给定长度为N≥2的数组arr [] 。任务是从给定的数组中删除一个元素，以使删除后的数组的GCD最大化。
例子：
Input: arr[] = {12, 15, 18}
Output: 6
Remove 12: GCD(15, 18) = 3
Remove 15: GCD(12, 18) = 6
Remove 18: GCD(12, 15) = 3
Input: arr[] = {14, 17, 28, 70}
Output: 14
方法：
- 想法是找到所有长度(N – 1)子序列的GCD值，并删除该GCD子序列中不存在的元素。找到的最大GCD就是答案。
- 为了最佳地找到子序列的GCD，请使用单状态动态编程维护一个prefixGCD []和一个suffixGCD []数组。
- GCD的最大值(前缀GCD [i – 1]，后缀GCD [i + 1])是必需的答案。
下面是上述方法的实现：
CPP
// C++ implementation of the above approach
#include
using namespace std;
// Function to return the maximized gcd
// after removing a single element
// from the given array
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
// 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 removed
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[] = { 14, 17, 28, 70 };
int n = sizeof(a) / sizeof(a[0]);
cout << MaxGCD(a, n);
return 0;
} Java
// Java implementation of the above approach
class Test
{
// 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 return the maximized gcd
// after removing a single element
// from the given array
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
// 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 removed
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;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 14, 17, 28, 70 };
int n = a.length;
System.out.println(MaxGCD(a, n));
}
}
// This code is contributed by AnkitRai01Python3
# Python3 implementation of the above approach
import math as mt
# Function to return the maximized gcd
# after removing a single element
# from the given array
def MaxGCD(a, n):
# Prefix and Suffix arrays
Prefix=[0 for i in range(n + 2)]
Suffix=[0 for i in range(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] = mt.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
# greater than or equal to i in Suffix[i]
for i in range(n-1,0,-1):
Suffix[i] =mt.gcd(Suffix[i + 1], a[i - 1])
# If first or last element of
# the array has to be removed
ans = max(Suffix[2], Prefix[n - 1])
# If any other element is replaced
for i in range(2,n):
ans = max(ans, mt.gcd(Prefix[i - 1], Suffix[i + 1]))
# Return the maximized gcd
return ans
# Driver code
a=[14, 17, 28, 70]
n = len(a)
print(MaxGCD(a, n))
# This code is contributed by mohit kumar 29C#
// C# implementation of 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 return the maximized gcd
// after removing a single element
// from the given array
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
// 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 removed
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;
}
// Driver code
static public void Main ()
{
int []a = { 14, 17, 28, 70 };
int n = a.Length;
Console.Write(MaxGCD(a, n));
}
}
// This code is contributed by ajit.输出：
14
时间复杂度： O(N * log(M))其中M是数组中的最大元素。