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📜  最多替换给定数组中的一个元素后最大可能的 GCD

📅  最后修改于: 2021-09-22 09:39:17             🧑  作者: Mango

给定一个大小为N > 1的数组arr[] 。任务是通过最多替换一个元素来找到数组的最大可能 GCD。
例子:

方法:

  • 想法是找到所有长度为(N – 1)的子序列的 GCD 值,并删除子序列中必须替换的元素,因为它可以被子序列中已有的任何其他元素替换。找到的最大 GCD 就是答案。
  • 为了最优地找到子序列的 GCD,使用单状态动态规划维护一个prefixGCD[]和一个suffixGCD[]数组。
  • GCD(prefixGCD[i – 1], suffixGCD[i + 1])的最大值就是要求的答案。另请注意,如果必须替换第一个或最后一个元素,还需要比较suffixGCD[1]prefixGCD[N – 2]

下面是上述方法的实现:

C++
// C++ implementation of the approach
#include 
using namespace std;
 
// Function to return the maximum
// possible gcd after replacing
// a single element
int MaxGCD(int a[], int n)
{
 
    // Prefix and Suffix arrays
    int Prefix[n + 2];
    int Suffix[n + 2];
 
    // Single state dynamic programming relation
    // for storing gcd of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (int i = 2; i <= n; i += 1) {
        Prefix[i] = __gcd(Prefix[i - 1], a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing gcd of all the elements having
    // index greater than or equal to i in Suffix[i]
    for (int i = n - 1; i >= 1; i -= 1) {
        Suffix[i] = __gcd(Suffix[i + 1], a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be replaced
    int ans = max(Suffix[2], Prefix[n - 1]);
 
    // If any other element is replaced
    for (int i = 2; i < n; i += 1) {
        ans = max(ans, __gcd(Prefix[i - 1], Suffix[i + 1]));
    }
 
    // Return the maximized gcd
    return ans;
}
 
// Driver code
int main()
{
    int a[] = { 6, 7, 8 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << MaxGCD(a, n);
 
    return 0;
}


Java
// Java implementation of the approach
class GFG
{
 
// Function to return the maximum
// possible gcd after replacing
// a single element
static int MaxGCD(int a[], int n)
{
 
    // Prefix and Suffix arrays
    int []Prefix = new int[n + 2];
    int []Suffix = new int[n + 2];
 
    // Single state dynamic programming relation
    // for storing gcd of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (int i = 2; i <= n; i += 1)
    {
        Prefix[i] = __gcd(Prefix[i - 1],
                               a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing gcd of all the elements having
    // index greater than or equal to i in Suffix[i]
    for (int i = n - 1; i >= 1; i -= 1)
    {
        Suffix[i] = __gcd(Suffix[i + 1],
                               a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be replaced
    int ans = Math.max(Suffix[2], Prefix[n - 1]);
 
    // If any other element is replaced
    for (int i = 2; i < n; i += 1)
    {
        ans = Math.max(ans, __gcd(Prefix[i - 1],
                                  Suffix[i + 1]));
    }
 
    // Return the maximized gcd
    return ans;
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
public static void main(String[] args)
{
    int a[] = { 6, 7, 8 };
    int n = a.length;
 
    System.out.println(MaxGCD(a, n));
}
}
 
// This code is contributed by PrinciRaj1992


Python3
# Python3 implementation of the approach
from math import gcd as __gcd
 
# Function to return the maximum
# possible gcd after replacing
# a single element
def MaxGCD(a, n) :
 
    # Prefix and Suffix arrays
    Prefix = [0] * (n + 2);
    Suffix = [0] * (n + 2);
 
    # Single state dynamic programming relation
    # for storing gcd of first i elements
    # from the left in Prefix[i]
    Prefix[1] = a[0];
     
    for i in range(2, n + 1) :
        Prefix[i] = __gcd(Prefix[i - 1], a[i - 1]);
 
    # Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    # Single state dynamic programming relation
    # for storing gcd of all the elements having
    # index greater than or equal to i in Suffix[i]
    for i in range(n - 1, 0, -1) :
        Suffix[i] = __gcd(Suffix[i + 1], a[i - 1]);
 
    # If first or last element of
    # the array has to be replaced
    ans = max(Suffix[2], Prefix[n - 1]);
 
    # If any other element is replaced
    for i in range(2, n) :
        ans = max(ans, __gcd(Prefix[i - 1],
                             Suffix[i + 1]));
 
    # Return the maximized gcd
    return ans;
 
# Driver code
if __name__ == "__main__" :
 
    a = [ 6, 7, 8 ];
    n = len(a);
 
    print(MaxGCD(a, n));
 
# This code is contributed by AnkitRai01


C#
// C# implementation of the approach
using System;
     
class GFG
{
 
// Function to return the maximum
// possible gcd after replacing
// a single element
static int MaxGCD(int []a, int n)
{
 
    // Prefix and Suffix arrays
    int []Prefix = new int[n + 2];
    int []Suffix = new int[n + 2];
 
    // Single state dynamic programming relation
    // for storing gcd of first i elements
    // from the left in Prefix[i]
    Prefix[1] = a[0];
    for (int i = 2; i <= n; i += 1)
    {
        Prefix[i] = __gcd(Prefix[i - 1],
                            a[i - 1]);
    }
 
    // Initializing Suffix array
    Suffix[n] = a[n - 1];
 
    // Single state dynamic programming relation
    // for storing gcd of all the elements having
    // index greater than or equal to i in Suffix[i]
    for (int i = n - 1; i >= 1; i -= 1)
    {
        Suffix[i] = __gcd(Suffix[i + 1],
                            a[i - 1]);
    }
 
    // If first or last element of
    // the array has to be replaced
    int ans = Math.Max(Suffix[2], Prefix[n - 1]);
 
    // If any other element is replaced
    for (int i = 2; i < n; i += 1)
    {
        ans = Math.Max(ans, __gcd(Prefix[i - 1],
                                Suffix[i + 1]));
    }
 
    // Return the maximized gcd
    return ans;
}
 
static int __gcd(int a, int b)
{
    return b == 0 ? a : __gcd(b, a % b);    
}
 
// Driver code
public static void Main(String[] args)
{
    int []a = { 6, 7, 8 };
    int n = a.Length;
 
    Console.WriteLine(MaxGCD(a, n));
}
}
 
// This code is contributed by 29AjayKumar


Javascript


输出:
2

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