Divide by 2 or 3

Divide by 2 or 3

Problem Statement

You are given a sequence of positive integers: A=(a_1,a_2,\ldots,a_N)A=(a1​,a2​,…,aN​).
You can choose and perform one of the following operations any number of times, possibly zero.

• Choose an integer ii such that 1 \leq i \leq N1≤i≤N and a_iai​ is a multiple of 22, and replace a_iai​ with \frac{a_i}{2}2ai​​.
• Choose an integer ii such that 1 \leq i \leq N1≤i≤N and a_iai​ is a multiple of 33, and replace a_iai​ with \frac{a_i}{3}3ai​​.

Your objective is to make AA satisfy a_1=a_2=\ldots=a_Na1​=a2​=…=aN​.
Find the minimum total number of times you need to perform an operation to achieve the objective. If there is no way to achieve the objective, print -1 instead.

Constraints

• 2 \leq N \leq 10002≤N≤1000
• 1 \leq a_i \leq 10^91≤ai​≤109
• All values in the input are integers.

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Input

The input is given from Standard Input in the following format:

NN
a_1a1​ a_2a2​ \ldots… a_NaN​

Output

Print the answer.

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Sample Input 1 Copy

Copy

3
1 4 3

Sample Output 1 Copy

Copy

3

Here is a way to achieve the objective in three operations, which is the minimum needed.

• Choose an integer i=2i=2 such that a_iai​ is a multiple of 22, and replace a_2a2​ with \frac{a_2}{2}2a2​​. AA becomes (1,2,3)(1,2,3).
• Choose an integer i=2i=2 such that a_iai​ is a multiple of 22, and replace a_2a2​ with \frac{a_2}{2}2a2​​. AA becomes (1,1,3)(1,1,3).
• Choose an integer i=3i=3 such that a_iai​ is a multiple of 33, and replace a_3a3​ with \frac{a_3}{3}3a3​​. AA becomes (1,1,1)(1,1,1).

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Sample Input 2 Copy

Copy

3
2 7 6

Sample Output 2 Copy

Copy

-1

There is no way to achieve the objective.

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Sample Input 3 Copy

Copy

6
1 1 1 1 1 1

Sample Output 3 Copy

Copy

0

题解：要想所有的数能变成同一个数，那它们的因素只可能有2 ， 3 以及 它们的最大公因数 ，注意最大公因数也可能是2，3的倍数。

因此： 1.所有数的求最大公因数

            2.数组元素除最大公因数，此时不数组中不可能存在一个元素（不等于1），可以使得数组中每个数都等于该元素.

           3.因此最小的操作次数是除去最大公因数后到1 的操作次数之和。

#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;
 
ll n, p, ans;
ll a[1100];
 
long long gcd(long long a, long long b)
{
	while (b ^= a ^= b ^= a %= b);
	return a;
}
 
int main() {
	cin >> n;
	for (ll i = 1; i <= n; i++) {
		cin >> a[i];
	}
	p = a[1];
	for (ll i = 2; i <= n; i++) {
		p = gcd(a[i], p);//找最大公因数
	}
	for (ll i = 1; i <= n; i++) {
		ll x = a[i] / p;
		while (x % 2 == 0) {
			x /= 2;
			ans++;
		}
		while (x % 3 == 0) {
			x /= 3;
			ans++;
		}
		if (x != 1) {
			ans = -1;
			break;
		}
	}
	printf("%lld\n", ans);
	return 0;
}