在给定的LCM下最小化K个正整数的总和

给定两个正整数K和X ，任务是找到具有LCM X的K个正整数(允许重复)的最小和。

例子：

Input: K = 2, X = 6 
Output: 5 
Explanation: 
K(= 2) positive integers of minimum possible sum having LCM X(= 6) are { 2, 3 }. 
Therefore, the minimum possible sum of K(= 2) positive integers = (2 + 3) = 5. 
Therefore, the required output is 5.

Input: K = 3 X = 11 
Output: 13 
Explanation: 
K(= 3) positive integers of minimum possible sum having LCM X(= 11) are { 1, 1, 11 }. 
Therefore, the minimum possible sum of K(= 3) positive integers = (1 + 1 + 11) = 13. 
Therefore, the required output is 13.

方法：可以使用贪婪技术解决问题。这个想法是用素数乘积的形式表示X。以所有可能的方式从X的所有素数中选择K个素数，并计算它们各自的总和。最后，在所有这些中打印最小可能的总和。请按照以下步骤解决问题：

• 初始化一个数组，例如primePow [] ，以存储X的所有素数。
• 如果阵列primePow []的长度小于或等于K，则包括primePow []以K的正整数和所有的数组元素中的k的正整数必须是1的剩余元件。最后，打印出K个正整数的总和

Illustration: 
If K = 5, X = 240 
primePow[] = { 2⁴, 3¹, 5¹ } = { 16, 3, 5 } 
K positive integers will be { 16, 3, 5, 1, 1 } 
Therefore, the sum of K positive integers will be 26 
 

• 否则，以所有可能的方式将primePow []数组划分为K个组，并计算它们各自的总和。最后，打印K个正整数的最小和。

If K = 3, X = 210 
primePow[] = { 2¹, 3¹, 5¹, 5¹ } = { 2, 3, 5, 7 } 
Partition primePow[] array into { { 2, 3 }, { 5 }, { 7 } } 
210 = (2 * 3) * (5) * 7 = 6 * 5 * 7 
K positive integers will be { 6, 5, 7 } 
Therefore, the sum of K(= 3) positive integers will be 18 
 

下面是上述实现的实现：

C++

// C++ program to implement
// the above approach
#include 
using namespace std;
 
// Function to find the prime
// power of X
vector primePower(int X)
{
 
    // Stores prime powers of X
    vector primePow;
 
    // Iterate over the range [2, sqrt(X)]
    for (int i = 2; i * i <= X; i++) {
 
        // If X is divisible by i
        if (X % i == 0) {
 
            // Stores prime power
            int p = 1;
 
            // Calculate prime power
            // of X
            while (X % i == 0) {
 
                // Update X
                X /= i;
 
                // Update p
                p *= i;
            }
 
            // Insert prime powers
            // into primePow[]
            primePow.push_back(p);
        }
    }
 
    // If X exceeds 1
    if (X > 1) {
        primePow.push_back(X);
    }
 
    return primePow;
}
 
// Function to calculate the
// sum of array elements
int getSum(vector& ar)
{
    // Stores sum of
    // array elements
    int sum = 0;
 
    // Traverse the array
    for (int i : ar) {
 
        // Update sum
        sum += i;
    }
 
    return sum;
}
 
// Function to partition array into K groups such
// that sum of elements of the K groups is minimum
int getMinSum(int pos, vector& arr,
              vector& primePow)
{
 
    // If count of prime powers
    // is equal to pos
    if (pos == primePow.size()) {
        return getSum(arr);
    }
 
    // Stores minimum sum of of each
    // partition of arr[] into K groups
    int res = INT_MAX;
 
    // Traverse the array arr[]
    for (int i = 0; i < arr.size();
         i++) {
 
        // Include primePow[pos] into
        // i-th groups
        arr[i] *= primePow[pos];
 
        // Update res
        res = min(res, getMinSum(pos + 1,
                                 arr, primePow));
 
        // Remove factors[pos] from
        // i-th groups
        arr[i] /= primePow[pos];
    }
 
    return res;
}
 
// Utility function to calculate minimum sum
// of K positive integers whose LCM is X
int minimumSumWithGivenLCM(int k, int x)
{
    // Stores all prime powers of X
    vector primePow = primePower(x);
 
    // Stores count of prime powers
    int n = primePow.size();
 
    // Stores minimum sum of K positive
    // integers whose LCM is X
    int sum = 0;
 
