作者：Roger_丨8_8 | 来源：互联网 | 2023-10-11 11:44
计算乘积等于K的所有不同对原文:https://www.
计算乘积等于 K 的所有不同对
原文:https://www . geesforgeks . org/count-all-distinct-pairs-with-product-equal-to-k/
给定一个大小为 N 的整数数组arr【】和一个正整数 K ，任务是用等于 K 的乘积对数组中所有不同的对进行计数。
示例:
输入: arr[] = {1，5，3，4，2}，K = 3
输出: 1
说明:
只有一对(1，3)积= K = 3。
输入: arr[] = {1，2，16，4，4}，K = 16
输出: 2
说明:
有两对(1，16)和(4，4)，乘积= K = 16。
高效方法: 想法是使用哈希。
- 最初，将数组中的所有元素插入到散列表中。插入时，如果 hashmap 中已经存在特定元素，则忽略。
- 现在，我们在散列中有所有唯一的元素。因此，对于数组 arr[i]中的每个元素，我们检查 hashmap 中是否存在 arr[i] / K 。
- 如果该值存在，那么我们增加计数并从散列中移除该特定元素(以获得唯一的对)。
以下是上述方法的实现:
C++
// C++ program to count the number of pairs
// whose product is equal to K
#include <algorithm>
#include <iostream>
using namespace std;
#define MAX 100000
// Function to count the number of pairs
// whose product is equal to K
int countPairsWithProductK(int arr[], int n, int k)
{
// Initialize the count
int count = 0;
// Initialize empty hashmap.
bool hashmap[MAX] = { false };
// Insert array elements to hashmap
for (int i = 0; i < n; i++)
hashmap[arr[i]] = true;
for (int i = 0; i < n; i++) {
int x = arr[i];
double index = 1.0 * k / arr[i];
// Checking if the index is a whole number
// and present in the hashmap
if (index >= 0
&& ((index - (int)(index)) == 0)
&& hashmap[k / x])
count++;
hashmap[x] = false;
}
return count;
}
// Driver code
int main()
{
int arr[] = { 1, 5, 3, 4, 2 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 3;
cout << countPairsWithProductK(arr, N, K);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to count the number of pairs
// whose product is equal to K
class GFG
{
static int MAX = 100000;
// Function to count the number of pairs
// whose product is equal to K
static int countPairsWithProductK(int arr[], int n, int k)
{
// Initialize the count
int count = 0;
int i;
// Initialize empty hashmap.
boolean hashmap[] = new boolean[MAX];
// Insert array elements to hashmap
for (i = 0; i < n; i++)
hashmap[arr[i]] = true;
for (i = 0; i < n; i++) {
int x = arr[i];
double index = 1.0 * k / arr[i];
// Checking if the index is a whole number
// and present in the hashmap
if (index >= 0
&& ((index - (int)(index)) == 0)
&& hashmap[k / x])
count++;
hashmap[x] = false;
}
return count;
}
// Driver code
public static void main(String []args)
{
int arr[] = { 1, 5, 3, 4, 2 };
int N = arr.length;
int K = 3;
System.out.print(countPairsWithProductK(arr, N, K));
}
}
Python 3
# Python3 program to count the number of pairs
# whose product is equal to K
MAX = 100000;
# Function to count the number of pairs
# whose product is equal to K
def countPairsWithProductK(arr, n, k) :
# Initialize the count
count = 0;
# Initialize empty hashmap.
hashmap = [False]*MAX ;
# Insert array elements to hashmap
for i in range(n) :
hashmap[arr[i]] = True;
for i in range(n) :
x = arr[i];
index = 1.0 * k / arr[i];
# Checking if the index is a whole number
# and present in the hashmap
if (index >= 0
and ((index - int(index)) == 0)
and hashmap[k // x]) :
count += 1;
hashmap[x] = False;
return count;
# Driver code
if __name__ == "__main__" :
arr = [ 1, 5, 3, 4, 2 ];
N = len(arr);
K = 3;
print(countPairsWithProductK(arr, N, K));
# This code is contributed by AnkitRai01
C
// C# program to count the number of pairs
// whose product is equal to K
using System;
class GFG
{
static int MAX = 100000;
// Function to count the number of pairs
// whose product is equal to K
static int countPairsWithProductK(int []arr, int n, int k)
{
// Initialize the count
int count = 0;
int i;
// Initialize empty hashmap.
bool []hashmap = new bool[MAX];
// Insert array elements to hashmap
for (i = 0; i < n; i++)
hashmap[arr[i]] = true;
for (i = 0; i < n; i++) {
int x = arr[i];
double index = 1.0 * k / arr[i];
// Checking if the index is a whole number
// and present in the hashmap
if (index >= 0
&& ((index - (int)(index)) == 0)
&& hashmap[k / x])
count++;
hashmap[x] = false;
}
return count;
}
// Driver code
public static void Main(String []args)
{
int []arr = { 1, 5, 3, 4, 2 };
int N = arr.Length;
int K = 3;
Console.Write(countPairsWithProductK(arr, N, K));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript program to count the number of pairs
// whose product is equal to K
let MAX = 100000;
// Function to count the number of pairs
// whose product is equal to K
function countPairsWithProductK(arr,n,k)
{
// Initialize the count
let count = 0;
let i;
// Initialize empty hashmap.
let hashmap = new Array(MAX);
// Insert array elements to hashmap
for (i = 0; i < n; i++)
hashmap[arr[i]] = true;
for (i = 0; i < n; i++) {
let x = arr[i];
let index = 1.0 * k / arr[i];
// Checking if the index is a whole number
// and present in the hashmap
if (index >= 0
&& ((index - Math.floor(index)) == 0)
&& hashmap[k / x])
count++;
hashmap[x] = false;
}
return count;
}
// Driver code
let arr=[1, 5, 3, 4, 2];
let N = arr.length;
let K = 3;
document.write(countPairsWithProductK(arr, N, K));
// This code is contributed by rag2127
script>
Output:
时间复杂度: O(N * log(N))
辅助空间: O(MAX)
![[大整数乘法] java代码实现](https://img1.php1.cn/3cd4a/24c6f/9f3/0133bb25da242824.jpeg)
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