1141A Game 23

cf传送门:Game 23
vj传送门:Game 23

题目描述

A. Game 23

Polycarp plays “Game 23”. Initially he has a number n and his goal is to transform it to m. In one move, he can multiply n by 2 or multiply n by 3. He can perform any number of moves.

Print the number of moves needed to transform n to m. Print -1 if it is impossible to do so.

It is easy to prove that any way to transform n to m contains the same number of moves (i.e. number of moves doesn’t depend on the way of transformation).

Input
The only line of the input contains two integers n and m (1≤n≤m≤5⋅108).

Output
Print the number of moves to transform n to m, or -1 if there is no solution.

Examples
Input
120 51840
Output
7
Input
42 42
Output
0
Input
48 72
Output
-1
Note
In the first example, the possible sequence of moves is: 120→240→720→1440→4320→12960→25920→51840. The are 7 steps in total.

In the second example, no moves are needed. Thus, the answer is 0.

In the third example, it is impossible to transform 48 to 72.

题意分析

因为是全英文的题面,我选择了先看样例,又看到了提示,因为紧张加小激动我把题意理解错了。。。以为是个水题(对大佬是真水题吧)。题目要求就是输入两个数,问你从第一个数变成第二个数需要几步,变化的方法是乘以2或者乘以3,如果可以,输出步数,不可以的话就输出“-1”,听学长说是质因子分解,可是我不会(哈哈哈哈,以后会了再来试试。经过我跟小伙伴的一波探讨发现可以用第二个数对第一个数先求余,如果余数不为0,直接输出“-1”。余数如果为0,就对两数的商进行处理,除以三能除尽就一直除直到除不尽,再判断,如果除以二能除尽就一直除直到除不尽,每循环一次次数加一,最后都除不尽的时候需要在进行一次判断了看现在的商是不是1,如果不是则说明变化的过程中只有二和三是不够的,所以需要输出“-1”,反之输出次数即可。

代码实现

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll n,m,k,a,b;
    cin>>n>>m;
    if(m%n!=0)
        cout<<"-1";
    ll s=0;
    if(m%n==0)
    {
        k=m/n;
        while(k%3==0)
        {
            k/=3;
            s++;
        }
        while(k%2==0)
        {
            k/=2;
            s++;
        }
        if(k!=1)
        {
            cout<<"-1"<<endl;
            return 0;
        }
        cout<<s;
    }
    cout<<endl;
    return 0;
 }