    // If n is less than
    // or equal to k
    if (n <= k) {
 
        // Traverse primePow[] array
        for (int i : primePow) {
 
            // Update sum
            sum += i;
        }
 
        // Update sum
        sum += k - n;
    }
 
    else {
 
        // arr[i]: Stores element in i-th group
        // by partitioning the primePow[] array
        vector arr(k, 1);
 
        // Update sum
        sum = getMinSum(0, arr, primePow);
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int k = 3, x = 210;
 
    cout << minimumSumWithGivenLCM(k, x);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
// Function to find the prime
// power of X
static Vector primePower(int X)
{
 
    // Stores prime powers of X
    Vector primePow = new Vector();
 
    // Iterate over the range [2, Math.sqrt(X)]
    for (int i = 2; i * i <= X; i++) {
 
        // If X is divisible by i
        if (X % i == 0) {
 
            // Stores prime power
            int p = 1;
 
            // Calculate prime power
            // of X
            while (X % i == 0) {
 
                // Update X
                X /= i;
 
                // Update p
                p *= i;
            }
 
            // Insert prime powers
            // into primePow[]
            primePow.add(p);
        }
    }
 
    // If X exceeds 1
    if (X > 1) {
        primePow.add(X);
    }
 
    return primePow;
}
 
// Function to calculate the
// sum of array elements
static int getSum(int []ar)
{
    // Stores sum of
    // array elements
    int sum = 0;
 
    // Traverse the array
    for (int i : ar) {
 
        // Update sum
        sum += i;
    }
 
    return sum;
}
 
// Function to partition array into K groups such
// that sum of elements of the K groups is minimum
static int getMinSum(int pos, int []arr,
              Vector primePow)
{
 
    // If count of prime powers
    // is equal to pos
    if (pos == primePow.size()) {
        return getSum(arr);
    }
 
    // Stores minimum sum of of each
    // partition of arr[] into K groups
    int res = Integer.MAX_VALUE;
 
    // Traverse the array arr[]
    for (int i = 0; i < arr.length;
         i++) {
 
        // Include primePow[pos] into
        // i-th groups
        arr[i] *= primePow.get(pos);
 
        // Update res
        res = Math.min(res, getMinSum(pos + 1,
                                 arr, primePow));
 
        // Remove factors[pos] from
        // i-th groups
        arr[i] /= primePow.get(pos);
    }
 
    return res;
}
 
// Utility function to calculate minimum sum
// of K positive integers whose LCM is X
static int minimumSumWithGivenLCM(int k, int x)
{
    // Stores all prime powers of X
    Vector primePow = primePower(x);
 
    // Stores count of prime powers
    int n = primePow.size();
 
    // Stores minimum sum of K positive
    // integers whose LCM is X
    int sum = 0;
 
    // If n is less than
    // or equal to k
    if (n <= k) {
 
        // Traverse primePow[] array
        for (int i : primePow) {
 
            // Update sum
            sum += i;
        }
 
        // Update sum
        sum += k - n;
    }
 
    else {
 
        // arr[i]: Stores element in i-th group
        // by partitioning the primePow[] array
        int []arr = new int[k];
        Arrays.fill(arr, 1);
 
        // Update sum
        sum = getMinSum(0, arr, primePow);
    }
 
    return sum;
}
 
// Driver Code
public static void main(String[] args)
{
    int k = 3, x = 210;
 
    System.out.print(minimumSumWithGivenLCM(k, x));
}
}
 
// This code is contributed by 29AjayKumar

Python3

# Python3 program to implement
# the above approach
 
# Function to find the prime
# power of X
def primePower(X):
 
    # Stores prime powers of X
    primePow = []
 
    # Iterate over the range [2, sqrt(X)]
    for i in range(2, X + 1):
 
        if i * i > X + 1:
            break
 
        # If X is divisible by i
        if (X % i == 0):
 
            # Stores prime power
            p = 1
 
            # Calculate prime power
            # of X
            while (X % i == 0):
 
                # Update X
                X //= i
 
                # Update p
                p *= i
 
            # Insert prime powers
            # into primePow[]
            primePow.append(p)
 
    # If X exceeds 1
    if (X > 1):
        primePow.append(X)
 
    return primePow
 
# Function to calculate the
# sum of array elements
def getSum(ar):
     
    # Stores sum of
    # array elements
    sum = 0
 
    # Traverse the array
    for i in ar:
 
        # Update sum
        sum += i
 
    return sum
 
# Function to partition array into K groups such
# that sum of elements of the K groups is minimum
def getMinSum(pos, arr, primePow):
 
