最小化添加的质数总和以使数组不减少

给定数组arr [] ，任务是通过将质数添加到数组元素以使其增加的质数之和为最小来将其转换为非递减数组。

例子：

Input: arr[] = {2, 1, 5, 4, 3} 
Output: 7 
Explanation: 
{2, 1, 5, 4, 3} -> {2, 3, 5, 6, 6} 
By changing the 2^nd element of array to 3(1 + 2), the 4^th element of the array to 6(4 + 2) and the 5^th element to 6( 3 + 3). 
So, the total cost is 2 + 2 + 3 = 7

Input: arr[] = {3, 3, 3, 3} 
Output: 10

方法：这个想法是在数组元素上添加尽可能最少的质数，以使数组不减少。步骤如下：

1. 初始化变量以存储最小成本，例如res 。
2. 使用Eratosthenes筛网生成并存储所有最多10 ^7的质数。
3. 现在，遍历数组并执行以下步骤：
   • 如果当前元素小于前一个元素，则找到最小的质数(例如最近的质数)，可以添加该质数以使数组不减少。
   • 更新当前元素arr [i] = arr [i] +最近的prime 。
   • 将最接近的原始数添加到res 。
4. 打印获得的res的最终值。

下面是上述方法的实现：

C++

// C++ Program to implement
// the above approach
 
#include 
using namespace std;
 
#define MAX 10000000
 
// Stores if an index is a
// prime / non-prime value
bool isPrime[MAX];
 
// Stores the prime
vector primes;
 
// Function to generate all
// prime numbers
void SieveOfEratosthenes()
{
    memset(isPrime, true, sizeof(isPrime));
 
    for (int p = 2; p * p <= MAX; p++) {
 
        // If current element is prime
        if (isPrime[p] == true) {
 
            // Set all its multiples non-prime
            for (int i = p * p; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
 
    // Store all prime numbers
    for (int p = 2; p <= MAX; p++)
        if (isPrime[p])
            primes.push_back(p);
}
 
// Function to find the closest
// prime to a particular number
int prime_search(vector primes,
                 int diff)
{
 
    // Applying binary search
    // on primes vector
    int low = 0;
    int high = primes.size() - 1;
 
    int res;
 
    while (low <= high) {
        int mid = (low + high) / 2;
 
        // If the prime added makes
        // the elements equal
        if (primes[mid] == diff) {
 
            // Return this as the
            // closest prime
            return primes[mid];
        }
 
        // If the array remains
        // non-decreasing
        else if (primes[mid] < diff) {
 
            // Search for a bigger
            // prime number
            low = mid + 1;
        }
 
        // Otherwise
        else {
 
            res = primes[mid];
 
            // Check if a smaller prime  can
            // make array non-decreasing or not
            high = mid - 1;
        }
    }
 
    // Return closest number
    return res;
}
 
// Function to find the minimum cost
int minCost(int arr[], int n)
{
 
    // Find all primes
    SieveOfEratosthenes();
 
    // Store the result
    int res = 0;
 
    // Iterate over the array
    for (int i = 1; i < n; i++) {
 
        // Current element is less
        // than the previous element
        if (arr[i] < arr[i - 1]) {
            int diff = arr[i - 1] - arr[i];
 
            // Find the closest prime which
            // makes the array non decreasing
            int closest_prime
                = prime_search(primes, diff);
 
            // Add to overall cost
            res += closest_prime;
 
            // Update current element
            arr[i] += closest_prime;
        }
    }
 
    // Return the minimum cost
    return res;
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 2, 1, 5, 4, 3 };
    int n = 5;
 
    // Function Call
    cout << minCost(arr, n);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
 
static final int MAX = 10000000;
 
// Stores if an index is a
// prime / non-prime value
static boolean []isPrime = new boolean[MAX + 1];
 
// Stores the prime
static Vector primes = new Vector();
 
// Function to generate all
// prime numbers
static void SieveOfEratosthenes()
{
    Arrays.fill(isPrime, true);
 
    for(int p = 2; p * p <= MAX; p++)
    {
         
        // If current element is prime
        if (isPrime[p] == true)
        {
             
            // Set all its multiples non-prime
            for(int i = p * p; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
 
    // Store all prime numbers
    for(int p = 2; p <= MAX; p++)
        if (isPrime[p])
            primes.add(p);
}
 
// Function to find the closest
// prime to a particular number
static int prime_search(Vector primes,
                        int diff)
{
 
    // Applying binary search
    // on primes vector
    int low = 0;
    int high = primes.size() - 1;
 
    int res = -1;
 
    while (low <= high)
    {
        int mid = (low + high) / 2;
 
        // If the prime added makes
        // the elements equal
        if (primes.get(mid) == diff)
        {
             
            // Return this as the
            // closest prime
            return primes.get(mid);
        }
 
        // If the array remains
        // non-decreasing
        else if (primes.get(mid) < diff)
        {
             
            // Search for a bigger
            // prime number
            low = mid + 1;
        }
 
        // Otherwise
        else
        {
            res = primes.get(mid);
 
            // Check if a smaller prime can
            // make array non-decreasing or not
            high = mid - 1;
        }
    }
 
    // Return closest number
    return res;
}
 
// Function to find the minimum cost
static int minCost(int arr[], int n)
{
     
    // Find all primes
    SieveOfEratosthenes();
 
    // Store the result
    int res = 0;
 
    // Iterate over the array
    for(int i = 1; i < n; i++)
    {
         
        // Current element is less
        // than the previous element
        if (arr[i] < arr[i - 1])
        {
            int diff = arr[i - 1] - arr[i];
 
