通过将 2 的幂相加来形成数字 X 的最小成本

给定一个由N 个整数组成的数组arr[]和一个整数X 。数组中的元素arr[i]表示使用2 ⁱ的成本。任务是找到选择加起来为X的数字的最小成本。
例子：

Input: arr[] = { 20, 50, 60, 90 }, X = 7
Output: 120
2² + 2¹ + 2⁰ = 4 + 2 + 1 = 7 with cost = 60 + 50 + 20 = 130
But we can use 2² + 3 * 2⁰ = 4 + 3 * 1 = 7 with cost = 60 + 3 * 20 = 120 which is minimum possible.
Input: arr[] = { 10, 5, 50 }, X = 4
Output: 10

编程需要懂一点英语

方法：这个问题可以使用基本的动态规划来解决。这里使用了每个数字都可以使用 2 的幂形成的事实。我们可以初步计算形成 2 的幂本身所需的最小成本。重复将如下：

a[i] = min(a[i], 2 * a[i – 1])

编程需要懂一点英语

重新计算数组后，我们可以简单地根据数字X 中的设置位继续添加成本。
下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the minimum cost
int MinimumCost(int a[], int n, int x)
{
 
    // Re-compute the array
    for (int i = 1; i < n; i++) {
        a[i] = min(a[i], 2 * a[i - 1]);
    }
 
    int ind = 0;
 
    int sum = 0;
 
    // Add answers for set bits
    while (x) {
 
        // If bit is set
        if (x & 1)
            sum += a[ind];
 
        // Increase the counter
        ind++;
 
        // Right shift the number
        x = x >> 1;
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int a[] = { 20, 50, 60, 90 };
    int x = 7;
    int n = sizeof(a) / sizeof(a[0]);
    cout << MinimumCost(a, n, x);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG
{
     
// Function to return the minimum cost
static int MinimumCost(int a[], int n, int x)
{
 
    // Re-compute the array
    for (int i = 1; i < n; i++)
    {
        a[i] = Math.min(a[i], 2 * a[i - 1]);
    }
 
    int ind = 0;
 
    int sum = 0;
 
    // Add answers for set bits
    while (x > 0)
    {
 
        // If bit is set
        if (x != 0 )
            sum += a[ind];
 
        // Increase the counter
        ind++;
 
        // Right shift the number
        x = x >> 1;
    }
 
    return sum;
}
 
// Driver code
public static void main (String[] args)
{
 
    int a[] = { 20, 50, 60, 90 };
    int x = 7;
    int n =a.length;
    System.out.println (MinimumCost(a, n, x));
}
}
 
// This Code is contributed by akt_mit

Python3

# Python 3 implementation of the approach
 
# Function to return the minimum cost
def MinimumCost(a, n, x):
     
    # Re-compute the array
    for i in range(1, n, 1):
        a[i] = min(a[i], 2 * a[i - 1])
 
    ind = 0
 
    sum = 0
 
    # Add answers for set bits
    while (x):
         
        # If bit is set
        if (x & 1):
            sum += a[ind]
 
        # Increase the counter
        ind += 1
 
        # Right shift the number
        x = x >> 1
 
    return sum
 
# Driver code
if __name__ == '__main__':
    a = [20, 50, 60, 90]
    x = 7
    n = len(a)
    print(MinimumCost(a, n, x))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Function to return the minimum cost
public static int MinimumCost(int []a, int n, int x)
{
 
    // Re-compute the array
    for (int i = 1; i < n; i++)
    {
        a[i] = Math.Min(a[i], 2 * a[i - 1]);
    }
 
    int ind = 0;
 
    int sum = 0;
 
    // Add answers for set bits
    while (x > 0)
    {
 
        // If bit is set
        if (x != 0 )
            sum += a[ind];
 
        // Increase the counter
        ind++;
 
        // Right shift the number
        x = x >> 1;
    }
 
    return sum;
}
 
// Driver code
public static void Main ()
{
 
    int []a = { 20, 50, 60, 90 };
    int x = 7;
    int n =a.Length;
    Console.WriteLine(MinimumCost(a, n, x));
}
}
 
// This Code is contributed by SoM15242

PHP

> 1;
    }
 
    return $sum;
}
 
    // Driver code
    $a = array( 20, 50, 60, 90 );
    $x = 7;
    $n = sizeof($a) / sizeof($a[0]);
    echo MinimumCost($a, $n, $x);
 
// This code is contributed by ajit.
?>

Javascript

输出：