📜  查找数字乘以给定数字n的最小数字

📅  最后修改于: 2021-04-29 08:55:47             🧑  作者: Mango

给定数字“ n”,找到最小的数字“ p”,这样,如果我们将“ p”的所有数字相乘,便得到“ n”。结果“ p”应至少包含两位数。
例子:

Input:  n = 36
Output: p = 49 
// Note that 4*9 = 36 and 49 is the smallest such number

Input:  n = 100
Output: p = 455
// Note that 4*5*5 = 100 and 455 is the smallest such number

Input: n = 1
Output:p = 11
// Note that 1*1 = 1

Input: n = 13
Output: Not Possible

对于给定的n,以下是要考虑的两种情况。
情况1:n <10当n小于10时,输出始终为n + 10。例如,对于n = 7,输出为17。对于n = 9,输出为19。
情况2:n> = 10求n的所有因子都在2到9之间(包括2和9)。想法是从9开始搜索,以使结果中的位数最小化。例如,9比33更可取,8比24更可取。
将所有找到的因子存储在数组中。该数组将以非递增顺序包含数字,因此最终以相反顺序打印该数组。
以下是上述概念的实现。

C++
#include
using namespace std;
 
// Maximum number of digits in output
#define MAX 50
 
// prints the smallest number
// whose digits multiply to n
void findSmallest(int n)
{
    int i, j = 0;
     
    // To store digits of result
    // in reverse order
    int res[MAX];
 
    // Case 1: If number is smaller than 10
    if (n < 10)
    {
        cout << n + 10;
        return;
    }
 
    // Case 2: Start with 9 and
    // try every possible digit
    for (i = 9; i > 1; i--)
    {
        // If current digit divides n, then store all
        // occurrences of current digit in res
        while (n % i == 0)
        {
            n = n / i;
            res[j] = i;
            j++;
        }
    }
 
    // If n could not be broken
    // in form of digits (prime factors
    // of n are greater than 9)
    if (n > 10)
    {
        cout << "Not possible";
        return;
    }
 
    // Print the result array in reverse order
    for (i = j - 1; i >= 0; i--)
        cout << res[i];
}
 
// Driver Code
int main()
{
    findSmallest(7);
    cout << "\n";
 
    findSmallest(36);
    cout << "\n";
 
    findSmallest(13);
    cout << "\n";
 
    findSmallest(100);
    return 0;
}
 
// This code is contributed by Code_Mech


C
#include
 
// Maximum number of digits in output
#define MAX 50
 
// prints the smallest number whose digits multiply to n
void findSmallest(int n)
{
    int i, j=0;
    int res[MAX]; // To sore digits of result in reverse order
 
    // Case 1: If number is smaller than 10
    if (n < 10)
    {
        printf("%d", n+10);
        return;
    }
 
    // Case 2: Start with 9 and try every possible digit
    for (i=9; i>1; i--)
    {
        // If current digit divides n, then store all
        // occurrences of current digit in res
        while (n%i == 0)
        {
            n = n/i;
            res[j] = i;
            j++;
        }
    }
 
    // If n could not be broken in form of digits (prime factors of n
    // are greater than 9)
    if (n > 10)
    {
        printf("Not possible");
        return;
    }
 
    // Print the result array in reverse order
    for (i=j-1; i>=0; i--)
        printf("%d", res[i]);
}
 
// Driver program to test above function
int main()
{
    findSmallest(7);
    printf("\n");
 
    findSmallest(36);
    printf("\n");
 
    findSmallest(13);
    printf("\n");
 
    findSmallest(100);
    return 0;
}


Java
// Java program to find the smallest number whose
// digits multiply to a given number n
 
import java.io.*;
 
class Smallest
{
    // Function to prints the smallest number whose
    // digits multiply to n
    static void findSmallest(int n)
    {
        int i, j=0;
        int MAX = 50;
        // To sore digits of result in reverse order
        int[] res = new int[MAX];
  
        // Case 1: If number is smaller than 10
        if (n < 10)
        {
            System.out.println(n+10);
            return;
        }
  
