检查数组中是否存在每对的乘积
原文:https://www . geesforgeks . org/check-product-ever-pair-exists-array/
给定一个 n 个整数的数组，我们需要检查每对数字 a[i] & a[j]是否存在 a[k]，使得 a[k] = a[i]a[j]，其中 k 也可以等于 I 或 j。 例:*
Input : arr[] = {0\. 1}
Output : Yes
Here a[0]*a[1] is equal to a[0]
Input : arr[] = {5, 6}
Output : No
如果数组遵循下面提到的所有条件: 条件 1 : 数组必须具有小于或等于 1 的除 1、0、-1 之外的元素数，因为如果它具有多于 1 个这样的元素，则数组中不存在乘积等于这两个(或更多)元素中最大值的元素。假设这种元素的数量是 2，它们的值是 5，6，所以数组中没有等于 56 = 30 的元素。 条件 2 : 如果数组中还有其他数字，比如 x(除了 0、1 和-1 之外)和-1 也存在，那么同样回答为假。因为-1 的存在使得要求 x 和-x 都应该存在，但是这违反了条件 1。 条件 3 :* 如果数组中有一个以上的“-1”，并且没有人，那么答案也将是否定的，因为两个“-1”的乘积等于 1。 以下是上述条件的执行情况。
C++
// C++ program to find if
// product of every pair
// is present in array.
#include<bits/stdc++.h>
using namespace std;
// Returns true if product
// of every pair in arr[]
// is present in arr[]
bool checkArray(int arr[] , int n)
{
// variable to store number
// of zeroes, ones, minus
// one and other numbers.
int zero = 0, one = 0,
minusone = 0, other=0;
for (int i = 0; i < n; i++)
{
// incrementing the
// variable values
if (arr[i] == 0)
zero++;
else if (arr[i] == 1)
one++;
else if (arr[i] == -1)
minusone++;
else
other++;
}
// checking the conditions
if (other > 1)
return false;
else if (other != 0 &&
minusone != 0)
return false;
else if (minusone > 1 &&
one == 0)
return false;
return true;
}
// Driver Code
int main()
{
int arr[] = {0, 1, 1, 10};
int n = sizeof(arr) / sizeof(arr[0]);
if (checkArray(arr, n))
cout << "Yes";
else
cout << "No";
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program to find
// if product of every pair
// is present in array.
class GFG
{
// Returns true if product
// of every pair in arr[]
// is present in arr[]
static boolean checkArray(int arr[] ,
int n)
{
// variable to store number
// of zeroes, ones, minus
// one and other numbers.
int zero = 0, one = 0,
minusone = 0, other=0;
for (int i = 0; i < n; i++)
{
// incrementing the
// variable values
if (arr[i] == 0)
zero++;
else if (arr[i] == 1)
one++;
else if (arr[i] == -1)
minusone++;
else
other++;
}
// checking the conditions
if (other > 1)
return false;
else if (other != 0 &&
minusone != 0)
return false;
else if (minusone > 1 &&
one == 0)
return false;
return true;
}
// Driver Code
public static void main (String[] args)
{
int arr[] = {0, 1, 1, 10};
int n = arr.length;
if (checkArray(arr, n))
System.out.println("Yes");
else
System.out.println("No");
}
}
// This code is contributed by Harsh Agarwal
Python 3
# Python3 program to find
# if product of every pair
# is present in array.
# Returns True if product
# of every pair in arr[] is
# present in arr[]
def checkArray(arr, n):
# variable to store number
# of zeroes, ones, minus
# one and other numbers.
zero = 0; one = 0;
minusone = 0; other = 0
for i in range(0, n):
# incrementing the
# variable values
if (arr[i] == 0):
zero += 1
elif (arr[i] == 1):
one += 1
elif (arr[i] == -1):
minusone += 1
else:
other += 1
# checking the conditions
if (other > 1):
return false
elif (other != 0 and
minusone != 0):
return false
elif (minusone > 1 and
one == 0):
return false
return True
# Driver Code
arr = [0, 1, 1, 10]
n = len(arr)
if (checkArray(arr, n)):
print("Yes")
else:
print("No")
# This code is contributed
# by Smitha Dinesh Semwal.
C
// C# program to find if
// product of every pair
// is present in array.
using System;
class GFG
{
// Returns true if product
// of every pair in arr[]
// is present in arr[]
static Boolean checkArray(int []arr ,
int n)
{
// variable to store number
// of zeroes, ones, minus
// one and other numbers.
int zero = 0, one = 0,
minusone = 0, other=0;
for (int i = 0; i < n; i++)
{
// incrementing the
// variable values
if (arr[i] == 0)
zero++;
else if (arr[i] == 1)
one++;
else if (arr[i] == -1)
minusone++;
else
other++;
}
// checking the conditions
if (other > 1)
return false;
else if (other != 0 &&
minusone != 0)
return false;
else if (minusone > 1 &&
one == 0)
return false;
return true;
}
// Driver Code
public static void Main (String[] args)
{
int []arr = {0, 1, 1, 10};
int n = arr.Length;
if (checkArray(arr, n))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by parashar....
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// PHP program to find if
// product of every pair
// is present in array.
// Returns true if product
// of every pair in arr[]
// is present in arr[]
function checkArray($arr , $n)
{
// variable to store number
// of zeroes, ones, minus
// one and other numbers.
$zero = 0; $one = 0;
$minusone = 0; $other=0;
for ($i = 0; $i < $n; $i++)
{
// incrementing the
// variable values
if ($arr[$i] == 0)
$zero++;
else if ($arr[$i] == 1)
$one++;
else if ($arr[$i] == -1)
$minusone++;
else
$other++;
}
// checking the conditions
if ($other > 1)
return false;
else if ($other != 0 &&
$minusone != 0)
return false;
else if ($minusone > 1 &&
$one == 0)
return false;
return true;
}
// Driver Code
{
$arr = array(0, 1, 1, 10);
$n = sizeof($arr) / sizeof($arr[0]);
if (checkArray($arr, $n))
echo "Yes";
else
echo "No";
return 0;
}
//This code is contributed
// by nitin mittal.
?>
java 描述语言
<script>
//javascript program to find if
// product of every pair
// is present in array.
// Returns true if product
// of every pair in arr[]
// is present in arr[]
function checkArray(arr,n)
{
// variable to store number
// of zeroes, ones, minus
// one and other numbers.
let zero = 0, one = 0,
minusone = 0, other=0;
for (let i = 0; i < n; i++)
{
// incrementing the
// variable values
if (arr[i] == 0)
zero++;
else if (arr[i] == 1)
one++;
else if (arr[i] == -1)
minusone++;
else
other++;
}
// checking the conditions
if (other > 1)
return false;
else if (other != 0 &&
minusone != 0)
return false;
else if (minusone > 1 &&
one == 0)
return false;
return true;
}
let arr = [0, 1, 1, 10];
let n = arr.length;
if (checkArray(arr, n))
document.write("Yes");
else
document.write("No");
// This code is contributed by vaibhavrabadiya117.
</script>
输出:
yes
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