数组中总和等于乘积的对的数量
给定一个数组 arr[] ，任务是找出数组中的对的数量 (arr[i]，arr[j]) ，使得arr[I]+arr[j]= arr[I]* arr[j] 示例:
输入: arr[] = {2，2，3，4，6} 输出: 1 (2，2)是唯一可能的对，因为(2 + 2) = (2 * 2) = 4。 输入: arr[] = {1，2，3，4，5} 输出: 0
方法:唯一可能满足给定条件的整数对是 (0，0) 和 (2，2) 。所以现在的任务是统计数组中0和 2s 的个数，分别存储在 cnt0 和 cnt2 中，那么需要的个数就是(CNT 0 (CNT 0–1))/2+(CNT 2 (CNT 2–1))/2。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count
// of the required pairs
int sumEqualProduct(int a[], int n)
{
int zero = 0, two = 0;
// Find the count of 0s
// and 2s in the array
for (int i = 0; i < n; i++) {
if (a[i] == 0) {
zero++;
}
if (a[i] == 2) {
two++;
}
}
// Find the count of required pairs
int cnt = (zero * (zero - 1)) / 2
+ (two * (two - 1)) / 2;
// Return the count
return cnt;
}
// Driver code
int main()
{
int a[] = { 2, 2, 3, 4, 2, 6 };
int n = sizeof(a) / sizeof(a[0]);
cout << sumEqualProduct(a, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG {
// Function to return the count
// of the required pairs
static int sumEqualProduct(int a[], int n)
{
int zero = 0, two = 0;
// Find the count of 0s
// and 2s in the array
for (int i = 0; i < n; i++) {
if (a[i] == 0) {
zero++;
}
if (a[i] == 2) {
two++;
}
}
// Find the count of required pairs
int cnt = (zero * (zero - 1)) / 2
+ (two * (two - 1)) / 2;
// Return the count
return cnt;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 2, 2, 3, 4, 2, 6 };
int n = a.length;
System.out.print(sumEqualProduct(a, n));
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python 3 implementation of the approach
# Function to return the count
# of the required pairs
def sumEqualProduct(a, n):
zero = 0
two = 0
# Find the count of 0s
# and 2s in the array
for i in range(n):
if a[i] == 0:
zero += 1
if a[i] == 2:
two += 1
# Find the count of required pairs
cnt = (zero * (zero - 1)) // 2 + \
(two * (two - 1)) // 2
# Return the count
return cnt
# Driver code
a = [ 2, 2, 3, 4, 2, 6 ]
n = len(a)
print(sumEqualProduct(a, n))
# This code is contributed by Ankit kumar
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the count
// of the required pairs
static int sumEqualProduct(int []a, int n)
{
int zero = 0, two = 0;
// Find the count of 0s
// and 2s in the array
for (int i = 0; i < n; i++)
{
if (a[i] == 0)
{
zero++;
}
if (a[i] == 2)
{
two++;
}
}
// Find the count of required pairs
int cnt = (zero * (zero - 1)) / 2 +
(two * (two - 1)) / 2;
// Return the count
return cnt;
}
// Driver code
public static void Main(String[] args)
{
int []a = { 2, 2, 3, 4, 2, 6 };
int n = a.Length;
Console.Write(sumEqualProduct(a, n));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count
// of the required pairs
function sumEqualProduct(a, n)
{
var zero = 0, two = 0;
// Find the count of 0s
// and 2s in the array
for (var i = 0; i < n; i++) {
if (a[i] == 0) {
zero++;
}
if (a[i] == 2) {
two++;
}
}
// Find the count of required pairs
var cnt = (zero * (zero - 1)) / 2
+ (two * (two - 1)) / 2;
// Return the count
return cnt;
}
// Driver code
var a = [2, 2, 3, 4, 2, 6];
var n = a.length;
document.write( sumEqualProduct(a, n));
// This code is contributed by importantly.
</script>
Output:
3
时间复杂度: O(N)
辅助空间: O(1)
版权属于：月萌API www.moonapi.com，转载请注明出处
