Codeforces Round 726 (Div. 2) |
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Finished |
An array $$$b$$$ of length $$$k$$$ is called good if its arithmetic mean is equal to $$$1$$$. More formally, if $$$$$$\frac{b_1 + \cdots + b_k}{k}=1.$$$$$$
Note that the value $$$\frac{b_1+\cdots+b_k}{k}$$$ is not rounded up or down. For example, the array $$$[1,1,1,2]$$$ has an arithmetic mean of $$$1.25$$$, which is not equal to $$$1$$$.
You are given an integer array $$$a$$$ of length $$$n$$$. In an operation, you can append a non-negative integer to the end of the array. What's the minimum number of operations required to make the array good?
We have a proof that it is always possible with finitely many operations.
The first line contains a single integer $$$t$$$ ($$$1 \leq t \leq 1000$$$) — the number of test cases. Then $$$t$$$ test cases follow.
The first line of each test case contains a single integer $$$n$$$ ($$$1 \leq n \leq 50$$$) — the length of the initial array $$$a$$$.
The second line of each test case contains $$$n$$$ integers $$$a_1,\ldots,a_n$$$ ($$$-10^4\leq a_i \leq 10^4$$$), the elements of the array.
For each test case, output a single integer — the minimum number of non-negative integers you have to append to the array so that the arithmetic mean of the array will be exactly $$$1$$$.
4 3 1 1 1 2 1 2 4 8 4 6 2 1 -2
0 1 16 1
In the first test case, we don't need to add any element because the arithmetic mean of the array is already $$$1$$$, so the answer is $$$0$$$.
In the second test case, the arithmetic mean is not $$$1$$$ initially so we need to add at least one more number. If we add $$$0$$$ then the arithmetic mean of the whole array becomes $$$1$$$, so the answer is $$$1$$$.
In the third test case, the minimum number of elements that need to be added is $$$16$$$ since only non-negative integers can be added.
In the fourth test case, we can add a single integer $$$4$$$. The arithmetic mean becomes $$$\frac{-2+4}{2}$$$ which is equal to $$$1$$$.
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