查找给定方程的解数

给定三个整数A ， B和C ，任务是找到X的值的计数，以便满足以下条件，
X = B * Sm(X) ^A + C其中Sm(X)表示X的数字总和，且1 9 。

例子：

Input: A = 3, B = 2, C = 8
Output: 3
For X = 10, 2 * (1)³ + 8 = 10
For X = 2008, 2 * (10)³ + 8 = 2008
For X = 13726, 2 * (19)³ + 8 = 13726

Input: A = 2, B = 3, C = 10
Output: 1

方法：在这里可以得出一个重要的观察结果：数字的总和最多为81(即X = 999999999)，并且与每个数字的总和相对应，我们得到单个X值。因此，我们可以遍历每个数字的总和，并检查X的结果值是否有效。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to return the count of
// valid values of X
int getCount(int a, int b, int c)
{
    int count = 0;
  
    // Iterate through all possible
    // sum of digits of X
    for (int i = 1; i <= 81; i++) {
  
        // Get current value of X for sum of digits i
        int cr = b * pow(i, a) + c;
  
        int tmp = cr;
        int sm = 0;
  
        // Find sum of digits of cr
        while (tmp) {
            sm += tmp % 10;
            tmp /= 10;
        }
  
        // If cr is a valid choice for X
        if (sm == i && cr < 1e9)
            count++;
    }
  
    // Return the count
    return count;
}
  
// Driver code
int main()
{
    int a = 3, b = 2, c = 8;
    cout << getCount(a, b, c);
  
    return 0;
}

Java

// Java implementation of the approach 
class GfG 
{ 
  
// Function to return the count of 
// valid values of X 
static int getCount(int a, int b, int c) 
{ 
    int count = 0; 
  
    // Iterate through all possible 
    // sum of digits of X 
    for (int i = 1; i <= 81; i++) 
    { 
  
        // Get current value of X for sum of digits i 
        int cr = b * (int)Math.pow(i, a) + c; 
  
        int tmp = cr; 
        int sm = 0; 
  
        // Find sum of digits of cr 
        while (tmp != 0) 
        { 
            sm += tmp % 10; 
            tmp /= 10; 
        } 
  
        // If cr is a valid choice for X 
        if (sm == i && cr < 1e9) 
            count++; 
    } 
  
    // Return the count 
    return count; 
} 
  
// Driver code 
public static void main(String[] args) 
{ 
    int a = 3, b = 2, c = 8; 
    System.out.println(getCount(a, b, c)); 
}
} 
  
// This code is contributed by Prerna Saini.

Python3

# Python3 implementation of the approach
  
# Function to return the count of
# valid values of X
def getCount(a, b, c):
  
    count = 0
  
    # Iterate through all possible
    # sum of digits of X
    for i in range(1, 82):
  
        # Get current value of X for
        # sum of digits i
        cr = b * pow(i, a) + c
  
        tmp = cr
        sm = 0
  
        # Find sum of digits of cr
        while (tmp):
            sm += tmp % 10
            tmp //= 10
          
        # If cr is a valid choice for X
        if (sm == i and cr < 10**9):
            count += 1
      
    # Return the count
    return count
  
# Driver code
a, b, c = 3, 2, 8
print(getCount(a, b, c))
  
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach 
using System;
  
class GFG 
{ 
  
// Function to return the count of 
// valid values of X 
static int getCount(int a, int b, int c) 
{ 
    int count = 0; 
  
    // Iterate through all possible 
    // sum of digits of X 
    for (int i = 1; i <= 81; i++) 
    { 
  
        // Get current value of X for sum 
        // of digits i 
        int cr = b * (int)Math.Pow(i, a) + c; 
  
        int tmp = cr; 
        int sm = 0; 
  
        // Find sum of digits of cr 
        while (tmp != 0) 
        { 
            sm += tmp % 10; 
            tmp /= 10; 
        } 
  
        // If cr is a valid choice for X 
        if (sm == i && cr < 1e9) 
            count++; 
    } 
  
    // Return the count 
    return count; 
} 
  
// Driver code 
public static void Main() 
{ 
    int a = 3, b = 2, c = 8; 
    Console.Write(getCount(a, b, c)); 
}
} 
  
// This code is contributed
// by Akanksha Rai

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输出：

3