从除数中找出数
原文:https://www . geesforgeks . org/find-number-from-its-dividers/
给定一个由 N 个整数组成的数组 arr[] 。整数表示一个数 X 除了 1 和 X 本身之外的所有除数。任务是找到编号 X 。如果没有这样的元素，则打印 -1 。
示例:
输入: arr[] = {2，10，5，4} 输出: 20
输入: arr[] = {2，10，5 } T3】输出: 20
输入: arr[] = {2，15} 输出: -1
方法:对给定的 N 除数进行排序，编号 X 将是排序数组中的第一个编号*最后一个编号。通过将除 1 和 X 之外的所有 X 的除数存储在另一个数组中，交叉检查 X 是否与给定语句矛盾，如果形成的数组与给定数组不相同，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 <bits/stdc++.h>
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<int> 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 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 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<Integer> vec = new Vector<Integer>();
// 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;
// Function call
System.out.print(findX(a, n));
}
}
// This code is contributed by 29AjayKumar
Python 3
# 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<int> vec = new List<int>();
// 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
java 描述语言
<script>
// Javascript implementation of the approach
// Function that returns X
function findX(a, n)
{
// Sort the given array
a.sort((x,y) => x - y);
// Get the possible X
let x = a[0] * a[n - 1];
// Container to store divisors
let vec = [];
// Find the divisors of x
for (let i = 2; i * i <= x; i++)
{
// Check if divisor
if (x % i == 0)
{
vec.push(i);
if (parseInt(x / i) != i)
vec.push(parseInt(x / i));
}
}
// sort the vec because a is sorted
// and we have to compare all the elements
vec.sort((x,y) => x - y);
// if size of both vectors is not same
// then we are sure that both vectors
// can't be equal
if (vec.length != n)
return -1;
else
{
// Check if a and vec have
// same elements in them
let i = 0;
for (let j = 0; j < vec.length; j++)
{
if (a[i++] != vec[j])
return -1;
}
}
return x;
}
// Driver code
let a = [ 2, 5, 4, 10 ];
let n = a.length;
// Function call
document.write(findX(a, n));
</script>
Output
20
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