Arithmetic Array.cpp

Task: An array b of length k is called good if its arithmetic mean is equal to 1. More formally, if
b1+⋯+bkk=1.

Note that the value b1+⋯+bkk
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.
Input

The first line contains a single integer t
(1≤t≤1000) — the number of test cases. Then t

test cases follow.

The first line of each test case contains a single integer n
(1≤n≤50) — the length of the initial array a

.

The second line of each test case contains n
integers a1,…,an (−104≤ai≤104

), the elements of the array.
Output

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

.
Example
Input
Copy

4
3
1 1 1
2
1 2
4
8 4 6 2
1
-2

Output
Copy

0
1
16
1

Note

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 −2+42 which is equal to 1.




Solution:



#include <bits/stdc++.h>
#define ll long long
using namespace std;    


int main(){
    
    ll t;
    cin>>t;
    while(t--){
        ll n,a,b=0;
        cin>>n;
        for(ll i=0;i<n;i++){
            cin>>a;
            b+=a;
        }
        if(b<n){
            cout<<1<<endl;
        }
        else{
            cout<<b-n<<endl;
        }
    }
    return 0;
}