得到0。求+1的次数。
#include <iostream>
#include <cstdio>
using namespace std;
int main()
{
int a[201];
int sum=0;
int n,m,flag,i,s=0;
cin>>n;
while(n--){
flag=0;
sum=0;
cin>>m;
for(i=0;i<m;i++){
cin>>a[i];
if(a[i]==0){
a[i]++;
flag++;
}
sum=sum+a[i];
}
if(sum==0){
flag++;
}
printf("%d\n",flag);
}
}
