将 X 数字附加到 N 的末尾以使其可被 M 整除

给定三个正整数N 、 M和X ，任务是通过在N的右侧附加X位来生成一个数字，以便该数字可以被M整除。如果存在多个解决方案，则打印其中任何一个。否则，打印-1 。

例子：

Input: N = 10, M = 5, X = 4 
Output: 105555 
Explanation: One of possible values of N (= 10) by appending X(= 4) digits on the right side of N is 105555, which is divisible by M (= 5).

Input: N = 4, M = 50, X = 2 
Output: 400 
 

方法：我们的想法是对N的右侧追加X位数通过来自区间[0，9]尝试所有可能的数字和上的N，检查右侧附加X数字后，如果数字是整除M或不是。如果发现是真的，则打印数字。否则，打印-1 。以下是递推关系：

$isDiv(N, X) = \Sigma^{9}_{i = 0} isDiv(N * 10 + i, X - 1)$

请按照以下步骤解决问题：

• 使用上面的递推关系，通过在N的右侧附加X位来检查数字N是否可以被M整除。如果发现为真，则通过附加X位来打印N的值。
• 否则，打印-1 。

下面是上述方法的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to check if the value of N by
// appending X digits on right side of N
// is divisible by M or not
bool isDiv(int N, int X, int M, int& res)
{
 
    // Base Case
    if (X == 0) {
 
        // If N is divisible
        // by M
        if (N % M == 0) {
 
            // Update res
            res = N;
            return true;
        }
 
        return false;
    }
 
    // Iterate over the range [0, 9]
    for (int i = 0; i <= 9; i++) {
 
        // If N is divisible by M by
        // appending X digits
        if (isDiv(N * 10 + i, X - 1, M, res)) {
 
            return true;
        }
    }
}
 
// Driver Code
int main()
{
    int N = 4, M = 50, X = 2;
 
    // Stores the number by appending
    // X digits on the right side of N
    int res = -1;
 
    isDiv(N, X, M, res);
    cout << res;
}

Java

// Java program for the above approach
import java.io.*;
import java.util.*;
 
class GFG
{
 
  // Function to check if the value of N by
  // appending X digits on right side of N
  // is divisible by M or not
  static int isDiv(int N, int X, int M, int res)
  {
 
    // Base Case
    if (X == 0)
    {
 
      // If N is divisible
      // by M
      if (N % M == 0)
      {
 
        // Update res
        res = N;
        return res;
      }
 
      return res;
    }
 
    // Iterate over the range [0, 9]
    for (int i = 0; i < 9; i++)
    {
 
      // If N is divisible by M by
      // appending X digits
      int temp = isDiv(N * 10 + i, X - 1, M, res);
      if (temp != -1)
      {
        return temp;
      }
    }
    return res;
  }
 
  // Driver code
  public static void main(String[] args)
    throws java.lang.Exception
  {
    int N = 4, M = 50, X = 2;
 
    // Stores the number by appending
    // X digits on the right side of N
    int res = -1;
 
    res = isDiv(N, X, M, res);
    System.out.println(res);
  }
}
 
// This code is contributed by 18bhupenderyadav18.

Python3

# Python3 program to implement
# the above approach
 
# Function to check if the value of N by
# appending X digits on right side of N
# is divisible by M or not
 
# global variable to store result
res = -1
 
 
def isDiv(N, X, M):
    # Base case
    if(X == 0):
        # If N is divisible
        # by M
        if(N % M == 0):
            global res
            res = N
            return True
 
        return False
 
    # Iterate over the range [0, 9]
    for i in range(10):
        # if N is Divisible by M upon appending X digits
        if(isDiv(N*10+i, X-1, M)):
            return True
 
 
# Driver Code
if __name__ == "__main__":
    N, M, X = 4, 50, 2
 
    if(isDiv(N, X, M)):
        print(res)
    else:
        print("-1")

C#

// C# program to implement
// the above approach 
using System;
  
class GFG
{
  
  // Function to check if the value of N by
  // appending X digits on right side of N
  // is divisible by M or not
  static int isDiv(int N, int X, int M, int res)
  {
  
    // Base Case
    if (X == 0)
    {
  
      // If N is divisible
      // by M
      if (N % M == 0)
      {
  
        // Update res
        res = N;
        return res;
      }
  
      return res;
    }
  
    // Iterate over the range [0, 9]
    for (int i = 0; i < 9; i++)
    {
  
      // If N is divisible by M by
      // appending X digits
      int temp = isDiv(N * 10 + i, X - 1, M, res);
      if (temp != -1)
      {
        return temp;
      }
    }
    return res;
  }
  
  // Driver code
  public static void Main()
  {
    int N = 4, M = 50, X = 2;
  
    // Stores the number by appending
    // X digits on the right side of N
    int res = -1;
  
    res = isDiv(N, X, M, res);
    Console.WriteLine(res);
  }
}
 
// This code is contributed by sanjoy_62

Javascript

 
// Javascript program to implement
// the above approach
 
var res = -1;
 
// Function to check if the value of N by
// appending X digits on right side of N
// is divisible by M or not
function isDiv(N, X, M)
{
 
    // Base Case
    if (X == 0) {
 
        // If N is divisible
        // by M
        if (N % M == 0) {
 
            // Update res
            res = N;
            return true;
        }
 
        return false;
    }
 
    // Iterate over the range [0, 9]
    for (var i = 0; i <= 9; i++) {
 
        // If N is divisible by M by
        // appending X digits
        if (isDiv(N * 10 + i, X - 1, M)) {
 
            return true;
        }
    }
}
 
// Driver Code
 
var N = 4, M = 50, X = 2;
 
// Stores the number by appending
// X digits on the right side of N
isDiv(N, X, M, res);
 
document.write(res);
 
 
 

输出

400

时间复杂度： O(10 ^X )
辅助空间： O(1)