给定数组中所有对的乘积

给定N个整数的数组arr [] ，任务是从给定数组中查找所有可能的对的乘积，例如：

• (arr [i]，arr [i])也被视为有效对。
• (arr [i]，arr [j])和(arr [j]，arr [i])被视为两个不同的对。

打印结果答案模数10 ^ 9 + 7。

例子：

Input: arr[] = {1, 2} 
Output: 16 
Explanation: 
All valid pairs are (1, 1), (1, 2), (2, 1) and (2, 2). 
Hence, 1 * 1 * 1 * 2 * 2 * 1 * 2 * 2 = 16

Input: arr[] = {1, 2, 3} 
Output: 46656 
Explanation: 
All valid pairs are (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2) and (3, 3). 
Hence the product is 1*1*1**2*1*3*2*1*2*2*2*3*3*1*3*2*3*3 = 46656 

幼稚的方法：为了解决上述问题，幼稚的方法是找到所有可能的对，并计算每对元素的乘积。

下面是上述方法的实现：

C++

// C++ implementation to find the
// product of all the pairs from
// the given array
 
#include 
using namespace std;
#define mod 1000000007
 
// Function to return the product of
// the elements of all possible pairs
// from the array
int productPairs(int arr[], int n)
{
 
    // To store the required product
    int product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
 
            // Multiply the product of
            // the elements of the
            // current pair
            product *= (arr[i] % mod
                        * arr[j] % mod)
                       % mod;
            product = product % mod;
        }
    }
 
    // Return the final result
    return product % mod;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
 
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << productPairs(arr, n);
 
    return 0;
}

Java

// Java implementation to find the
// product of all the pairs from
// the given array
import java.util.*;
 
class GFG{
     
static final int mod = 1000000007;
 
// Function to return the product of
// the elements of all possible pairs
// from the array
static int productPairs(int arr[], int n)
{
 
    // To store the required product
    int product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for(int i = 0; i < n; i++)
    {
       for(int j = 0; j < n; j++)
       {
           
          // Multiply the product
          // of the elements of the
          // current pair
          product *= (arr[i] % mod *
                      arr[j] % mod) % mod;
          product = product % mod;
       }
    }
 
    // Return the final result
    return product % mod;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
 
    System.out.print(productPairs(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 implementation to find the
# product of all the pairs from
# the given array
mod = 1000000007;
 
# Function to return the product of
# the elements of all possible pairs
# from the array
def productPairs(arr, n):
   
    # To store the required product
    product = 1;
 
    # Nested loop to calculate all
    # possible pairs
    for i in range(n):
        for j in range(n):
           
            # Multiply the product
            # of the elements of the
            # current pair
            product *= (arr[i] % mod *
                        arr[j] % mod) % mod;
            product = product % mod;
 
    # Return the final result
    return product % mod;
 
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3];
    n = len(arr);
 
    print(productPairs(arr, n));
 
# This code is contributed by 29AjayKumar

C#

// C# implementation to find the
// product of all the pairs from
// the given array
using System;
class GFG{
     
static readonly int mod = 1000000007;
 
// Function to return the product of
// the elements of all possible pairs
// from the array
static int productPairs(int []arr, int n)
{
 
    // To store the required product
    int product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for(int i = 0; i < n; i++)
    {
        for(int j = 0; j < n; j++)
        {
                 
            // Multiply the product
            // of the elements of the
            // current pair
            product *= (arr[i] % mod *
                        arr[j] % mod) % mod;
            product = product % mod;
        }
    }
 
    // Return the readonly result
    return product % mod;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
 
    Console.Write(productPairs(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript

 
//Javascript implementation to find the
// product of all the pairs from
// the given array
 
mod = 1000000007
 
// Function to return the product of
// the elements of all possible pairs
// from the array
function productPairs(arr, n)
{
 
    // To store the required product
    let product = 1;
 
    // Nested loop to calculate all
    // possible pairs
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
 
            // Multiply the product of
            // the elements of the
            // current pair
            product *= (arr[i] % mod
                        * arr[j] % mod)
                    % mod;
            product = product % mod;
        }
    }
 
    // Return the final result
    return product % mod;
}
 
// Driver code
    let arr = [ 1, 2, 3 ];
 
    let n = arr.length;
 
    document.write(productPairs(arr, n));
 
// This code is contributed by Mayank Tyagi
 

输出：
46656

时间复杂度： O(N ² )

高效的方法：我们可以观察到每个元素作为对中的一个元素(X，Y)的出现正好是(2 * N)次。正好是X的N倍，正好是Y的N倍。

下面是上述方法的实现：

C++

// C++ implementation to Find the product
// of all the pairs from the given array
#include 
using namespace std;
#define mod 1000000007
#define ll long long int
 
