给定大小为N的数组arr ，其中包含范围为[1，M]的数字，任务是找到范围为[1，M]的元素，该元素可使LCM最大化。
例子：
Input: arr[]={3, 4, 2, 7}, M = 8
Output: 5
Explanation:
The LCM of existing array (3, 4, 2, 7) = 84
Adding the remaining numbers in 1 to 8 and check the corresponding LCM of the resulting array.
1: LCM of(1, 3, 4, 2, 7) is 84
5: LCM of(5, 3, 4, 2, 7) is 420
6: LCM of(6, 3, 4, 2, 7) is 84
8: LCM of(5, 3, 4, 2, 7) is 168
Clearly, adding 5 maximizes the LCM.
Input: arr[]={2, 5, 3, 8, 1}, M = 9
Output: 7
天真的方法：
- 计算给定阵列的LCM。
- 将数组中不存在的[1，M]范围内的每个元素相加后，计算LCM并返回最大的元素。
高效方法：
- 使用筛子预先计算直到1000的素数。
- 存储给定阵列LCM的每个素数的频率。
- 从值[1，M]进行迭代，对于数组中不存在的每个值，计算该数量的质数因子频率与给定数组的LCM的频率差。
- 返回提供最大积的元素。
下面的代码是上述方法的实现：
C++
// C++ program to find the element
// to be added to maximize the LCM
#include
using namespace std;
// Vector which stores the prime factors
// of all the numbers upto 10000
vector primeFactors[10001];
set s;
// Function which finds prime
// factors using sieve method
void findPrimeFactors()
{
// Boolean array which stores
// true if the index is prime
bool primes[10001];
memset(primes, true, sizeof(primes));
// Sieve of Eratosthenes
for (int i = 2; i < 10001; i++) {
if (primes[i]) {
for (int j = i; j < 10001; j += i) {
if (j != i) {
primes[j] = false;
}
primeFactors[j].push_back(i);
}
}
}
}
// Function which stores frequency of every
// prime factor of LCM of the initial array
void primeFactorsofLCM(int* frequecyOfPrimes,
int* arr, int n)
{
for (int i = 0; i < n; i++) {
for (auto a : primeFactors[arr[i]]) {
int k = 0;
// While the prime factor
// divides the number
while ((arr[i] % a) == 0) {
arr[i] /= a;
k++;
}
frequecyOfPrimes[a]
= max(frequecyOfPrimes[a], k);
}
}
}
// Function which returns the element
// which should be added to array
int elementToBeAdded(int* frequecyOfPrimes, int m)
{
int product = 1;
// To store the final answer
int ans = 1;
for (int i = 2; i <= m; i++) {
if (s.find(i) != s.end())
continue;
int j = i;
int current = 1;
for (auto a : primeFactors[j]) {
int k = 0;
// While the prime factor
// divides the number
while ((j % a) == 0) {
j /= a;
k++;
if (k > frequecyOfPrimes[a]) {
current *= a;
}
}
}
// Check element which provides
// the maximum product
if (current > product) {
product = current;
ans = i;
}
}
return ans;
}
void findElement(int* arr, int n, int m)
{
for (int i = 0; i < n; i++)
s.insert(arr[i]);
int frequencyOfPrimes[10001] = { 0 };
primeFactorsofLCM(frequencyOfPrimes, arr, n);
cout << elementToBeAdded(frequencyOfPrimes, m)
<< endl;
}
// Driver code
int main()
{
// Precomputing the prime factors
// of all numbers upto 10000
findPrimeFactors();
int N = 5;
int M = 9;
int arr[] = { 2, 5, 3, 8, 1 };
findElement(arr, N, M);
return 0;
} Java
// Java program to find the element
// to be added to maximize the LCM
import java.util.*;
class GFG{
// Vector which stores the prime factors
// of all the numbers upto 10000
static Vector []primeFactors = new Vector[10001];
static HashSet s = new HashSet();
// Function which finds prime
// factors using sieve method
static void findPrimeFactors()
{
// Boolean array which stores
// true if the index is prime
boolean []primes = new boolean[10001];
Arrays.fill(primes, true);
// Sieve of Eratosthenes
for (int i = 2; i < 10001; i++) {
if (primes[i]) {
for (int j = i; j < 10001; j += i) {
if (j != i) {
primes[j] = false;
}
primeFactors[j].add(i);
}
}
}
}
// Function which stores frequency of every
// prime factor of LCM of the initial array
static void primeFactorsofLCM(int []frequecyOfPrimes,
int[] arr, int n)
{
for (int i = 0; i < n; i++) {
for (int a : primeFactors[arr[i]]) {
int k = 0;
// While the prime factor
