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
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 00. If we add 11 to the first element, the array will be [3,?1,?1][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][?1,1,1,1], the sum will be equal to 22 and the product will be equal to ?1?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][2,?2,1], the sum will be 11 and the product will be ?4?4.
一個思維題，只要注意一點，在乘法為0的時候，就代表著肯定有0存在，這樣的話就只能+1了，但是在這一過程中需要注意加法不能變為0，所以需要特判一下。
代碼如下：

#include<bits/stdc++.h> #define ll long long using namespace std;const int maxx=110; int a[maxx]; int n;int main() {int t;scanf("%d",&t);while(t--){scanf("%d",&n);int sum0=0;int sum=0,csum=1;for(int i=1;i<=n;i++){scanf("%d",&a[i]);sum+=a[i];csum*=a[i];sum0+=(a[i]==0);}if(sum&&csum) cout<<0<<endl;else if(sum&&!csum){if(sum+sum0!=0) cout<<sum0<<endl;else cout<<sum0+1<<endl;}else if(!sum&&csum) cout<<1<<endl;else {if(sum+sum0!=0) cout<<sum0<<endl;else cout<<sum0+1<<endl;}}return 0; }

努力加油a啊，(^o)/~

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