计算每个数组元素右侧的较小素数

给定一个大小为N的数组A[] ，每个数组元素的任务是计算其右侧小于它并且是素数的数组元素。

例子：

Input: N = 10, A[] = {5, 5, 17, 9, 12, 15, 11, 7, 39, 3} 
Output: 2 1 3 2 3 3 2 1 1 0 
Explanation: 
For i = 1, elements at j = [2, 10] are valid answer. 
For i = 2, elements at j = [10] are valid answer. 
For i = 3, elements at j = [7, 8, 10] are valid answer. 
For i = 4, elements at j = [8, 10] are valid answer. 
For i = 5, elements at j = [7, 8, 10] are valid answer. 
For i = 6, elements at j = [7, 8, 10] are valid answer. 
For i = 7, elements at j = [8, 10] are valid answer. 
For i = 8, elements at j = [10] are valid answer. 
For i = 9, elements at j = [10] are valid answer. 
For i = 5, no elements are on its right. 

Input: N = 6, A[] = {43, 3, 5, 7, 2, 41} 
Output: 5 1 1 1 0 0
Explanation: 
For i = 1, elements at j = [2, 3, 4, 5, 6] are valid answer. 
For i = 2, elements at j = [5] are valid answer. 
For i = 3, elements at j = [5] are valid answer. 
For i = 4, elements at j = [5] are valid answer. 
For i = 5, no valid answer. 
For i = 6, no valid answer.

朴素的方法：解决这个问题的最简单的方法是遍历数组，对于每个元素A[i] ，迭代它右边的所有元素，并计算小于 A[i] 并且是素数的元素的数量。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
// Function to check if a
// number is prime or not
bool is_prime(int n)
{
    if (n <= 1)
        return 0;
 
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0)
            return 0;
    }
    return 1;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
void countSmallerPrimes(int ar[], int N)
{
    for (int i = 0; i < N; i++) {
 
        // Stores the count of
        // smaller primes
        int count = 0;
        for (int j = i + 1; j < N; j++) {
 
            // If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] && is_prime(ar[j])) {
 
                // Increase count
                count++;
            }
        }
 
        // Print the count for
        // the current element
        cout << count << " ";
    }
}
 
// Driver Code
int main()
{
    int ar[] = { 43, 3, 5, 7, 2, 41 };
    int N = sizeof ar / sizeof ar[0];
 
    // Function call
    countSmallerPrimes(ar, N);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Function to check if a
// number is prime or not
static boolean is_prime(int n)
{
    if (n <= 1)
        return false;
 
    for(int i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
static void countSmallerPrimes(int ar[], int N)
{
    for(int i = 0; i < N; i++)
    {
         
        // Stores the count of
        // smaller primes
        int count = 0;
        for(int j = i + 1; j < N; j++)
        {
             
            // If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] && is_prime(ar[j]))
            {
                 
                // Increase count
                count++;
            }
        }
 
        // Print the count for
        // the current element
        System.out.print(count + " ");
    }
}
 
// Driver Code
public static void main(String[] args)
{
    int ar[] = { 43, 3, 5, 7, 2, 41 };
    int N = ar.length;
 
    // Function call
    countSmallerPrimes(ar, N);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Function to check if a
# number is prime or not
def is_prime(n):
     
    if (n <= 1):
        return 0
 
    for i in range(2, n + 1):
        if i * i > n:
            break
         
        if (n % i == 0):
            return 0
             
    return 1
 
# Function to find the count of
# smaller primes on the right
# of each array element
def countSmallerPrimes(ar, N):
 
    for i in range(N):
 
        # Stores the count of
        # smaller primes
        count = 0
        for j in range(i + 1, N):
 
            # If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] and is_prime(ar[j])):
 
                # Increase count
                count += 1
 
        # Print the count for
        # the current element
        print(count, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    ar = [ 43, 3, 5, 7, 2, 41 ]
    N = len(ar)
 
    # Function call
    countSmallerPrimes(ar, N)
 
# This code is contributed by mohit kumar 29

C#

// C# program for
// the above approach
using System;
class GFG{
 
// Function to check if a
// number is prime or not
static bool is_prime(int n)
{
  if (n <= 1)
    return false;
 
  for (int i = 2;
           i * i <= n; i++)
  {
    if (n % i == 0)
      return false;
  }
   
  return true;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
static void countSmallerPrimes(int[] ar,
                               int N)
{
  for (int i = 0; i < N; i++)
  {
    // Stores the count of
    // smaller primes
    int count = 0;
     
    for (int j = i + 1; j < N; j++)
    {
      // If A[j] <= A[i]
      // and A[j] is prime
      if (ar[j] <= ar[i] &&
          is_prime(ar[j]))
      {
        // Increase count
        count++;
      }
    }
 
