打印所有数字，直到N，其数字乘积等于K

给定两个整数N和K ，任务是打印范围为[1，N]的所有数字，这些数字的乘积等于K。如果找不到这样的数字，则打印“ -1” 。

例子：

Input: N =100, K = 25
Output: 55
Explanation: There is only a single number 55 whose product of digits is equal to K.

Input: N = 500, K = 10
Output: 25 52 125 152 215 251 

方法：请按照以下步骤解决问题：

• 初始化一个变量，例如flag，以存储是否存在满足给定条件的任何数字。
• 声明一个函数productOfDigits()，以查找数字的乘积。
• 迭代范围[1，N] ：
  • 如果arr [i]的数字乘积等于K ，则打印该数字并设置flag = 1 。
• 如果flag等于0 ，则意味着在[1，N]范围内找不到该数字。因此，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 find the product
// of digits of a number
int productOfDigits(int N)
{
    // Stores the product of
    // digits of a number
    int product = 1;
 
    while (N != 0) {
        product = product * (N % 10);
        N = N / 10;
    }
 
    // Return the product
    return product;
}
 
// Function to print all numbers upto
// N having product of digits equal to K
void productOfDigitsK(int N, int K)
{
    // Stores whether any number satisfying
    // the given conditions exists or not
    int flag = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= N; ++i) {
 
        // If product of digits of
        // arr[i] is equal to K or not
        if (K == productOfDigits(i)) {
 
            // Print that number
            cout << i << " ";
            flag = 1;
        }
    }
 
    // If no numbers are found
    if (flag == 0)
        cout << "-1";
}
 
// Driver Code
int main()
{
    // Given value of N & K
    int N = 500, K = 10;
 
    // Function call to print all numbers
    // from [1, N] with product of digits K
    productOfDigitsK(N, K);
}

Java

// Java Program to implement
// the above approach
import java.io.*;
import java.util.*;
 
class GFG {
 
  // Function to find the product
  // of digits of a number
  static int productOfDigits(int N)
  {
     
    // Stores the product of
    // digits of a number
    int product = 1;
 
    while (N != 0) {
      product = product * (N % 10);
      N = N / 10;
    }
 
    // Return the product
    return product;
  }
 
  // Function to print all numbers upto
  // N having product of digits equal to K
  static void productOfDigitsK(int N, int K)
  {
    // Stores whether any number satisfying
    // the given conditions exists or not
    int flag = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= N; ++i) {
 
      // If product of digits of
      // arr[i] is equal to K or not
      if (K == productOfDigits(i)) {
 
        // Print that number
        System.out.print(i + " ");
        flag = 1;
      }
    }
 
    // If no numbers are found
    if (flag == 0)
      System.out.println(-1);
  }
 
  // Driver Code
  public static void main(String[] args)
  {
     
    // Given value of N & K
    int N = 500, K = 10;
 
    // Function call to print all numbers
    // from [1, N] with product of digits K
    productOfDigitsK(N, K);
  }
}
 
// This code is contribute by Kingash.

Python3

# Python3 program for the above approach
 
# Function to find the product
# of digits of a number
def productOfDigits(N) :
 
    # Stores the product of
    # digits of a number
    product = 1;
 
    while (N != 0) :
        product = product * (N % 10);
        N = N // 10;
 
    # Return the product
    return product;
 
# Function to print all numbers upto
# N having product of digits equal to K
def productOfDigitsK(N, K) :
 
    # Stores whether any number satisfying
    # the given conditions exists or not
    flag = 0;
 
    # Iterate over the range [1, N]
    for i in range(1, N + 1) :
 
        # If product of digits of
        # arr[i] is equal to K or not
        if (K == productOfDigits(i)) :
 
            # Print that number
            print(i, end =" ");
            flag = 1;
 
    # If no numbers are found
    if (flag == 0) :
        print("-1");
 
# Driver Code
if __name__ == "__main__" :
     
    # Given value of N & K
    N = 500; K = 10;
 
    # Function call to print all numbers
    # from [1, N] with product of digits K
    productOfDigitsK(N, K);
     
    # This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
class GFG
{
 
  // Function to find the product
  // of digits of a number
  static int productOfDigits(int N)
  {
 
    // Stores the product of
    // digits of a number
    int product = 1;
 
    while (N != 0) {
      product = product * (N % 10);
      N = N / 10;
    }
 
    // Return the product
    return product;
  }
 
  // Function to print all numbers upto
  // N having product of digits equal to K
  static void productOfDigitsK(int N, int K)
  {
 
    // Stores whether any number satisfying
    // the given conditions exists or not
    int flag = 0;
 
    // Iterate over the range [1, N]
    for (int i = 1; i <= N; ++i) {
 
      // If product of digits of
      // arr[i] is equal to K or not
      if (K == productOfDigits(i)) {
 
        // Print that number
        Console.Write(i + " ");
        flag = 1;
      }
    }
 
    // If no numbers are found
    if (flag == 0)
      Console.WriteLine(-1);
  }
 
 
  // Driver Code
  static public void Main()
  {
 
    // Given value of N & K
    int N = 500, K = 10;
 
    // Function call to print all numbers
    // from [1, N] with product of digits K
    productOfDigitsK(N, K);
  }
}
 
// This code is contributed by jana_sayantan.

输出

25 52 125 152 215 251 

时间复杂度： O(N * logN)
辅助空间： O(1)