// divides the number
while ((arr[i] % a) == 0) {
arr[i] /= a;
k++;
}
frequecyOfPrimes[a]
= Math.max(frequecyOfPrimes[a], k);
}
}
}
// Function which returns the element
// which should be added to array
static int elementToBeAdded(int []frequecyOfPrimes, int m)
{
int product = 1;
// To store the final answer
int ans = 1;
for (int i = 2; i <= m; i++) {
if (s.contains(i))
continue;
int j = i;
int current = 1;
for (int a : primeFactors[j]) {
int k = 0;
// While the prime factor
// divides the number
while ((j % a) == 0) {
j /= a;
k++;
if (k > frequecyOfPrimes[a]) {
current *= a;
}
}
}
// Check element which provides
// the maximum product
if (current > product) {
product = current;
ans = i;
}
}
return ans;
}
static void findElement(int[] arr, int n, int m)
{
for (int i = 0; i < n; i++)
s.add(arr[i]);
int frequencyOfPrimes[] = new int[10001];
primeFactorsofLCM(frequencyOfPrimes, arr, n);
System.out.print(elementToBeAdded(frequencyOfPrimes, m)
+"\n");
}
// Driver code
public static void main(String[] args)
{
for (int i = 0; i < 10001; i++)
primeFactors[i] = new Vector();
// Precomputing the prime factors
// of all numbers upto 10000
findPrimeFactors();
int N = 5;
int M = 9;
int arr[] = { 2, 5, 3, 8, 1 };
findElement(arr, N, M);
}
}
// This code is contributed by Rajput-Ji C#
// C# program to find the element
// to be added to maximize the LCM
using System;
using System.Collections.Generic;
class GFG{
// List which stores the prime factors
// of all the numbers upto 10000
static List []primeFactors = new List[10001];
static HashSet s = new HashSet();
// Function which finds prime
// factors using sieve method
static void findPrimeFactors()
{
// Boolean array which stores
// true if the index is prime
bool []primes = new bool[10001];
for (int i = 0; i < 10001; i++)
primes[i] = true;
// Sieve of Eratosthenes
for (int i = 2; i < 10001; i++) {
if (primes[i]) {
for (int j = i; j < 10001; j += i) {
if (j != i) {
primes[j] = false;
}
primeFactors[j].Add(i);
}
}
}
}
// Function which stores frequency of every
// prime factor of LCM of the initial array
static void primeFactorsofLCM(int []frequecyOfPrimes,
int[] arr, int n)
{
for (int i = 0; i < n; i++) {
foreach (int a in primeFactors[arr[i]]) {
int k = 0;
// While the prime factor
// divides the number
while ((arr[i] % a) == 0) {
arr[i] /= a;
k++;
}
frequecyOfPrimes[a]
= Math.Max(frequecyOfPrimes[a], k);
}
}
}
// Function which returns the element
// which should be added to array
static int elementToBeAdded(int []frequecyOfPrimes, int m)
{
int product = 1;
// To store the readonly answer
int ans = 1;
for (int i = 2; i <= m; i++) {
if (s.Contains(i))
continue;
int j = i;
int current = 1;
foreach (int a in primeFactors[j]) {
int k = 0;
// While the prime factor
// divides the number
while ((j % a) == 0) {
j /= a;
k++;
if (k > frequecyOfPrimes[a]) {
current *= a;
}
}
}
// Check element which provides
// the maximum product
if (current > product) {
product = current;
ans = i;
}
}
return ans;
}
static void findElement(int[] arr, int n, int m)
{
for (int i = 0; i < n; i++)
s.Add(arr[i]);
int []frequencyOfPrimes = new int[10001];
primeFactorsofLCM(frequencyOfPrimes, arr, n);
Console.Write(elementToBeAdded(frequencyOfPrimes, m)
+"\n");
}
// Driver code
public static void Main(String[] args)
{
for (int i = 0; i < 10001; i++)
primeFactors[i] = new List();
// Precomputing the prime factors
// of all numbers upto 10000
findPrimeFactors();
int N = 5;
int M = 9;
int []arr = { 2, 5, 3, 8, 1 };
findElement(arr, N, M);
}
}
// This code is contributed by PrinciRaj1992 输出：
7
时间复杂度： O(N * log N + M * log M)