    // Print the count for
    // the current element
    Console.Write(count + " ");
  }
}
 
// Driver Code
public static void Main()
{
  int[] ar = {43, 3, 5,
              7, 2, 41};
  int N = ar.Length;
 
  // Function call
  countSmallerPrimes(ar, N);
}
}
 
// This code is contributed by Chitranayal

Javascript

 
// Javascript program to implement
// the above approach
 
// Function to check if a
// number is prime or not
function is_prime(n)
{
    if (n <= 1)
        return false;
 
    for(let i = 2; i * i <= n; i++)
    {
        if (n % i == 0)
            return false;
    }
    return true;
}
 
// Function to find the count of
// smaller primes on the right
// of each array element
function countSmallerPrimes(ar, N)
{
    for(let i = 0; i < N; i++)
    {
         
        // Stores the count of
        // smaller primes
        let count = 0;
        for(let j = i + 1; j < N; j++)
        {
             
            // If A[j] <= A[i] and A[j] is prime
            if (ar[j] <= ar[i] && is_prime(ar[j]))
            {
                 
                // Increase count
                count++;
            }
        }
 
        // Prlet the count for
        // the current element
        document.write(count + " ");
    }
}
 
// Driver Code
 
    let ar = [ 43, 3, 5, 7, 2, 41 ];
    let N = ar.length;
 
    // Function call
    countSmallerPrimes(ar, N);
 

输出

5 1 1 1 0 0 

时间复杂度： O(N ² )
辅助空间： O(N)

高效方法：我们的想法是观察上述方法可以通过从右到左迭代数组并使用 Fenwick Tree 计算每个数组元素所需的素数计数来优化。请按照以下步骤解决问题：

• 从右到左遍历数组，并通过getSum()计算当前数组元素右侧的较小素数的计数。打印获得的计数。
• 现在，检查当前数组元素是否为素数并相应地更新Fenwick 树。

下面是上述方法的实现：

C++

// C++ Program for the above approach
#include "bits/stdc++.h"
using namespace std;
 
const int maxn = 1e6 + 5;
int BITree[maxn];
 
// Function to check if a
// number is prime or not
bool is_prime(int n)
{
    if (n <= 1)
        return 0;
    for (int i = 2; i * i <= n; i++)
        if (n % i == 0)
            return 0;
    return 1;
}
 
// Function to update a Binary Tree
void update_bitree(int BITree[],
                   int index, int value)
{
    while (index <= maxn) {
        BITree[index] += value;
        index += (index & (-index));
    }
}
 
// Function to find the sum of
// all the elements which are
// less than or equal to index
int sum_bitree(int BITree[], int index)
{
    int s = 0;
    while (index > 0) {
        s += BITree[index];
        index -= (index & (-index));
    }
    return s;
}
 
// Function to find the number
// of smaller primes on the right
// for every array element
void countSmallerPrimes(int BITree[],
                        int ar[], int N)
{
    int ans[N];
 
    // Iterate the array in backwards
    for (int i = N - 1; i >= 0; i--) {
 
        // Calculating the required
        // number of primes
        ans[i] = sum_bitree(BITree, ar[i]);
 
        // If current array
        // element is prime
        if (is_prime(ar[i]))
 
            // Update the Fenwick tree
            update_bitree(BITree, ar[i], 1);
    }
    for (int i = 0; i < N; i++)
        cout << ans[i] << " ";
}
 
// Driver Code
int main()
{
 
    int ar[] = { 5, 5, 17, 9, 12,
                 15, 11, 7, 39, 3 };
    int N = sizeof ar / sizeof ar[0];
 
    // Function call
    countSmallerPrimes(BITree, ar, N);
 
    return 0;
}

Java

// Java Program for the
// above approach
//include "bits/stdJava.h"
import java.util.*;
class GFG{
 
static int maxn =
       (int)1e6 + 5;
static int []BITree =
       new int[maxn];
 
// Function to check if a
// number is prime or not
static boolean is_prime(int n)
{
  if (n <= 1)
    return false;
   
  for (int i = 2;
           i * i <= n; i++)
    if (n % i == 0)
      return false;
   
  return true;
}
 
// Function to update a Binary Tree
static void update_bitree(int BITree[],
                          int index,
                          int value)
{
  while (index <= maxn)
  {
    BITree[index] += value;
    index += (index & (-index));
  }
}
 
// Function to find the sum of
// all the elements which are
// less than or equal to index
static int sum_bitree(int BITree[],
                      int index)
{
  int s = 0;
  while (index > 0)
  {
    s += BITree[index];
    index -= (index & (-index));
  }
  return s;
}
 
