CF 1300.A——Non-zero【签到题】

题目传送门

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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 $a1+a2+ … +an≠0 \ and\ a1⋅a2⋅ … ⋅an≠0.$

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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 .

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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.

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input

4
3
2 -1 -1
4
-1 0 0 1
2
-1 2
3
0 -2 1

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output

1
2
0
2

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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.

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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

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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

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AC-Code

#include<bits/stdc++.h>
using 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;
}