最小除以10并乘以2即可将给定数减少为1

给定一个整数N ，任务是将N减为1的最小数量的运算(包括乘以2和除以10) 。如果无法获得1 ，则打印“ -1” 。

例子：

Input: N = 5
Output: 2
Explanation:
Below are the operations performed:
1st operation: Multiply N by 2. Therefore, N = 5 * 2 = 10.
2nd operation: Divide N by 10. Therefore, N = 10/10 = 1.
Therefore, minimum number of operations required is 2.

Input: N = 4
Output: -1

方法：想法是检查给定数M的素数。如果给定数具有2和5以外的素数，则无法通过给定操作将给定数减少为1 。如果2的数量在其素数上超过5的数量，则不可能将N减小为1，因为无法将2的所有幂都减小。
请按照以下步骤解决问题：

• 计算存在于N的素因子中的2 s的数量，并将其存储在变量中，例如cnt2 ，并将N更新为N / 2 ^cnt2 。
• 计算存在于N的素因子中的5 s的数目，并将其存储在变量中，例如cnt5 ，并将N更新为N / 5 ^cnt5 。
• 完成上述步骤后，如果N为1且cnt2≤cnt5 ，则所需的最小步骤数为2 * cnt5 – cnt2 。
• 否则，打印“ -1”，因为使用给定的操作无法将N减小为1 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum number
// operations required to reduce N to 1
int minimumMoves(int n)
{
 
    // Stores count of powers of 2 and 5
    int cnt2 = 0, cnt5 = 0;
 
    // Calculating the primefactors 2
    while (n % 2 == 0) {
        n /= 2;
        cnt2++;
    }
 
    // Calculating the primefactors 5
    while (n % 5 == 0) {
        n /= 5;
        cnt5++;
    }
 
    // If n is 1 and cnt2 <= cnt5
    if (n == 1 && cnt2 <= cnt5) {
 
        // Return the minimum operations
        return 2 * cnt5 - cnt2;
    }
 
    // Otherwise, n can't be reduced
    else
        return -1;
}
 
// Driver Code
int main()
{
    // Given Number N
    int N = 25;
 
    // Function Call
    cout << minimumMoves(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
 
// Function to find the minimum number
// operations required to reduce N to 1
static int minimumMoves(int n)
{
     
    // Stores count of powers of 2 and 5
    int cnt2 = 0, cnt5 = 0;
 
    // Calculating the primefactors 2
    while (n % 2 == 0)
    {
        n /= 2;
        cnt2++;
    }
 
    // Calculating the primefactors 5
    while (n % 5 == 0)
    {
        n /= 5;
        cnt5++;
    }
 
    // If n is 1 and cnt2 <= cnt5
    if (n == 1 && cnt2 <= cnt5)
    {
         
        // Return the minimum operations
        return 2 * cnt5 - cnt2;
    }
 
    // Otherwise, n can't be reduced
    else
        return -1;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given Number N
    int N = 25;
 
    // Function Call
    System.out.print(minimumMoves(N));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to find the minimum number
# operations required to reduce N to 1
def minimumMoves(n):
 
    # Stores count of powers of 2 and 5
    cnt2 = 0
    cnt5 = 0
 
    # Calculating the primefactors 2
    while (n % 2 == 0):
        n //= 2
        cnt2 += 1
 
    # Calculating the primefactors 5
    while (n % 5 == 0):
        n //= 5
        cnt5 += 1
 
    # If n is 1 and cnt2 <= cnt5
    if (n == 1 and cnt2 <= cnt5):
         
        # Return the minimum operations
        return 2 * cnt5 - cnt2
 
    # Otherwise, n can't be reduced
    else:
        return -1
 
# Driver Code
if __name__ == '__main__':
     
    # Given Number N
    N = 25
 
    # Function Call
    print(minimumMoves(N))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Function to find the minimum number
// operations required to reduce N to 1
static int minimumMoves(int n)
{
     
    // Stores count of powers of 2 and 5
    int cnt2 = 0, cnt5 = 0;
 
    // Calculating the primefactors 2
    while (n % 2 == 0)
    {
        n /= 2;
        cnt2++;
    }
 
    // Calculating the primefactors 5
    while (n % 5 == 0)
    {
        n /= 5;
        cnt5++;
    }
 
    // If n is 1 and cnt2 <= cnt5
    if (n == 1 && cnt2 <= cnt5)
    {
         
        // Return the minimum operations
        return 2 * cnt5 - cnt2;
    }
 
    // Otherwise, n can't be reduced
    else
        return -1;
}
 
// Driver Code
public static void Main()
{
     
    // Given Number N
    int N = 25;
 
    // Function Call
    Console.WriteLine(minimumMoves(N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR

输出：

4

时间复杂度： O(log N)
辅助空间： O(1)