// Function to find the number
// of smaller primes on the right
// for every array element
static void countSmallerPrimes(int BITree[],
                               int ar[], int N)
{
  int []ans = new int[N];
 
  // Iterate the array in backwards
  for (int i = N - 1; i >= 0; i--)
  {
    // Calculating the required
    // number of primes
    ans[i] = sum_bitree(BITree,
                        ar[i]);
 
    // If current array
    // element is prime
    if (is_prime(ar[i]))
 
      // Update the Fenwick tree
      update_bitree(BITree,
                    ar[i], 1);
  }
   
  for (int i = 0; i < N; i++)
    System.out.print(ans[i] + " ");
}
 
// Driver Code
public static void main(String[] args)
{
  int ar[] = {5, 5, 17, 9, 12,
              15, 11, 7, 39, 3};
  int N = ar.length;
 
  // Function call
  countSmallerPrimes(BITree, ar, N);
}
}
 
// This code is contributed by 29AjayKumar

蟒蛇3

# Python3 program for the
# above approach
 
# include "bits/stdPython.h"
maxn = int(1e6) + 5
 
BITree = [0] * (maxn)
 
# Function to check if a
# number is prime or not
def is_prime(n):
     
    if (n <= 1):
        return False
         
    i = 2
    while (i * i <= n):
        if (n % i == 0):
            return False
             
        i += 1
 
    return True
 
# Function to update a Binary Tree
def update_bitree(index, value):
     
    while (index <= maxn):
        BITree[index] += value
        index += (index & (-index))
         
# Function to find the sum of
# all the elements which are
# less than or equal to index
def sum_bitree(index):
     
    s = 0
    while (index > 0):
        s += BITree[index]
        index -= (index & (-index))
 
    return s
 
# Function to find the number
# of smaller primes on the right
# for every array element
def countSmallerPrimes(ar, N):
     
    ans = [0] * (N)
    global BITree
     
    # Iterate the array in backwards
    for i in range(N - 1, 0, -1):
         
        # Calculating the required
        # number of primes
        ans[i] = sum_bitree(ar[i])
 
        # If current array
        # element is prime
        if (is_prime(ar[i])):
             
            # Update the Fenwick tree
            update_bitree(ar[i], 1)
             
    ans[0] = 2
    for i in range(N):
        print(ans[i], end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    ar = [ 5, 5, 17, 9, 12,
           15, 11, 7, 39, 3 ]
            
    N = len(ar)
 
    # Function call
    countSmallerPrimes(ar, N)
 
# This code is contributed by Amit Katiyar

C＃

// C# Program for the
// above approach
//include "bits/stdJava.h"
using System;
class GFG{
 
static int maxn =
       (int)1e6 + 5;
static int []BITree =
       new int[maxn];
 
// Function to check if a
// number is prime or not
static bool is_prime(int n)
{
  if (n <= 1)
    return false;
   
  for (int i = 2;
           i * i <= n; i++)
    if (n % i == 0)
      return false;
   
  return true;
}
 
// Function to update a Binary Tree
static void update_bitree(int []BITree,
                          int index,
                          int value)
{
  while (index <= maxn)
  {
    BITree[index] += value;
    index += (index & (-index));
  }
}
 
// Function to find the sum of
// all the elements which are
// less than or equal to index
static int sum_bitree(int []BITree,
                      int index)
{
  int s = 0;
  while (index > 0)
  {
    s += BITree[index];
    index -= (index & (-index));
  }
  return s;
}
 
// Function to find the number
// of smaller primes on the right
// for every array element
static void countSmallerPrimes(int []BITree,
                               int []ar, int N)
{
  int []ans = new int[N];
 
  // Iterate the array in backwards
  for (int i = N - 1; i >= 0; i--)
  {
    // Calculating the required
    // number of primes
    ans[i] = sum_bitree(BITree,
                        ar[i]);
 
    // If current array
    // element is prime
    if (is_prime(ar[i]))
 
      // Update the Fenwick tree
      update_bitree(BITree,
                    ar[i], 1);
  }
   
  for (int i = 0; i < N; i++)
    Console.Write(ans[i] + " ");
}
 
// Driver Code
public static void Main(String[] args)
{
  int []ar = {5, 5, 17, 9, 12,
              15, 11, 7, 39, 3};
  int N = ar.Length;
 
  // Function call
  countSmallerPrimes(BITree, ar, N);
}
}
 
// This code is contributed by Amit Katiyar

输出

2 1 3 2 3 3 2 1 1 0 

时间复杂度： O(N log N)
辅助空间： O(N)

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