查找所有整除最大数组元素的数字
原文：https://www.geeksforgeeks.org/find-all-numbers-that-divide-maximum-array-elements/
给定一个由N个数字组成的数组，任务是打印所有大于 1 的数字，这些数字整除数组元素的最大值。
示例：
输入：
a[] = {6, 6, 12, 12, 18, 13}输出：
2 3 6所有数字均整除数组元素的最大值，即 4
输入：
a[] = {12, 15, 27, 20, 40}输出：
2 3 4 5
方法：
-
使用散列存储每个数组元素的所有因数的计数。 我们可以在
O(sqrt N)中找出所有因数。 -
遍历所有因素，并找到被数字整除的最大数组元素的数量。
-
再次重新遍历所有因数，并打印出现次数最多的因数。
下面是上述方法的实现。
C++
// C++ program to print all the numbers
// that divides maximum of array elements
#include <bits/stdc++.h>
using namespace std;
// Function that prints all the numbers
// which divides maximum of array elements
void printNumbers(int a[], int n)
{
// hash to store the number of times
// a factor is there
unordered_map<int, int> mpp;
for (int i = 0; i < n; i++) {
int num = a[i];
// find all the factors
for (int j = 1; j * j <= num; j++) {
// if j is factor of num
if (num % j == 0) {
if (j != 1)
mpp[j]++;
if ((num / j) != j)
mpp[num / j]++;
}
}
}
// find the maximum times
// it can divide
int maxi = 0;
for (auto it : mpp) {
maxi = max(it.second, maxi);
}
// print all the factors of
// numbers which divides the
// maximum array elements
for (auto it : mpp) {
if (it.second == maxi)
cout << it.first << " ";
}
}
// Driver Code
int main()
{
int a[] = { 12, 15, 27, 20, 40 };
int n = sizeof(a) / sizeof(a[0]);
printNumbers(a, n);
}
Java
// Java program to print all the numbers
// that divides maximum of array elements
import java.util.*;
class GFG
{
// Function that prints all the numbers
// which divides maximum of array elements
static void printNumbers(int a[], int n)
{
// hash to store the number of times
// a factor is there
Map<Integer,
Integer> mpp = new HashMap<>();
for (int i = 0; i < n; i++)
{
int num = a[i];
// find all the factors
for (int j = 1; j * j <= num; j++)
{
// if j is factor of num
if (num % j == 0)
{
if (j != 1)
{
if(mpp.containsKey(j))
{
mpp.put(j, mpp.get(j) + 1);
}
else
{
mpp.put(j, 1);
}
}
if ((num / j) != j)
{
if(mpp.containsKey(num / j))
{
mpp.put(num / j, mpp.get(num / j) + 1);
}
else
{
mpp.put(num / j, 1);
}
}
}
}
}
// find the maximum times
// it can divide
int maxi = 0;
for (Map.Entry<Integer,
Integer> it : mpp.entrySet())
{
maxi = Math.max(it.getValue(), maxi);
}
// print all the factors of
// numbers which divides the
// maximum array elements
for (Map.Entry<Integer,
Integer> it : mpp.entrySet())
{
if (it.getValue() == maxi)
System.out.print(it.getKey() + " ");
}
}
// Driver Code
public static void main(String[] args)
{
int a[] = { 12, 15, 27, 20, 40 };
int n = a.length;
printNumbers(a, n);
}
}
// This code is contributed by Princi Singh
Python3
# Python3 program to print all the numbers
# that divides maximum of array elements
# Function that prints all the numbers
# which divides maximum of array elements
def printNumbers(a, n):
# hash to store the number of times
# a factor is there
mpp = dict()
for i in range(n):
num = a[i]
# find all the factors
for j in range(1, num + 1):
if j * j > num:
break
# if j is factor of num
if (num % j == 0):
if (j != 1):
mpp[j]=mpp.get(j, 0) + 1
if ((num // j) != j):
mpp[num // j]=mpp.get(num//j, 0) + 1
# find the maximum times
# it can divide
maxi = 0
for it in mpp:
maxi = max(mpp[it], maxi)
# print all the factors of
# numbers which divides the
# maximum array elements
for it in mpp:
if (mpp[it] == maxi):
print(it,end=" ")
# Driver Code
a = [12, 15, 27, 20, 40 ]
n = len(a)
printNumbers(a, n)
# This code is contributed by mohit kumar
C
// C# program to print all the numbers
// that divides maximum of array elements
using System;
using System.Collections.Generic;
class GFG
{
// Function that prints all the numbers
// which divides maximum of array elements
static void printNumbers(int []a, int n)
{
// hash to store the number of times
// a factor is there
Dictionary<int,int> mpp = new Dictionary<int,int>();
for (int i = 0; i < n; i++)
{
int num = a[i];
// find all the factors
for (int j = 1; j * j <= num; j++)
{
// if j is factor of num
if (num % j == 0)
{
if (j != 1)
{
if(mpp.ContainsKey(j))
{
var v = mpp[j];
mpp.Remove(j);
mpp.Add(j, v + 1);
}
else
{
mpp.Add(j, 1);
}
}
if ((num / j) != j)
{
if(mpp.ContainsKey(num / j))
{
var v = mpp[num/j];
mpp.Remove(num/j);
mpp.Add(num / j, v + 1);
}
else
{
mpp.Add(num / j, 1);
}
}
}
}
}
// find the maximum times
// it can divide
int maxi = 0;
foreach(KeyValuePair<int, int> it in mpp)
{
maxi = Math.Max(it.Value, maxi);
}
// print all the factors of
// numbers which divides the
// maximum array elements
foreach(KeyValuePair<int, int> it in mpp)
{
if (it.Value == maxi)
Console.Write(it.Key + " ");
}
}
// Driver Code
public static void Main(String[] args)
{
int []a = { 12, 15, 27, 20, 40 };
int n = a.Length;
printNumbers(a, n);
}
}
// This code is contributed by 29AjayKumar
输出：
5 2 3 4
时间复杂度：O(N * sqrt(max(arr)))。
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