找到可用于称量[1，X]范围内所有重量的最佳重量

给定一个整数X ，任务是找到一组最佳的权重{w ₁ ，w ₂ ，w ₃ ，…，w _n }，这样我们就可以使用双面加权来加权/确定从1到X的所有权重平衡盘。请注意，所有权重必须唯一，并且n应该尽可能小。

例子：

Input: X = 7
Output: 1 3 9

Weights	Left Side	Right Side
1	1	1
2	2 + 1	3
3	3	3
4	4	1 + 3
5	5 + 1 + 3	9
6	6 + 3	9
7	7 + 3	1 + 9

Input: X = 20
Output: 1 3 9 27

方法：

1. 一种最佳方法是使用权重为3的幂，即{1、3、9、27、81、243，…}
2. 通过归纳可以证明，使用{1，3，9，27，81，…，pow(3，n)} ，我们可以平衡从1到(pow(3，n + 1)– 1)的所有权重。 / 2 。
3. 因此，找到n的值，以便可以平衡从1到X的所有值并打印结果。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
  
// Function to find the optimal weights
void findWeights(int X)
{
    int sum = 0;
  
    // Number of weights required
    int power = 0;
  
    // Finding the value of required powers of 3
    while (sum < X) {
        sum = pow(3, power + 1) - 1;
        sum /= 2;
        power++;
    }
  
    // Optimal Weights are powers of 3
    int ans = 1;
    for (int i = 1; i <= power; i++) {
        cout << ans << " ";
        ans = ans * 3;
    }
}
  
// Driver code
int main()
{
    int X = 2;
  
    findWeights(X);
  
    return 0;
}

Java

// Java implementation of the approach
import java.util.*; 
  
public class GFG 
{ 
      
    // Function to find the optimal weights 
    static void findWeights(int X) 
    { 
        int sum = 0; 
  
        // Number of weights required 
        int power = 0; 
        int number = 3;
          
        // Finding the value of required powers of 3 
        while (sum < X)
        { 
            sum = number - 1; 
            sum /= 2; 
            power++;
            number *= 3;
        } 
  
        // Optimal Weights are powers of 3 
        int ans = 1; 
        for (int i = 1; i <= power; i++)
        { 
            System.out.print(ans + " "); 
            ans = ans * 3; 
        } 
    } 
  
      
    // Driver code 
    public static void main (String[] args) 
    { 
        int X = 2; 
  
        findWeights(X); 
  
    } 
} 
  
// This code is contributed by Sam007.

Python3

# Python3 implementation of the approach
  
# Function to find the optimal weights
def findWeights(X):
  
    sum = 0
  
    # Number of weights required
    power = 0
  
    # Finding the value of required powers of 3
    while (sum < X):
        sum = pow(3, power + 1) - 1
        sum //= 2
        power += 1
  
    # Optimal Weights are powers of 3
    ans = 1
    for i in range(1, power + 1):
        print(ans, end = " ")
        ans = ans * 3
  
# Driver code
X = 2
  
findWeights(X)
  
# This code is contributed by Mohit Kumar

C#

// C# implementation of the approach
using System;
  
class GFG
{
      
    // Function to find the optimal weights 
    static void findWeights(int X) 
    { 
        int sum = 0; 
  
        // Number of weights required 
        int power = 0; 
        int number = 3;
          
        // Finding the value of required powers of 3 
        while (sum < X)
        { 
            sum = number - 1; 
            sum /= 2; 
            power++;
            number *= 3;
        } 
  
        // Optimal Weights are powers of 3 
        int ans = 1; 
        for (int i = 1; i <= power; i++)
        { 
            Console.Write(ans + " "); 
            ans = ans * 3; 
        } 
    } 
  
    // Driver code 
    static public void Main ()
    {
        int X = 2; 
        findWeights(X); 
    } 
} 
  
// This code is contributed by ajit.

输出：

1 3