用其他权力代表数字

给定两个数字w和m，我们需要确定是否有可能用w的幂表示m。可以将w的幂相加或减去以获得m，而w的每个幂只能使用一次。

例子：

Input : 3 7
Output : Yes
As 7 = 9 - 3 + 1 (3^2 - 3^1 + 3^0 )
so it is possible .

Input : 100 50
Output : No
As 50 is less than 100 so we can never
represent it in the powers of 100 .

在这里，我们必须用仅使用一次的w的幂来表示m，以便可以通过以下等式表示它。
c0 + c1 * w ^ 1 + c2 * w ^ 2 +…= m-(等式1)

其中每个c0，c1，c2…分别为-1 (用于减去w的幂)， 0 (不使用w的幂)， 1 (用于加w的幂)。

=> c1 * w ^ 1 + c2 * w ^ 2 +…= m – c0
=> w(c1 + c2 * w ^ 1 + c3 * w ^ 2 +…)= m – c0
=> c1 + c2 * w ^ 1 + c3 * w ^ 2 +…=(m – c0)/ w-(等式2)

现在，请注意方程式1和方程式2-我们正在尝试再次解决同一问题。因此，我们必须递归直到m> 0。为了使这种解决方案存在(m_ci)必须是w的倍数，其中ci是方程式的系数。 ci可以为-1，0，1。因此，我们必须检查所有三种可能性((m – 1)％w == 0)，((m + 1)％w == 0)和(m％w == 0) 。如果不是，那么将没有任何解决方案。

C++

// CPP program to check if m can be represented
// as powers of w.
#include 
using namespace std;
  
bool asPowerSum(int w, int m)
{
    while (m) {
        if ((m - 1) % w == 0) 
            m = (m - 1) / w;
       else if ((m + 1) % w == 0) 
            m = (m + 1) / w;
          
        else if (m % w == 0) 
            m = m / w;
          
        else
            break; // None of 3 worked.
    }
  
    // If m is not zero means, it can't be 
    // represented in terms of powers of w.
    return (m == 0);
}
  
// Driver code
int main()
{
    int w = 3, m = 7;
    if (asPowerSum(w, m))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
   return 0;
}

Java

// Java program to check if m can 
// be represented as powers of w.
  
class GFG
{
    static boolean asPowerSum(int w, int m)
    {
        while (m > 0) 
        {
            if ((m - 1) % w == 0) 
                m = (m - 1) / w;
              
            else if ((m + 1) % w == 0) 
                m = (m + 1) / w;
              
            else if (m % w == 0) 
                m = m / w;
              
            else
                break; // None of 3 worked.
        }
      
        // If m is not zero means, it can't be 
        // represented in terms of powers of w.
        return (m == 0);
    } 
      
    // Driver function
    public static void main (String[] args)
    {
        int w = 3, m = 7;
        if (asPowerSum(w, m))
            System.out.println("Yes");
        else
            System.out.println("No"); 
    }
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python3 program to check if m can 
# be represented as powers of w.
def asPowerSum(w, m):
    while (m > 0):
        if ((m - 1) % w == 0):
            m = (m - 1) / w;
          
        elif ((m + 1) % w == 0):
            m = (m + 1) / w;
          
        elif (m % w == 0):
            m = m / w;
          
        else:
            break; # None of 3 worked.
      
    # If m is not zero means, it can't be
    # represented in terms of powers of w.
    return (m == 0);
  
# Driver code
w = 3; 
m = 7;
if (asPowerSum(w, m)):
    print("Yes");
else:
    print("No");
  
# This code is contributed by mits

C#

// C# program to check if 
// m can be represented
// as powers of w.
using System;
  
class GFG
{
    static bool asPowerSum(int w, 
                           int m)
    {
        while (m > 0) 
        {
            if ((m - 1) % w == 0) 
                m = (m - 1) / w;
              
            else if ((m + 1) % w == 0) 
                m = (m + 1) / w;
              
            else if (m % w == 0) 
                m = m / w;
              
            else
                break; // None of 3 worked.
        }
      
        // If m is not zero means, 
        // it can't be represented
        // in terms of powers of w.
        return (m == 0);
    } 
      
    // Driver Code
    static public void Main ()
    {
        int w = 3, m = 7;
        if (asPowerSum(w, m))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No"); 
    }
}
  
// This code is contributed 
// by akt_mit.

输出：

Yes