将N位数字添加到A，以便每次添加后都能被B整除

给定三个整数A ， B和N ，将以下过程重复N次：

1. 向A添加一个数字，以便在添加之后， A被B整除。
2. 在上述操作的N次迭代之后，打印可能的最小值A。
3. 如果操作失败，则打印-1 。

注意：每添加一位数字后，我们需要检查可除性。

例子：

Input: A = 10, B = 11, N = 1
Output: -1
No matter what digit you add, 10X will never be divisible by 11.

Input: A = 5, B = 3, N = 3
Output: 5100

方法：将第一个数字从0到9进行暴力破解，如果没有一个数字使A被B整除，则答案为-1 。否则，添加满足条件的第一个数字，然后在(n-1)次之后添加0 ，因为如果A被B整除，那么A * 10，A * 100，…也将被B整除。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
int addNDigits(int a, int b, int n)
{
    int num = a;
  
    // Try all digits from (0 to 9)
    for (int i = 0; i <= 9; i++) {
        int tmp = a * 10 + i;
        if (tmp % b == 0) {
            a = tmp;
            break;
        }
    }
  
    // Fails in the first move itself
    if (num == a)
        return -1;
  
    // Add (n-1) 0's
    for (int j = 0; j < n - 1; j++)
        a *= 10;
  
    return a;
}
  
// Driver Program to test above function
int main()
{
    int a = 5, b = 3, n = 3;
    cout << addNDigits(a, b, n);
    return 0;
}

Java

//Java implementation of the approach
  
import java.io.*;
  
class GFG {
  
  
static int addNDigits(int a, int b, int n)
{
    int num = a;
  
    // Try all digits from (0 to 9)
    for (int i = 0; i <= 9; i++) {
        int tmp = a * 10 + i;
        if (tmp % b == 0) {
            a = tmp;
            break;
        }
    }
  
    // Fails in the first move itself
    if (num == a)
        return -1;
  
    // Add (n-1) 0's
    for (int j = 0; j < n - 1; j++)
        a *= 10;
  
    return a;
}
  
// Driver Program to test above function
  
    public static void main (String[] args) {
    int a = 5, b = 3, n = 3;
    System.out.print( addNDigits(a, b, n));
    }
}
// This code is contributed by anuj_67..

Python3

# Python3 implementation of the approach
def addNDigits(a, b, n) :
  
    num = a
      
    # Try all digits from (0 to 9)
    for i in range(10) :
        tmp = a * 10 + i
          
        if (tmp % b == 0) :
            a = tmp
            break
          
    # Fails in the first move itself
    if (num == a) :
        return -1
  
    # Add (n-1) 0's
    for j in range(n - 1) :
        a *= 10
  
    return a
  
# Driver Code
if __name__ == "__main__" :
      
    a = 5
    b = 3
    n = 3
  
    print(addNDigits(a, b, n))
  
# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
  
class GFG 
{
  
static int addNDigits(int a, 
                      int b, int n)
{
    int num = a;
  
    // Try all digits from (0 to 9)
    for (int i = 0; i <= 9; i++)
    {
        int tmp = a * 10 + i;
        if (tmp % b == 0) 
        {
            a = tmp;
            break;
        }
    }
  
    // Fails in the first move itself
    if (num == a)
        return -1;
  
    // Add (n-1) 0's
    for (int j = 0; j < n - 1; j++)
        a *= 10;
  
    return a;
}
  
// Driver Code
public static void Main () 
{
    int a = 5, b = 3, n = 3;
    Console.WriteLine(addNDigits(a, b, n));
}
}
  
// This code is contributed 
// by anuj_67..

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输出：

5100