            // Find the closest prime which
            // makes the array non decreasing
            int closest_prime = prime_search(primes,
                                             diff);
 
            // Add to overall cost
            res += closest_prime;
 
            // Update current element
            arr[i] += closest_prime;
        }
    }
 
    // Return the minimum cost
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array
    int arr[] = { 2, 1, 5, 4, 3 };
    int n = 5;
 
    // Function call
    System.out.print(minCost(arr, n));
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Pthon3 Program to implement
# the above approach
MAX = 10000000
 
# Stores if an index is a
# prime / non-prime value
isPrime = [True] * (MAX + 1)
 
# Stores the prime
primes = []
 
# Function to generate all
# prime numbers
def SieveOfEratosthenes():
 
    global isPrime
 
    p = 2
    while p * p <= MAX:
 
        # If current element is prime
        if (isPrime[p] == True):
 
            # Set all its multiples non-prime
            for i in range (p * p, MAX + 1, p):
                isPrime[i] = False
               
        p += 1
       
    # Store all prime numbers
    for p in range (2, MAX + 1):
        if (isPrime[p]):
            primes.append(p)
   
# Function to find the closest
# prime to a particular number
def prime_search(primes, diff):
 
    # Applying binary search
    # on primes vector
    low = 0
    high = len(primes) - 1
    
    while (low <= high):
        mid = (low + high) // 2
 
        # If the prime added makes
        # the elements equal
        if (primes[mid] == diff):
 
            # Return this as the
            # closest prime
            return primes[mid]
        
        # If the array remains
        # non-decreasing
        elif (primes[mid] < diff):
 
            # Search for a bigger
            # prime number
            low = mid + 1
      
        # Otherwise
        else:
            res = primes[mid]
 
            # Check if a smaller prime  can
            # make array non-decreasing or not
            high = mid - 1
        
    # Return closest number
    return res
   
# Function to find the minimum cost
def minCost(arr, n):
 
    # Find all primes
    SieveOfEratosthenes()
 
    # Store the result
    res = 0
 
    # Iterate over the array
    for i in range (1, n):
 
        # Current element is less
        # than the previous element
        if (arr[i] < arr[i - 1]):
            diff = arr[i - 1] - arr[i]
 
            # Find the closest prime which
            # makes the array non decreasing
            closest_prime = prime_search(primes, diff)
 
            # Add to overall cost
            res += closest_prime
 
            # Update current element
            arr[i] += closest_prime
       
    # Return the minimum cost
    return res
 
# Driver Code
if __name__ == "__main__":
   
    # Given array
    arr = [2, 1, 5, 4, 3]
    n = 5
 
    #Function Call
    print (minCost(arr, n))
    
# This code is contributed by Chitranayal

C#

// C# program to implement 
// the above approach 
using System;
using System.Collections;
using System.Collections.Generic; 
  
class GFG{
    
static int MAX = 10000000;
   
// Stores if an index is a
// prime / non-prime value
static bool []isPrime = new bool[MAX + 1];
   
// Stores the prime
static ArrayList primes = new ArrayList();
   
// Function to generate all
// prime numbers
static void SieveOfEratosthenes()
{
    Array.Fill(isPrime, true);
   
    for(int p = 2; p * p <= MAX; p++)
    {
           
        // If current element is prime
        if (isPrime[p] == true)
        {
               
            // Set all its multiples non-prime
            for(int i = p * p; i <= MAX; i += p)
                isPrime[i] = false;
        }
    }
   
    // Store all prime numbers
    for(int p = 2; p <= MAX; p++)
        if (isPrime[p])
            primes.Add(p);
}
   
// Function to find the closest
// prime to a particular number
static int prime_search(ArrayList primes,
                        int diff)
{
   
    // Applying binary search
    // on primes vector
    int low = 0;
    int high = primes.Count - 1;
   
    int res = -1;
   
    while (low <= high)
    {
        int mid = (low + high) / 2;
   
        // If the prime added makes
        // the elements equal
        if ((int)primes[mid] == diff)
        {
               
            // Return this as the
            // closest prime
            return (int)primes[mid];
        }
   
        // If the array remains
        // non-decreasing
        else if ((int)primes[mid] < diff)
        {
               
            // Search for a bigger
            // prime number
            low = mid + 1;
        }
   
        // Otherwise
        else
        {
            res = (int)primes[mid];
   
            // Check if a smaller prime can
            // make array non-decreasing or not
            high = mid - 1;
        }
    }
   
    // Return closest number
    return res;
}
   
// Function to find the minimum cost
static int minCost(int []arr, int n)
{
       
    // Find all primes
    SieveOfEratosthenes();
   
    // Store the result
    int res = 0;
   
    // Iterate over the array
    for(int i = 1; i < n; i++)
    {
           
        // Current element is less
        // than the previous element
        if (arr[i] < arr[i - 1])
        {
            int diff = arr[i - 1] - arr[i];
   
            // Find the closest prime which
            // makes the array non decreasing
            int closest_prime = prime_search(primes,
                                             diff);
   
            // Add to overall cost
            res += closest_prime;
   
            // Update current element
            arr[i] += closest_prime;
        }
    }
   
    // Return the minimum cost
    return res;
}
  
// Driver Code
public static void Main(string[] args)
{
     
    // Given array
    int []arr = { 2, 1, 5, 4, 3 };
    int n = 5;
   
    // Function call
    Console.Write(minCost(arr, n));
}
}
 
// This code is contributed by rutvik_56

输出：

7

时间复杂度： O(N * log(logN))
辅助空间： O(N)