    # If count of prime powers
    # is equal to pos
    if (pos == len(primePow)):
        return getSum(arr)
 
    # Stores minimum sum of of each
    # partition of arr[] into K groups
    res = 10**9
 
    # Traverse the array arr[]
    for i in range(len(arr)):
         
        # Include primePow[pos] into
        # i-th groups
        arr[i] *= primePow[pos]
 
        # Update res
        res = min(res, getMinSum(pos + 1, arr, primePow))
 
        #Remove factors[pos] from
        #i-th groups
        arr[i] //= primePow[pos]
 
    return res
 
# Utility function to calculate minimum sum
# of K positive integers whose LCM is X
def minimumSumWithGivenLCM(k, x):
     
    # Stores all prime powers of X
    primePow = primePower(x)
 
    # Stores count of prime powers
    n = len(primePow)
 
    # Stores minimum sum of K positive
    # integers whose LCM is X
    sum = 0
 
    # If n is less than
    # or equal to k
    if (n <= k):
 
        # Traverse primePow[] array
        for i in primePow:
 
            # Update sum
            sum += i
 
        # Update sum
        sum += k - n
    else:
 
        # arr[i]: Stores element in i-th group
        # by partitioning the primePow[] array
        arr = [1] * (k)
 
        # Update sum
        sum = getMinSum(0, arr, primePow)
 
    return sum
 
# Driver Code
if __name__ == '__main__':
    k = 3
    x = 210
 
    print(minimumSumWithGivenLCM(k, x))
 
    # This code is contributed by mohit kumar 29

C#

// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to find the prime
// power of X
static List primePower(int X)
{
     
    // Stores prime powers of X
    List primePow = new List();
 
    // Iterate over the range [2, Math.Sqrt(X)]
    for(int i = 2; i * i <= X; i++)
    {
         
        // If X is divisible by i
        if (X % i == 0)
        {
             
            // Stores prime power
            int p = 1;
 
            // Calculate prime power
            // of X
            while (X % i == 0)
            {
                 
                // Update X
                X /= i;
 
                // Update p
                p *= i;
            }
 
            // Insert prime powers
            // into primePow[]
            primePow.Add(p);
        }
    }
 
    // If X exceeds 1
    if (X > 1)
    {
        primePow.Add(X);
    }
    return primePow;
}
 
// Function to calculate the
// sum of array elements
static int getSum(int []ar)
{
     
    // Stores sum of
    // array elements
    int sum = 0;
 
    // Traverse the array
    foreach(int i in ar)
    {
         
        // Update sum
        sum += i;
    }
    return sum;
}
 
// Function to partition array into K groups such
// that sum of elements of the K groups is minimum
static int getMinSum(int pos, int []arr,
                List primePow)
{
     
    // If count of prime powers
    // is equal to pos
    if (pos == primePow.Count)
    {
        return getSum(arr);
    }
 
    // Stores minimum sum of of each
    // partition of []arr into K groups
    int res = int.MaxValue;
 
    // Traverse the array []arr
    for(int i = 0; i < arr.Length; i++)
    {
         
        // Include primePow[pos] into
        // i-th groups
        arr[i] *= primePow[pos];
 
        // Update res
        res = Math.Min(res, getMinSum(
            pos + 1, arr, primePow));
 
        // Remove factors[pos] from
        // i-th groups
        arr[i] /= primePow[pos];
    }
    return res;
}
 
// Utility function to calculate minimum sum
// of K positive integers whose LCM is X
static int minimumSumWithGivenLCM(int k, int x)
{
     
    // Stores all prime powers of X
    List primePow = primePower(x);
 
    // Stores count of prime powers
    int n = primePow.Count;
 
    // Stores minimum sum of K positive
    // integers whose LCM is X
    int sum = 0;
 
    // If n is less than
    // or equal to k
    if (n <= k)
    {
         
        // Traverse primePow[] array
        foreach(int i in primePow)
        {
             
            // Update sum
            sum += i;
        }
 
        // Update sum
        sum += k - n;
    }
 
    else
    {
         
        // arr[i]: Stores element in i-th group
        // by partitioning the primePow[] array
        int []arr = new int[k];
        for(int l = 0; l < arr.Length; l++)
            arr[l] = 1;
 
        // Update sum
        sum = getMinSum(0, arr, primePow);
    }
    return sum;
}
 
// Driver Code
public static void Main(String[] args)
{
    int k = 3, x = 210;
 
    Console.Write(minimumSumWithGivenLCM(k, x));
}
}
 
// This code is contributed by aashish1995

输出：

18

时间复杂度： O(sqrt(X)+ 3 ^Y )，其中Y是素数的最大数量
辅助空间： O(K + Y)