找到可用于对 [1, X] 范围内的所有权重进行加权的最佳权重

给定一个整数X ，任务是找到一组最佳权重{w ₁ , w ₂ , w ₃ , …, w _n } ，这样我们就可以使用双边称重来衡量/确定从1到X的所有权重平衡盘。请注意，所有权重必须是唯一的，并且n应尽可能小。
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 3 的幂的权重，即{1, 3, 9, 27, 81, 243, …}
2. 可以通过归纳证明，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，使得从1到X 的所有值都可以平衡并打印结果。

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.

Javascript

 
// JavaScript implementation of the approach
 
 // Function to find the optimal weights
function findWeights(X)
{
    let sum = 0;
   
        // Number of weights required
        let power = 0;
        let number = 3;
           
        // Finding the value of required powers of 3
        while (sum < X)
        {
            sum = number - 1;
            sum = Math.floor(sum/2);
            power++;
            number *= 3;
        }
   
        // Optimal Weights are powers of 3
        let ans = 1;
        for (let i = 1; i <= power; i++)
        {
            document.write(ans + " ");
            ans = ans * 3;
        }
}
 
// Driver code
let X = 2;
   
findWeights(X);
 
 
// This code is contributed by unknown2108
 

输出：

1 3

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