给定一个数组arr []，其中包含另一个数组的每个可能的元素对的GCD。任务是找到用于计算GCD阵列的原始编号。例如，如果原始数字为{4，6，8}，则给定的数组将为{4，2，4，2，6，2，2，4，2，8}。 %20every%20pair_0.jpg)
例子：
Input: arr[] = {5, 1, 1, 12}
Output: 12 5
gcd(12, 12) = 12
gcd(12, 5) = 1
gcd(5, 12) = 1
gcd(5, 5) = 5
Input: arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2}
Output: 12 10 7 5 1
方法：
- 以降序对数组进行排序。
- 最高元素将始终是原始数字之一。保留该编号并将其从阵列中删除。
- 计算上一步中元素的GCD，使当前元素从最大值开始计算，并丢弃给定数组中的GCD值。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Utility function to print
// the contents of an array
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to find the required numbers
void findNumbers(int arr[], int n)
{
// Sort array in decreasing order
sort(arr, arr + n, greater());
int freq[arr[0] + 1] = { 0 };
// Count frequency of each element
for (int i = 0; i < n; i++)
freq[arr[i]]++;
// Size of the resultant array
int size = sqrt(n);
int brr[size] = { 0 }, x, l = 0;
for (int i = 0; i < n; i++) {
if (freq[arr[i]] > 0) {
// Store the highest element in
// the resultant array
brr[l] = arr[i];
// Decrement the frequency of that element
freq[brr[l]]--;
l++;
for (int j = 0; j < l; j++) {
if (i != j) {
// Compute GCD
x = __gcd(arr[i], brr[j]);
// Decrement GCD value by 2
freq[x] -= 2;
}
}
}
}
printArr(brr, size);
}
// Driver code
int main()
{
int arr[] = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2 };
int n = sizeof(arr) / sizeof(arr[0]);
findNumbers(arr, n);
return 0;
} Java
// Java implementation of the approach
import java.util.Arrays;
class GFG
{
// Utility function to print
// the contents of an array
static void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
{
System.out.print(arr[i] + " ");
}
}
// Function to find the required numbers
static void findNumbers(int arr[], int n)
{
// Sort array in decreasing order
Arrays.sort(arr);
reverse(arr);
int freq[] = new int[arr[0] + 1];
// Count frequency of each element
for (int i = 0; i < n; i++)
{
freq[arr[i]]++;
}
// Size of the resultant array
int size = (int) Math.sqrt(n);
int brr[] = new int[size], x, l = 0;
for (int i = 0; i < n; i++)
{
if (freq[arr[i]] > 0 && l < size)
{
// Store the highest element in
// the resultant array
brr[l] = arr[i];
// Decrement the frequency of that element
freq[brr[l]]--;
l++;
for (int j = 0; j < l; j++)
{
if (i != j)
{
// Compute GCD
x = __gcd(arr[i], brr[j]);
// Decrement GCD value by 2
freq[x] -= 2;
}
}
}
}
printArr(brr, size);
}
// reverse array
public static void reverse(int[] input)
{
int last = input.length - 1;
int middle = input.length / 2;
for (int i = 0; i <= middle; i++)
{
int temp = input[i];
input[i] = input[last - i];
input[last - i] = temp;
}
}
static int __gcd(int a, int b)
{
if (b == 0)
{
return a;
}
return __gcd(b, a % b);
}
// Driver code
public static void main(String[] args)
{
int arr[] = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2};
int n = arr.length;
findNumbers(arr, n);
}
}
// This code has been contributed by 29AjayKumarPython3
# Python 3 implementation of the approach
from math import sqrt, gcd
# Utility function to print
# the contents of an array
def printArr(arr, n):
for i in range(n):
print(arr[i], end = " ")
# Function to find the required numbers
def findNumbers(arr, n):
# Sort array in decreasing order
arr.sort(reverse = True)
freq = [0 for i in range(arr[0] + 1)]
# Count frequency of each element
for i in range(n):
freq[arr[i]] += 1
# Size of the resultant array
size = int(sqrt(n))
brr = [0 for i in range(len(arr))]
l = 0
for i in range(n):
if (freq[arr[i]] > 0):
# Store the highest element in
# the resultant array
brr[l] = arr[i]
# Decrement the frequency of that element
freq[brr[l]] -= 1
l += 1
for j in range(l):
if (i != j):
# Compute GCD
x = gcd(arr[i], brr[j])
# Decrement GCD value by 2
freq[x] -= 2
printArr(brr, size)
# Driver code
if __name__ == '__main__':
arr = [1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1,
1, 5, 5, 5, 7, 10, 12, 2, 2]
n = len(arr)
findNumbers(arr, n)
# This code is contributed by
# Surendra_GangwarC#
// C# implementation for above approach
using System;
class GFG
{
// Utility function to print
// the contents of an array
static void printArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
{
Console.Write(arr[i] + " ");
}
}
// Function to find the required numbers
static void findNumbers(int []arr, int n)
{
// Sort array in decreasing order
Array.Sort(arr);
reverse(arr);
int []freq = new int[arr[0] + 1];
// Count frequency of each element
for (int i = 0; i < n; i++)
{
freq[arr[i]]++;
}
// Size of the resultant array
int size = (int) Math.Sqrt(n);
int []brr = new int[size];int x, l = 0;
for (int i = 0; i < n; i++)
{
if (freq[arr[i]] > 0 && l < size)
{
// Store the highest element in
// the resultant array
brr[l] = arr[i];
// Decrement the frequency of that element
freq[brr[l]]--;
l++;
for (int j = 0; j < l; j++)
{
if (i != j)
{
// Compute GCD
x = __gcd(arr[i], brr[j]);
// Decrement GCD value by 2
freq[x] -= 2;
}
}
}
}
printArr(brr, size);
}
// reverse array
public static void reverse(int []input)
{
int last = input.Length - 1;
int middle = input.Length / 2;
for (int i = 0; i <= middle; i++)
{
int temp = input[i];
input[i] = input[last - i];
input[last - i] = temp;
}
}
static int __gcd(int a, int b)
{
if (b == 0)
{
return a;
}
return __gcd(b, a % b);
}
// Driver code
public static void Main(String[] args)
{
int []arr = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2};
int n = arr.Length;
findNumbers(arr, n);
}
}
/* This code contributed by PrinciRaj1992 */PHP
0)
{
// Store the highest element in
// the resultant array
$brr[$l] = $arr[$i];
// Decrement the frequency of that element
$freq[$brr[$l]]--;
$l++;
for ($j = 0; $j < $l; $j++)
{
if ($i != $j)
{
// Compute GCD
$x = gcd($arr[$i], $brr[$j]);
// Decrement GCD value by 2
$freq[$x] -= 2;
}
}
}
}
printArr($brr, $size);
}
// Driver code
$arr = array(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 5, 5, 5, 7, 10, 12, 2, 2 );
$n = count($arr) ;
findNumbers($arr, $n);
// This code is contributed by Ryuga
?>输出：
12 10 7 5 1
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