给定一个由N个数字组成的数组，该数组具有两个数字的所有除数，且顺序为任意数。任务是找到在数组中给出除数的两个数字。
例子：
Input: a[] = {10, 2, 8, 1, 2, 4, 1, 20, 4, 5}
Output: 20 8
The divisors of 20 and 8 are given in the array.
Input: a[] = {2, 2, 1, 1}
Output: 1 2
方法：按照以下步骤解决上述问题：
- 在数组中找到最大元素，这是两个数字之一。
- 保留一个频率数组，该频率对数组元素的出现进行计数。
- 减少频率阵列中最大元素的所有除数的数量。
- 下一个最大元素将是第二个数字。
下面是上述方法的实现：
C++
// C++ program to find the two numbers
// when divisors are given in a random order
#include
using namespace std;
// Function to get the two numbers
void getNumbers(int a[], int n)
{
// Mark the frequency of all elements
unordered_map freq;
for (int i = 0; i < n; i++)
freq[a[i]]++;
// Get the first maximum element
int maxi1 = *max_element(a, a+n);
// Decrease the frequency of all divisors
// of the maximum number that are present
// int a[]
for (int i = 1; i * i <= maxi1; i++) {
if (maxi1 % i == 0 &&
freq.find(i) != freq.end() &&
freq[i] != 0) {
freq[i]--;
if (i != (maxi1 / i) &&
freq.find(maxi1 / i) != freq.end() &&
freq[maxi1 / i] != 0)
freq[maxi1 / i]--;
}
}
// The second number is the largest number
// present in remaining numbers.
int maxi2 = -1;
for (int i = 0; i < n; i++) {
if (freq[a[i]] != 0)
maxi2 = max(maxi2, a[i]);
}
cout << maxi1 << " " << maxi2;
}
// Driver Code
int main()
{
int a[] = { 10, 2, 8, 1, 2, 4, 1, 20, 4, 5 };
int n = sizeof(a) / sizeof(a[0]);
getNumbers(a, n);
return 0;
} Java
// Java program to find the two numbers
// when divisors are given in a random order
import java.util.*;
class GFG{
// Function to get the two numbers
static void getNumbers(int a[], int n)
{
// Mark the frequency of all elements
HashMap freq = new HashMap();
for (int i = 0; i < n; i++)
if(freq.containsKey(a[i])){
freq.put(a[i], freq.get(a[i])+1);
}
else{
freq.put(a[i], 1);
}
// Get the first maximum element
int maxi1 = Arrays.stream(a).max().getAsInt();
// Decrease the frequency of all divisors
// of the maximum number that are present
// int a[]
for (int i = 1; i * i <= maxi1; i++) {
if (maxi1 % i == 0 &&
freq.containsKey(i)&&
freq.get(i)!= 0) {
freq.put(i, freq.get(i)- 1);
if (i != (maxi1 / i) &&
freq.containsKey(maxi1 / i) &&
freq.get(maxi1 / i)!= 0)
freq.put(maxi1 / i, freq.get(maxi1 / i) - 1);
}
}
// The second number is the largest number
// present in remaining numbers.
int maxi2 = -1;
for (int i = 0; i < n; i++) {
if (freq.get(a[i]) != 0)
maxi2 = Math.max(maxi2, a[i]);
}
System.out.print(maxi1+ " " + maxi2);
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 10, 2, 8, 1, 2, 4, 1, 20, 4, 5 };
int n = a.length;
getNumbers(a, n);
}
}
// This code contributed by Rajput-Ji Python3
# Python3 program to find the two numbers
# when divisors are given in a random order
# from math lib import sqrt method
from math import sqrt
# Function to get the two numbers
def getNumbers(a, n) :
# Mark the frequency of all elements
freq = {};
for i in range(n) :
if a[i] not in freq.keys() :
freq[a[i]] = 0
freq[a[i]] += 1
# Get the first maximum element
maxi1 = max(a)
# Decrease the frequency of all divisors
# of the maximum number that are present
# int a[]
for i in range(1, int(sqrt(maxi1)) + 1) :
if (maxi1 % i == 0 and
freq[i] in freq.keys() and
freq[i] != 0) :
freq[i] -= 1
if (i != (maxi1 // i) and
freq[maxi1 // i] in freq.keys() and
freq[maxi1 // i] != 0) :
freq[maxi1 // i] -= 1
# The second number is the largest number
# present in remaining numbers.
maxi2 = -1
for i in range(n) :
if (freq[a[i]] != 0) :
maxi2 = max(maxi2, a[i])
print(maxi1, maxi2)
# Driver Code
if __name__ == "__main__" :
a = [ 10, 2, 8, 1, 2, 4, 1, 20, 4, 5 ]
n = len(a)
getNumbers(a, n)
# This code is contributed by RyugaC#
// C# program to find the two numbers
// when divisors are given in a random order
using System;
using System.Collections.Generic;
using System.Linq;
public class GFG{
// Function to get the two numbers
static void getNumbers(int []a, int n)
{
// Mark the frequency of all elements
Dictionary freq = new Dictionary();
for (int i = 0; i < n; i++)
if(freq.ContainsKey(a[i])){
freq[a[i]] = freq[a[i]]+1;
}
else{
freq.Add(a[i], 1);
}
// Get the first maximum element
int maxi1 = a.Max();
// Decrease the frequency of all divisors
// of the maximum number that are present
// int []a
for (int i = 1; i * i <= maxi1; i++) {
if (maxi1 % i == 0 &&
freq.ContainsKey(i)&&
freq[i]!= 0) {
freq[i] = freq[i]- 1;
if (i != (maxi1 / i) &&
freq.ContainsKey(maxi1 / i) &&
freq[maxi1 / i]!= 0)
freq[maxi1 / i] = freq[maxi1 / i] - 1;
}
}
// The second number is the largest number
// present in remaining numbers.
int maxi2 = -1;
for (int i = 0; i < n; i++) {
if (freq[a[i]] != 0)
maxi2 = Math.Max(maxi2, a[i]);
}
Console.Write(maxi1+ " " + maxi2);
}
// Driver Code
public static void Main(String[] args)
{
int []a = { 10, 2, 8, 1, 2, 4, 1, 20, 4, 5 };
int n = a.Length;
getNumbers(a, n);
}
}
// This code is contributed by PrinciRaj1992 输出：
20 8
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