        // Case 2: Start with 9 and try every possible digit
        for (i=9; i>1; i--)
        {
            // If current digit divides n, then store all
            // occurrences of current digit in res
            while (n%i == 0)
            {
                n = n/i;
                res[j] = i;
                j++;
            }
        }
  
        // If n could not be broken in form of digits (prime factors of n
        // are greater than 9)
        if (n > 10)
        {
            System.out.println("Not possible");
            return;
        }
  
        // Print the result array in reverse order
        for (i=j-1; i>=0; i--)
            System.out.print(res[i]);
        System.out.println();
    }
     
    // Driver program
    public static void main (String[] args)
    {
        findSmallest(7);
        findSmallest(36);
        findSmallest(13);
        findSmallest(100);
    }
}
 
// Contributed by Pramod Kumar


Python
# Python code to find the smallest number
# whose digits multiply to give n
 
# function to print the smallest number whose
# digits multiply to n
def findSmallest(n):
    # Case 1 - If the number is smaller than 10
    if n < 10:
        print n+10
        return
     
    # Case 2 - Start with 9 and try every possible digit
    res = [] # to sort digits
    for i in range(9,1,-1):
        # If current digit divides n, then store all
        # occurrences of current digit in res
        while n % i == 0:
            n = n / i
            res.append(i)
     
    # If n could not be broken in the form of digits
    # prime factors of  n are greater than 9
     
    if n > 10:
        print "Not Possible"
        return
         
    # Print the number from result array in reverse order
    n = res[len(res)-1]
    for i in range(len(res)-2,-1,-1):
        n = 10 * n + res[i]
    print n
     
# Driver Code
findSmallest(7)
 
findSmallest(36)
 
findSmallest(13)
 
findSmallest(100)
 
# This code is contributed by Harshit Agrawal


C#
// C# program to find the smallest number whose
// digits multiply to a given number n
using System;
 
class GFG {
     
    // Function to prints the smallest number
    // whose digits multiply to n
    static void findSmallest(int n)
    {
         
        int i, j=0;
        int MAX = 50;
         
        // To sore digits of result in
        // reverse order
        int []res = new int[MAX];
 
        // Case 1: If number is smaller than 10
        if (n < 10)
        {
            Console.WriteLine(n + 10);
            return;
        }
 
        // Case 2: Start with 9 and try every
        // possible digit
        for (i = 9; i > 1; i--)
        {
             
            // If current digit divides n, then
            // store all occurrences of current
            // digit in res
            while (n % i == 0)
            {
                n = n / i;
                res[j] = i;
                j++;
            }
        }
 
        // If n could not be broken in form of
        // digits (prime factors of n
        // are greater than 9)
        if (n > 10)
        {
            Console.WriteLine("Not possible");
            return;
        }
 
        // Print the result array in reverse order
        for (i = j-1; i >= 0; i--)
            Console.Write(res[i]);
             
        Console.WriteLine();
    }
     
    // Driver program
    public static void Main ()
    {
        findSmallest(7);
        findSmallest(36);
        findSmallest(13);
        findSmallest(100);
    }
}
 
// This code is contributed by nitin mittal.


PHP
 1; $i--)
    {
         
        // If current digit divides
        // n, then store all
        // occurrences of current
        // digit in res
        while ($n % $i == 0)
        {
            $n = $n / $i;
            $res[$j] = $i;
            $j++;
        }
    }
 
    // If n could not be broken
    // in form of digits
    // (prime factors of n
    // are greater than 9)
    if ($n > 10)
    {
        echo "Not possible";
        return;
    }
 
    // Print the result
    // array in reverse order
    for ($i = $j - 1; $i >= 0; $i--)
        echo $res[$i];
}
 
    // Driver Code
    findSmallest(7);
    echo "\n";
 
    findSmallest(36);
     
    echo "\n";
 
    findSmallest(13);
    echo "\n";
 
    findSmallest(100);
 
// This code is contributed by ajit
?>


Javascript


输出:

17
49
Not possible
455