给定对乘积数组pair [] ,任务是找到原始数组。数组arr []的成对乘积数组是包含所有有序对的乘积的数组,即{(arr [0] * arr [1]),(arr [0] * arr [2]), …,(arr [1] * arr [2]),(arr [1] * arr [3]),…,(arr [n – 2] * arr [n – 1])}} 。
例子:
Input: pair[] = {2, 3, 6}
Output: 1 2 3
Input: pair[] = {48, 18, 24, 24, 32, 12}
Output: 6 8 3 4
方法:首先从给定的对乘积数组中找到所需数组的大小。假设原始数组的大小为N ,而配对乘积数组的大小为X。因此,通过求解(N *(N – 1))/ 2 = X ,可以将N的值计算为N =(1 +(int)sqrt(1 + 8 * X))/ 2 。
现在让我们看一个例子的解决方案,假设[A,B,C,D]的对积数组为arr [AB,AC,AD,BC,BD,CD],然后取sqrt((arr [0 ] * arr [1])/ arr [n – 1]) -> sqrt((AB * AC)/ BC)将得到值A。
一旦恢复了第一个元素的值,即可将对乘积数组的所有其余元素除以它,以获得原始数组的其余元素。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
#include
using namespace std;
// Utility function to print the array
void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Function to generate the original
// array from the pair-product array
void constructArr(int pair[], int n)
{
int size = (1 + (int)sqrt(1 + 8 * n)) / 2;
int arr[size];
// First element of the resulting array
arr[0] = sqrt((pair[0] * pair[1]) / pair[size - 1]);
// Find all the other elements
for (int i = 1; i < size; i++)
arr[i] = pair[i - 1] / arr[0];
// Print the elements of the generated array
printArr(arr, size);
}
// Driver code
int main()
{
int pair[] = { 48, 18, 24, 24, 32, 12 };
int n = sizeof(pair) / sizeof(int);
constructArr(pair, n);
return 0;
}
Java
// Java implementation of the approach
class GFG
{
// Utility function to print the array
static void printArr(int arr[], int n)
{
for (int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Function to generate the original
// array from the pair-product array
static void constructArr(int pair[], int n)
{
int size = (1 + (int)Math.sqrt(1 + 8 * n)) / 2;
int []arr = new int[size];
// First element of the resulting array
arr[0] = (int) Math.sqrt((pair[0] * pair[1]) /
pair[size - 1]);
// Find all the other elements
for (int i = 1; i < size; i++)
arr[i] = pair[i - 1] / arr[0];
// Print the elements of the generated array
printArr(arr, size);
}
// Driver code
public static void main(String[] args)
{
int pair[] = { 48, 18, 24, 24, 32, 12 };
int n = pair.length;
constructArr(pair, n);
}
}
// This code is contributed by PrinciRaj1992
Python3
# Python3 implementation of the approach
from math import sqrt
# Utility function to print the array
def printArr(arr, n) :
for i in range(n) :
print(arr[i], end = " ");
# Function to generate the original
# array from the pair-product array
def constructArr(pair, n) :
size = int((1 + sqrt(1 + 8 * n)) // 2);
arr = [0] * (size);
# First element of the resulting array
arr[0] = int(sqrt((pair[0] * pair[1]) /
pair[size - 1]));
# Find all the other elements
for i in range(1, size) :
arr[i] = pair[i - 1] // arr[0];
# Print the elements of the generated array
printArr(arr, size);
# Driver code
if __name__ == "__main__" :
pair = [ 48, 18, 24, 24, 32, 12 ];
n = len(pair);
constructArr(pair, n);
# This code is contributed by AnkitRai01
C#
// C# implementation of the approach
using System;
class GFG
{
// Utility function to print the array
static void printArr(int []arr, int n)
{
for (int i = 0; i < n; i++)
Console.Write(arr[i] + " ");
}
// Function to generate the original
// array from the pair-product array
static void constructArr(int []pair, int n)
{
int size = (1 + (int)Math.Sqrt(1 + 8 * n)) / 2;
int []arr = new int[size];
// First element of the resulting array
arr[0] = (int) Math.Sqrt((pair[0] * pair[1]) /
pair[size - 1]);
// Find all the other elements
for (int i = 1; i < size; i++)
arr[i] = pair[i - 1] / arr[0];
// Print the elements of the generated array
printArr(arr, size);
}
// Driver code
public static void Main(String[] args)
{
int []pair = { 48, 18, 24, 24, 32, 12 };
int n = pair.Length;
constructArr(pair, n);
}
}
// This code is contributed by Rajput-Ji
输出:
6 8 3 4