Luigi and Uniformity Codechef solutions Today | Codechef Starters 64✅ (100/100) FULL | AC Code ✅| DIVIDEMAKEEQ

IF ANYONE DONE THIS AND IS YOU WANT TO HELP SO JUST WRITE DOWN CODE IN COMMENTS

For Solution

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int t;

    cin>t;

    while(t--){

    cin>>n;

for(i=0;i<n;i++) cin>>a[i];

    x=a[0];

for(i=1;i<n;i++){

    x=gcd(x,a[i]);

}

cnt=0;

for(i=0;i<n;i++){

    if(a[i]==x) cnt++;

}

cout<<n-cnt<<endl;

}

}

Problem

Luigi has an array $A$ of $N$ positive integers. He wants to make all elements of the array equal.

In one move, he can:

• Choose an index $i$ $(1\le i \le N)$ and divide the element $A_i$ by any one of its divisors.
  In other words, he can choose a positive integer $X$ such that $X | A_i$ and set $A_i := \frac{A_i}{X}$.

Find the minimum number of moves required to make all the elements of the array equal.

Input Format

• The first line of input will contain a single integer $T$, denoting the number of test cases.
• Each test case consists of two lines of input.
  • The first line of each test case contains $N$, the size of array $A$.
  • The second line of each test case contains $N$ space-separated integers, the elements of array $A$.

Output Format

For each test case, output on a new line, the minimum number of moves required to make all elements of the array equal.

Constraints

Sample 1:

4
2
11 22
5
38 38 38 38 38
4
4 4 16 8
4
11 13 17 19

1
0
2
4

Explanation:

Test case $1$: We can change $22$ to $11$ using an operation (divide by $2$). Thus, using one operation, we made all the elements equal.

Test case $2$: All numbers are already same. Thus, we do not need to do any operations.

Test case $3$: We make the following operations:

• Choose $i = 3$ and $X = 4$, such that $4|16$. Thus, $A_3 := \frac{16}{4} = 4$.
• Choose $i = 4$ and $X = 2$, such that $2|8$. Thus, $A_4 := \frac{8}{2} = 4$.

Thus, using two operations, we made all the elements equal.

Test case $4$: We make the following operations:

• Choose $i = 1$ and $X = 11$, such that $11|11$. Thus, $A_1 := \frac{11}{11} = 1$.
• Choose $i = 2$ and $X = 13$, such that $13|13$. Thus, $A_2 := \frac{13}{13} = 1$.
• Choose $i = 3$ and $X = 17$, such that $17|17$. Thus, $A_3 := \frac{17}{17} = 1$.
• Choose $i = 4$ and $X = 19$, such that $19|19$. Thus, $A_4 := \frac{19}{19} = 1$.

Thus, using four operations, we made all the elements equal.