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.

Input

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

NN
a_1a1 a_2a2 \ldots… a_NaN

Output

Print the answer.

Sample Input 1 Copy

Copy

3
1 4 3

Sample Output 1 Copy

Copy

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).

Sample Input 2 Copy

Copy

3
2 7 6

Sample Output 2 Copy

Copy

-1

There is no way to achieve the objective.

Sample Input 3 Copy

Copy

6
1 1 1 1 1 1

Sample Output 3 Copy

Copy

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 <stdio.h>
#include <algorithm>
#include <math.h>
#include <iostream>
#include <cmath>
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;
}

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Divide by 2 or 3

Problem Statement

Constraints

Input

Output

Sample Input 1 Copy

Sample Output 1 Copy

Sample Input 2 Copy

Sample Output 2 Copy

Sample Input 3 Copy

Sample Output 3 Copy

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