给定数组中每对绝对差的乘积
给定一个由 N 个元素组成的数组 arr[] ，任务是找出给定数组中所有对的绝对差的乘积。
示例:
输入: arr[] = {1，2，3，4} 输出: 10 解释: 积| 2-1 | * | 3-1 | * | 4-1 | * | 3-2 | * | 4-2 | * | 4-3 | = 12 输入: arr[] = {1，8，9，15，16} 输出:
方法:思路是生成给定数组的每一个可能的对arr【】求所有对的绝对差的乘积。
以下是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the product of
// abs diff of all pairs (x, y)
int getProduct(int a[], int n)
{
// To store product
int p = 1;
// Iterate all possible pairs
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
// Find the product
p *= abs(a[i] - a[j]);
}
}
// Return product
return p;
}
// Driver Code
int main()
{
// Given array arr[]
int arr[] = { 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
// Function Call
cout << getProduct(arr, N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.util.*;
class GFG{
// Function to return the product of
// abs diff of all pairs (x, y)
static int getProduct(int a[], int n)
{
// To store product
int p = 1;
// Iterate all possible pairs
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
// Find the product
p *= Math.abs(a[i] - a[j]);
}
}
// Return product
return p;
}
// Driver Code
public static void main(String[] args)
{
// Given array arr[]
int arr[] = { 1, 2, 3, 4 };
int N = arr.length;
// Function call
System.out.println(getProduct(arr, N));
}
}
// This code is contributed by Ritik Bansal
Python 3
# Python3 program for
# the above approach
# Function to return the product of
# abs diff of all pairs (x, y)
def getProduct(a, n):
# To store product
p = 1
# Iterate all possible pairs
for i in range (n):
for j in range (i + 1, n):
# Find the product
p *= abs(a[i] - a[j])
# Return product
return p
# Driver Code
if __name__ == "__main__":
# Given array arr[]
arr = [1, 2, 3, 4]
N = len(arr)
# Function Call
print (getProduct(arr, N))
# This code is contributed by Chitranayal
C
// C# program for the above approach
using System;
class GFG{
// Function to return the product of
// abs diff of all pairs (x, y)
static int getProduct(int []a, int n)
{
// To store product
int p = 1;
// Iterate all possible pairs
for(int i = 0; i < n; i++)
{
for(int j = i + 1; j < n; j++)
{
// Find the product
p *= Math.Abs(a[i] - a[j]);
}
}
// Return product
return p;
}
// Driver Code
public static void Main(string[] args)
{
// Given array arr[]
int []arr = { 1, 2, 3, 4 };
int N = arr.Length;
// Function call
Console.Write(getProduct(arr, N));
}
}
// This code is contributed by rutvik_56
java 描述语言
<script>
// Javascript program for the above approach
// Function to return the product of
// abs diff of all pairs (x, y)
function getProduct( a, n)
{
// To store product
var p = 1;
// Iterate all possible pairs
for (var i = 0; i < n; i++) {
for (var j = i + 1; j < n; j++) {
// Find the product
p *= Math.abs(a[i] - a[j]);
}
}
// Return product
return p;
}
// Driver Code
// Given array arr[]
var arr = [ 1, 2, 3, 4 ];
var N = arr.length;
// Function Call
document.write( getProduct(arr, N));
// This code is contributed by itsok.
</script>
Output:
12
时间复杂度:O(N2) 辅助空间: O(1)
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