CF 1300.A——Non-zero【签到题】

CF 1300.A——Non-zero【签到题】

​​题目传送门​​

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.

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.

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n个数，要求这n个数相乘不能为0，相加不能为0每次操作你可以将任意一个数字+1问最少操作次数

题解

先看相乘不为0，把所有为0的数字都加1，操作数=0的个数然后如果现在的数是0，那就随便找一个数+1即可

AC-Code

#includeusing namespace std; int main() { int T; cin >> T; while (T--) { int n; cin >> n; int zero = 0, sum = 0; for (int i = 0; i < n; ++i) { int a; cin >> a; sum += a; zero += a == 0 ? 1 : 0; } cout << zero + (sum + zero == 0 ? 1 : 0) << endl; } return 0;}

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