通过在任何步骤将 N 更改为 2*N 或 N/10 来将 N 更改为 1 的最小步骤

给定一个整数N，找出将N更改为1的最少操作次数。如果不可能，则打印-1。一种操作定义为将N转换为数字2*N或将N转换为数字N/10(仅当 N 可被 10 整除时)。

例子：

Input: N = 50
Output: 3
Explanation: N can be converted to 1 in 3 steps as follows:
-> Firstly, multiply N by 2 and change N from 50 to 100 (2*N)
-> Then, divide N by 10 and change N from 100 to 10 (N/10)
-> And then, divide N by 10 and change N from 10 to 1 (N/10)

Input: N = 120
Output: -1
Explanation: There is no possible way for Geek to get to position 1.

编程需要懂一点英语

方法：仅当N可以通过在每一步乘以2或除以10减少到1 时， N才能转换为1 。以下步骤如果N具有比图2和5等的一个主要因素，现在n不能降低到1。此外，如果在N的质因数分解中2s的数量超过5s ，则N不能将其减少到1，因为不可能将所有2s中和。 2s和5s 的相等数可以通过将数字除以10来中和。额外的5s可以通过乘以额外的2s然后除以10来抵消。请按照以下步骤解决问题：

将变量twos和Fives初始化为0，以计算数字N的质因数分解中2的数量和5的数量。
在 while 循环中迭代直到N%2等于0并执行以下步骤：
- 将数字N除以2。
- 将twos的值增加1。
在 while 循环中迭代直到N%5等于0并执行以下步骤：
- 将数字N除以5。
- 增加1五岁以下儿童的值。
如果N等于1和二进制补码的数目小于等于击掌的数量，然后，打印2 *击掌两岁作为答案。
否则，打印-1。

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if N can be changed
// to 1 or not.
void check(int N)
{
    int twos = 0, fives = 0;
 
    // Count the number of 2 in the prime
    // factorisation of N
    while (N % 2 == 0) {
        N /= 2;
        twos++;
    }
 
    // Count the number of 5 in the prime
    // factorisation of N
    while (N % 5 == 0) {
        N /= 5;
        fives++;
    }
 
    if (N == 1 && twos <= fives) {
        cout << 2 * fives - twos;
    }
    else {
        cout << -1;
    }
}
 
// Driver Code
int main()
{
    int N = 50;
 
    check(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
 
class GFG{
     
// Function to check if N can be changed
// to 1 or not.
static void check(int N)
{
    int twos = 0, fives = 0;
 
    // Count the number of 2 in the prime
    // factorisation of N
    while (N % 2 == 0)
    {
        N /= 2;
        twos++;
    }
 
    // Count the number of 5 in the prime
    // factorisation of N
    while (N % 5 == 0)
    {
        N /= 5;
        fives++;
    }
 
    if (N == 1 && twos <= fives)
    {
        System.out.println( 2 * fives - twos);
    }
    else
    {
        System.out.println(-1);
    }
}
 
// Driver Code
public static void main (String[] args)
{
    int N = 50;
     
    check(N);
}
}
 
// This code is contributed by Potta Lokesh

Python3

# Python 3 program for the above approach
 
# Function to check if N can be changed
# to 1 or not.
def check(N):
    twos = 0
    fives = 0
 
    # Count the number of 2 in the prime
    # factorisation of N
    while (N % 2 == 0):
        N /= 2
        twos += 1
 
    # Count the number of 5 in the prime
    # factorisation of N
    while (N % 5 == 0):
        N /= 5
        fives += 1
 
    if (N == 1 and twos <= fives):
        print(2 * fives - twos)
 
    else:
        print(-1)
 
# Driver Code
if __name__ == '__main__':
    N = 50
    check(N)
     
    # This code is contributed by SURENDRA_GANGWAR.

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to check if N can be changed
// to 1 or not.
static void check(int N)
{
    int twos = 0, fives = 0;
 
    // Count the number of 2 in the prime
    // factorisation of N
    while (N % 2 == 0) {
        N /= 2;
        twos++;
    }
 
    // Count the number of 5 in the prime
    // factorisation of N
    while (N % 5 == 0) {
        N /= 5;
        fives++;
    }
 
    if (N == 1 && twos <= fives) {
        Console.Write( 2 * fives - twos);
    }
    else {
        Console.Write(-1);
    }
}
 
// Driver Code
public static void Main()
{
     int N = 50;
 
    check(N);
}
}
 
// This code is contributed by sanjoy_62.

Javascript

输出

时间复杂度： O(log N)
辅助空间： O(1)