总和等于n的最小功率项数

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📜 总和等于n的最小功率项数

📅 最后修改于: 2021-04-26 08:14:21 🧑 作者: Mango

给定两个正整数n和x 。任务是将n表示为x(x ^a1 + x ^a2 + ….. + x ^a3 )的幂的和，这样x(x ^a1 ，x ^a2 ，…..，x ^a3 )的幂数应为最低限度。打印用于使和等于n的x的最小乘方数。
例子：

Input : n = 5, x = 3
Output : 3
5 = 30 + 30 + 31.
We use only 3 power terms of x { 30, 30, 31} 

Input : n = 13, x = 4
Output : 4
13 = 40 + 41 + 41 + 41.
We use only four power terms of x.

Input : n = 6, x = 1
Output : 6

如果x = 1，则答案将仅为n(n = 1 +1 + …. n次)。
这个想法是使用霍纳法。任何数字n都可以表示为n = x * a + b，其中0 <= b <= x-1。现在，由于b在0到x – 1之间，因此b应该表示为x ⁰ b次的总和。
此外，a可以类似的方式分解等等。
解决此问题的算法：

1. Initialize a variable ans to 0.
2. While n > 0
   a) ans = ans + n % x
   b) n = n/x
3. Return ans.

下面是上述想法的实现：

C++

// C++ program to calculate minimum number
// of powers of x to make sum equal to n.
#include 
using namespace std;
 
// Return minimum power terms of x required
int minPower(int n, int x)
{
    // if x is 1, return n since any power
    // of 1 is 1 only.
    if (x==1)
        return n;
 
    // Consider n = a * x  + b where a = n/x
    // and b = n % x.
    int ans = 0;
    while (n > 0)
    {
        // Update count of powers for 1's added
        ans += (n%x);
 
        // Repeat the process for reduced n
        n /= x;
    }
 
    return ans;
}
 
// Driven Program
int main()
{
    int n = 5, x = 3;
    cout << minPower(n, x) << endl;
    return 0;
}

Java

// Java program to calculate
// minimum numberof powers of
// x to make sum equal to n.
 
class GFG
{
    // Return minimum power
    // terms of x required
    static int minPower(int n, int x)
    {
        // if x is 1, return n since
        // any power of 1 is 1 only.
        if (x==1)
            return n;
     
        // Consider n = a * x + b where
        // a = n/x and b = n % x.
        int ans = 0;
        while (n > 0)
        {
            // Update count of powers
            // for 1's added
            ans += (n % x);
     
            // Repeat the process for reduced n
            n /= x;
        }
     
        return ans;
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int n = 5, x = 3;
        System.out.println(minPower(n, x));
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# Python program to
# calculate minimum number
# of powers of x to make
# sum equal to n.
 
# Return minimum power
# terms of x required
def minPower(n,x):
 
    # if x is 1, return
    # n since any power
    # of 1 is 1 only.
    if (x==1):
        return n
  
    # Consider n = a * x  + b where a = n/x
    # and b = n % x.
    ans = 0
    while (n > 0):
 
        # Update count of powers for 1's added
        ans += (n%x)
  
        # Repeat the process for reduced n
        n //= x
 
  
    return ans
     
# Driver code
n = 5
x = 3
print(minPower(n, x))
 
# This code is contributed
# by Anant Agarwal.

C#

// C# program to calculate
// minimum numberof powers
// of x to make sum equal
// to n.
using System;
 
class GFG
{
     
    // Return minimum power
    // terms of x required
    static int minPower(int n, int x)
    {
        // if x is 1, return n since
        // any power of 1 is 1 only.
        if (x == 1)
            return n;
     
        // Consider n = a * x + b where
        // a = n / x and b = n % x.
        int ans = 0;
        while (n > 0)
        {
            // Update count of
            // powers for 1's
            // added
            ans += (n % x);
     
            // Repeat the process
            // for reduced n
            n /= x;
        }
     
        return ans;
    }
     
    // Driver code
    public static void Main ()
    {
        int n = 5, x = 3;
        Console.WriteLine(minPower(n, x));
    }
}
 
// This code is contributed by vt_m.

PHP

 0)
    {
        // Update count of powers
        // for 1's added
        $ans += ($n % $x);
 
        // Repeat the process
        // for reduced n
        $n /= $x;
    }
 
    return $ans;
}
 
// Driver Code
$n = 5; $x = 3;
echo(minPower($n, $x));
 
// This code is contributed by Ajit.
?>

Javascript

输出：