查找给定素数的位置

给定数字N为质数，任务是找到给定质数在质数系列中的位置。
例子 ：

Input: N = 11 
Output: 5 
Explanation: 
The prime numbers are 2, 3, 5, 7, 11, 13, 17, …. 
Therefore, the position of 11 in this series is 5.
Input: N = 13 
Output: 6 
 

朴素方法：针对给定输入，此问题的朴素方法是计算小于该数字的质数，并跟踪小于给定N的质数。如果计数为K ，则答案为K +1 。这种方法的时间复杂度是二次的。
高效方法：想法是对Eratosthenes筛子进行略微修改。可以计算所有最大值的质数，并将其位置与数组一起存储在数组中。显然，当质数存储在数组中时，存储该数字的索引是该数字在系列中的位置。预计算之后，可以在恒定时间内计算出答案。
下面是上述方法的实现：

C++

// C++ program to find the position
// of the given prime number
 
#include 
#define limit 10000000
using namespace std;
int position[limit + 1];
 
// Function to precompute the position
// of every prime number using Sieve
void sieve()
{
    // 0 and 1 are not prime numbers
    position[0] = -1, position[1] = -1;
 
    // Variable to store the position
    int pos = 0;
    for (int i = 2; i <= limit; i++) {
        if (position[i] == 0) {
 
            // Incrementing the position for
            // every prime number
            position[i] = ++pos;
            for (int j = i * 2; j <= limit; j += i)
                position[j] = -1;
        }
    }
}
 
// Driver code
int main()
{
    sieve();
 
    int n = 11;
    cout << position[n];
    return 0;
}

Java

// Java program to find the position
// of the given prime number
class GFG{   
     
static final int limit = 10000000;
static int []position = new int[limit + 1];
  
// Function to precompute the position
// of every prime number using Sieve
static void sieve()
{
    // 0 and 1 are not prime numbers
    position[0] = -1;
    position[1] = -1;
  
    // Variable to store the position
    int pos = 0;
    for (int i = 2; i <= limit; i++) {
        if (position[i] == 0) {
  
            // Incrementing the position for
            // every prime number
            position[i] = ++pos;
            for (int j = i * 2; j <= limit; j += i)
                position[j] = -1;
        }
    }
}
  
// Driver code
public static void main(String[] args)
{
    sieve();
  
    int n = 11;
    System.out.print(position[n]);
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to find the position
# of the given prime number
limit = 1000000
position = [0]*(limit + 1)
  
# Function to precompute the position
# of every prime number using Sieve
def sieve():
 
    # 0 and 1 are not prime numbers
    position[0] = -1
    position[1] = -1
  
    # Variable to store the position
    pos = 0
    for i in range(2, limit + 1):
        if (position[i] == 0):
  
            # Incrementing the position for
            # every prime number
            pos += 1
            position[i] = pos
            for j in range( i * 2, limit + 1 ,i):
                position[j] = -1
  
# Driver code
if __name__ == "__main__":
    sieve()
  
    n = 11
    print(position[n])
     
# This code is contributed by chitranayal

C#

// C# program to find the position
// of the given prime number
using System;
 
class GFG{   
      
static readonly int limit = 1000000;
static int []position = new int[limit + 1];
   
// Function to precompute the position
// of every prime number using Sieve
static void sieve()
{
    // 0 and 1 are not prime numbers
    position[0] = -1;
    position[1] = -1;
   
    // Variable to store the position
    int pos = 0;
    for (int i = 2; i <= limit; i++) {
        if (position[i] == 0) {
   
            // Incrementing the position for
            // every prime number
            position[i] = ++pos;
            for (int j = i * 2; j <= limit; j += i)
                position[j] = -1;
        }
    }
}
   
// Driver code
public static void Main(String[] args)
{
    sieve();
   
    int n = 11;
    Console.Write(position[n]);
}
}
  
// This code is contributed by Princi Singh

Javascript

 
// Javascript program to find the position
// of the given prime number
var limit = 10000000
var position = Array(limit+1).fill(0);
 
// Function to precompute the position
// of every prime number using Sieve
function sieve()
{
 
    // 0 and 1 are not prime numbers
    position[0] = -1, position[1] = -1;
 
    // Variable to store the position
    var pos = 0;
    for (var i = 2; i <= limit; i++)
    {
        if (position[i] == 0)
        {
 
            // Incrementing the position for
            // every prime number
            position[i] = ++pos;
            for (var j = i * 2; j <= limit; j += i)
                position[j] = -1;
        }
    }
}
 
// Driver code
sieve();
var n = 11;
document.write( position[n]);
 
// This code is contributed by noob2000.

输出：

5