通过选择对使得 arr[i] >= arr[j] 并将 arr[i] 替换为 arr[i] – arr[j] 来最小化 Array 的最后一个剩余元素

给定一个包含N个正整数的数组arr[] ，任务是在执行任意次数以下操作后，找到数组中最后一个剩余元素的最小可能值：

选择一对索引(i, j)使得arr[i] >= arr[j]并将arr[i]替换为arr[i] – arr[j] 。
如果数组元素arr[i] <= 0 ，则将其从数组中删除。

例子：

Input: arr[] = {2, 4, 8, 32}
Output: 2
Explanation: In first 4 operations, select (i, j) as (3, 2). Hence, the array after 4 operations will become arr[] = {2, 4, 8, 0}. Here, arr[3] can be removed as arr[3] = 0. Similarly, perform the operation twice for (i, j) = (2, 1). The array after operations is arr[] = {2, 4}. Now perform the given operation twice for (i, j) as (1, 0). The final array will be arr[] = {2}. Therefore, the last remaining element is 2, which is the minimum possible.

Input: arr[] = {5, 13, 8, 10}
Output: 1

编程需要懂一点英语

方法：给定的问题可以通过以下观察来解决：

根据基本欧几里得算法求两个整数 (x, y) 的 GCD，可以推导出GCD(x, y) = GCD(x, y – x) ，如果y > x ，否则， x和y可以简单地交换。直到y – x的值减小到 0，这种关系才会成立。
因此，可以得出结论，通过从较小的值中减去较大的值，对(x, y)的最小可达值是GCD(x, y) 。

因此，使用上述观察，所需的答案将是给定数组arr[]的所有元素的 GCD。

下面是上述方法的实现：

C++

// C++ Program os the above approach
#include 
using namespace std;
 
// Function to minimize the last
// remaining element of array
int minValue(int arr[], int n)
{
    // Stores the required value
    int ans;
 
    // Initialize answer
    ans = arr[0];
 
    // Loop to traverse arr[]
    for (int i = 1; i < n; i++) {
        // Calculate the GCD
        ans = __gcd(ans, arr[i]);
    }
 
    // Return Answer
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 5, 13, 8, 10 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << minValue(arr, N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
  // Recursive function to return gcd of a and b
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
 
  // Function to minimize the last
  // remaining element of array
  static int minValue(int arr[], int n)
  {
    // Stores the required value
    int ans;
 
    // Initialize answer
    ans = arr[0];
 
    // Loop to traverse arr[]
    for (int i = 1; i < n; i++) {
      // Calculate the GCD
      ans = gcd(ans, arr[i]);
    }
 
    // Return Answer
    return ans;
  }
 
  // Driver Code
  public static void main (String[] args) {
    int arr[] = { 5, 13, 8, 10 };
    int N = arr.length;
    System.out.println(minValue(arr, N));
  }
}
 
// This code is contributed by hrithikgarg03188.

Python3

# Python Program os the above approach
def __gcd (a, b):
    if (not b):
        return a;
    return __gcd(b, a % b);
 
# Function to minimize the last
# remaining element of array
def minValue (arr, n):
 
    # Stores the required value
    ans = None
 
    # Initialize answer
    ans = arr[0];
 
    # Loop to traverse arr[]
    for i in range(1, n):
        # Calculate the GCD
        ans = __gcd(ans, arr[i]);
     
 
    # Return Answer
    return ans;
 
# Driver Code
arr = [5, 13, 8, 10];
N = len(arr)
print(minValue(arr, N));
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
 
class GFG {
 
  // Recursive function to return gcd of a and b
  static int gcd(int a, int b)
  {
    if (b == 0)
      return a;
    return gcd(b, a % b);
  }
 
  // Function to minimize the last
  // remaining element of array
  static int minValue(int []arr, int n)
  {
    // Stores the required value
    int ans;
 
    // Initialize answer
    ans = arr[0];
 
    // Loop to traverse arr[]
    for (int i = 1; i < n; i++) {
      // Calculate the GCD
      ans = gcd(ans, arr[i]);
    }
 
    // Return Answer
    return ans;
  }
 
  // Driver Code
  public static void Main () {
    int []arr = { 5, 13, 8, 10 };
    int N = arr.Length;
    Console.Write(minValue(arr, N));
  }
}
 
// This code is contributed by Samim Hossain Mondal.

Javascript

输出

时间复杂度： O(N)
辅助空间： O(1)