// Function to calculate
// (x^y)%1000000007
int power(int x, unsigned int y)
{
    int p = 1000000007;
 
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    x = x % p;
 
    while (y > 0) {
        // If y is odd, multiply x
        // with result
        if (y & 1)
            res = (res * x) % p;
 
        y = y >> 1;
        x = (x * x) % p;
    }
 
    // Return the final result
    return res;
}
 
// Function to return the product
// of the elements of all possible
// pairs from the array
ll productPairs(ll arr[], ll n)
{
 
    // To store the required product
    ll product = 1;
 
    // Iterate for every element
    // of the array
    for (int i = 0; i < n; i++) {
 
        // Each element appears (2 * n) times
        product
            = (product
               % mod
               * (int)power(
                     arr[i], (2 * n))
               % mod)
              % mod;
    }
 
    return product % mod;
}
 
// Driver code
int main()
{
    ll arr[] = { 1, 2, 3 };
    ll n = sizeof(arr) / sizeof(arr[0]);
 
    cout << productPairs(arr, n);
 
    return 0;
}

Java

// Java implementation to Find the product
// of all the pairs from the given array
import java.util.*;
 
class GFG{
static final int mod = 1000000007;
 
// Function to calculate
// (x^y)%1000000007
static int power(int x, int y)
{
    int p = 1000000007;
 
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    x = x % p;
 
    while (y > 0)
    {
         
        // If y is odd, multiply x
        // with result
        if (y % 2 == 1)
            res = (res * x) % p;
 
        y = y >> 1;
        x = (x * x) % p;
    }
 
    // Return the final result
    return res;
}
 
// Function to return the product
// of the elements of all possible
// pairs from the array
static int productPairs(int arr[], int n)
{
 
    // To store the required product
    int product = 1;
 
    // Iterate for every element
    // of the array
    for (int i = 0; i < n; i++)
    {
 
        // Each element appears (2 * n) times
        product = (product % mod *
                  (int)power(arr[i],
                            (2 * n)) % mod) % mod;
    }
 
    return product % mod;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
 
    System.out.print(productPairs(arr, n));
}
}
 
// This code is contributed by amal kumar choubey

Python3

# Python3 implementation to Find the product
# of all the pairs from the given array
mod = 1000000007
 
# Function to calculate
# (x^y)%1000000007
def power(x, y):
 
    p = 1000000007
 
    # Initialize result
    res = 1
 
    # Update x if it is more than
    # or equal to p
    x = x % p
 
    while (y > 0):
         
        # If y is odd, multiply x
        # with result
        if ((y & 1) != 0):
            res = (res * x) % p
 
        y = y >> 1
        x = (x * x) % p
 
    # Return the final result
    return res
 
# Function to return the product
# of the elements of all possible
# pairs from the array
def productPairs(arr, n):
 
    # To store the required product
    product = 1
 
    # Iterate for every element
    # of the array
    for i in range(n):
 
        # Each element appears (2 * n) times
        product = (product % mod *
          (int)(power(arr[i], (2 * n))) %
                            mod) % mod
 
    return (product % mod)
     
# Driver code
arr = [ 1, 2, 3 ]
n = len(arr)
 
print(productPairs(arr, n))
 
# This code is contributed by divyeshrabadiya07

C＃

// C# implementation to Find the product
// of all the pairs from the given array
using System;
class GFG{
const int mod = 1000000007;
 
// Function to calculate
// (x^y)%1000000007
static int power(int x, int y)
{
    int p = 1000000007;
 
    // Initialize result
    int res = 1;
 
    // Update x if it is more than
    // or equal to p
    x = x % p;
 
    while (y > 0)
    {
         
        // If y is odd, multiply x
        // with result
        if (y % 2 == 1)
            res = (res * x) % p;
 
        y = y >> 1;
        x = (x * x) % p;
    }
 
    // Return the final result
    return res;
}
 
// Function to return the product
// of the elements of all possible
// pairs from the array
static int productPairs(int []arr, int n)
{
 
    // To store the required product
    int product = 1;
 
    // Iterate for every element
    // of the array
    for (int i = 0; i < n; i++)
    {
 
        // Each element appears (2 * n) times
        product = (product % mod *
                  (int)power(arr[i],
                            (2 * n)) % mod) % mod;
    }
 
    return product % mod;
}
 
// Driver code
public static void Main()
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
 
    Console.Write(productPairs(arr, n));
}
}
 
// This code is contributed by Code_Mech

输出：
46656

时间复杂度： O(N)