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:
给你n个数，每个数都可以+1，问你执行至少多少次操作，使得n个数之和 和 n个数之积不为0
解析:
我们求一个总和，遇到为0的地方+1.
最后判断总和是否为0.如果不为0，那么说明我们至少执行了k次
如果为0，说明我还需要在执行一次操作即可。输出k+1
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+1000;
int a[N];
int t;
int n;
int main()
{
cin>>t;
while(t--)
{
cin>>n;
int sum=0;
int suml=0;
int k=0;
for(int i=0;i<n;i++)
{
cin>>a[i];
if(a[i]==0) k++,sum+=1;
else sum+=a[i];
suml+=a[i];
}
if(suml!=0&&k==0) cout<<0<<endl;
else
{
if(sum!=0)cout<<k<<endl;
else cout<<k+1<<endl;
}
}
}
