给定N个整数的数组arr [] 。整数表示数字X的所有除数，除了1和X本身。任务是找到数字X。如果没有这样的元素，则打印-1 。
例子：
Input: arr[] = {2, 10, 5, 4}
Output: 20
Input: arr[] = {2, 10, 5}
Output: 20
Input: arr[] = {2, 15}
Output: -1
方法：对给定的N个除数进行排序，数字X将是排序数组中的第一个数字*最后一个数字。交叉检查如果X违背了给定语句或通过存储X的所有除数除在另一个阵列1和X和如果不是形成阵列和给定阵列不相同，则打印-1，否则打印X。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
// Function that returns X
int findX(int a[], int n)
{
// Sort the given array
sort(a, a + n);
// Get the possible X
int x = a[0] * a[n - 1];
// Container to store divisors
vector vec;
// Find the divisors of x
for (int i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0)
{
vec.push_back(i);
if ((x / i) != i)
vec.push_back(x / i);
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
sort(vec.begin(), vec.end());
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.size() != n)
return -1;
else
{
// Check if a and vec have
// same elements in them
int i = 0;
for (auto it : vec)
{
if (a[i++] != it)
return -1;
}
}
return x;
}
// Driver code
int main()
{
int a[] = { 2, 5, 4, 10 };
int n = sizeof(a) / sizeof(a[0]);
// Function call
cout << findX(a, n);
return 0;
} Java
// Java implementation of the approach
import java.util.*;
class GFG {
// Function that returns X
static int findX(int a[], int n)
{
// Sort the given array
Arrays.sort(a);
// Get the possible X
int x = a[0] * a[n - 1];
// Container to store divisors
Vector vec = new Vector();
// Find the divisors of x
for (int i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0)
{
vec.add(i);
if ((x / i) != i)
vec.add(x / i);
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
Collections.sort(vec);
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.size() != n)
return -1;
else {
// Check if a and vec have
// same elements in them
int i = 0;
for (int it : vec) {
if (a[i++] != it)
return -1;
}
}
return x;
}
// Driver code
public static void main(String[] args)
{
int a[] = { 2, 5, 4, 10 };
int n = a.length;
// Funciton call
System.out.print(findX(a, n));
}
}
// This code is contributed by 29AjayKumar Python3
# Python3 implementation of the approach
# Function that returns X
import math
def findX(list, int):
# Sort the given array
list.sort()
# Get the possible X
x = list[0]*list[int-1]
# Container to store divisors
vec = []
# Find the divisors of x
i = 2
while(i * i <= x):
# Check if divisor
if(x % i == 0):
vec.append(i)
if ((x//i) != i):
vec.append(x//i)
i += 1
# sort the vec because a is sorted
# and we have to compare all the elements
vec.sort()
# if size of both vectors is not same
# then we are sure that both vectors
# can't be equal
if(len(vec) != int):
return -1
else:
# Check if a and vec have
# same elements in them
j = 0
for it in range(int):
if(a[j] != vec[it]):
return -1
else:
j += 1
return x
# Driver code
a = [2, 5, 4, 10]
n = len(a)
# Function call
print(findX(a, n))C#
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG {
// Function that returns X
static int findX(int[] a, int n)
{
// Sort the given array
Array.Sort(a);
// Get the possible X
int x = a[0] * a[n - 1];
// Container to store divisors
List vec = new List();
// Find the divisors of a number
for (int i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0) {
vec.Add(i);
if ((x / i) != i)
vec.Add(x / i);
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
vec.Sort();
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.Count != n)
{
return -1;
}
else
{
// Check if a and vec have
// same elements in them
int i = 0;
foreach(int it in vec)
{
if (a[i++] != it)
return -1;
}
}
return x;
}
// Driver code
public static void Main(String[] args)
{
int[] a = { 2, 5, 4, 10 };
int n = a.Length;
// Function call
Console.Write(findX(a, n));
}
}
// This code is contributed by 29AjayKumar 输出
20