给定数字n，任务是计算其原始数。原始数(表示为P n ＃)是前n个素数的乘积。数字的本数与数字的阶乘相似。在原始数中，并非所有自然数都被相乘，只有素数被相乘才能计算数字的原始数。用P＃表示。
例子：
Input: n = 3
Output: 30
Priomorial = 2 * 3 * 5 = 30
As a side note, factorial is 2 * 3 * 4 * 5
Input: n = 5
Output: 2310
Primorial = 2 * 3 * 5 * 7 * 11
一个幼稚的方法是逐个检查从1到n的所有数字是否为质数，如果是，则将乘法结果存储在n中，类似地将质数乘法的结果存储为n。
一种有效的方法是使用Sundaram的Sieve查找所有素数至n的n，然后通过将它们全部相乘来计算本数。
C++
// C++ program to find Primorial of given numbers
#include
using namespace std;
const int MAX = 1000000;
// vector to store all prime less than and equal to 10^6
vector primes;
// Function for sieve of sundaram. This function stores all
// prime numbers less than MAX in primes
void sieveSundaram()
{
// In general Sieve of Sundaram, produces primes smaller
// than (2*x + 2) for a number given number x. Since
// we want primes smaller than MAX, we reduce MAX to half
// This array is used to separate numbers of the form
// i+j+2ij from others where 1 <= i <= j
bool marked[MAX/2 + 1] = {0};
// Main logic of Sundaram. Mark all numbers which
// do not generate prime number by doing 2*i+1
for (int i = 1; i <= (sqrt(MAX)-1)/2 ; i++)
for (int j = (i*(i+1))<<1 ; j <= MAX/2 ; j += 2*i +1)
marked[j] = true;
// Since 2 is a prime number
primes.push_back(2);
// Print other primes. Remaining primes are of the
// form 2*i + 1 such that marked[i] is false.
for (int i=1; i<=MAX/2; i++)
if (marked[i] == false)
primes.push_back(2*i + 1);
}
// Function to calculate primorial of n
int calculatePrimorial(int n)
{
// Multiply first n primes
int result = 1;
for (int i=0; i Java
// Java program to find Primorial of given numbers
import java.util.*;
class GFG{
public static int MAX = 1000000;
// vector to store all prime less than and equal to 10^6
static ArrayList primes = new ArrayList();
// Function for sieve of sundaram. This function stores all
// prime numbers less than MAX in primes
static void sieveSundaram()
{
// In general Sieve of Sundaram, produces primes smaller
// than (2*x + 2) for a number given number x. Since
// we want primes smaller than MAX, we reduce MAX to half
// This array is used to separate numbers of the form
// i+j+2ij from others where 1 <= i <= j
boolean[] marked = new boolean[MAX];
// Main logic of Sundaram. Mark all numbers which
// do not generate prime number by doing 2*i+1
for (int i = 1; i <= (Math.sqrt(MAX) - 1) / 2 ; i++)
{
for (int j = (i * (i + 1)) << 1 ; j <= MAX / 2 ; j += 2 * i + 1)
{
marked[j] = true;
}
}
// Since 2 is a prime number
primes.add(2);
// Print other primes. Remaining primes are of the
// form 2*i + 1 such that marked[i] is false.
for (int i = 1; i <= MAX / 2; i++)
{
if (marked[i] == false)
{
primes.add(2 * i + 1);
}
}
}
// Function to calculate primorial of n
static int calculatePrimorial(int n)
{
// Multiply first n primes
int result = 1;
for (int i = 0; i < n; i++)
{
result = result * primes.get(i);
}
return result;
}
// Driver code
public static void main(String[] args)
{
int n = 5;
sieveSundaram();
for (int i = 1 ; i <= n; i++)
{
System.out.println("Primorial(P#) of "+i+" is "+calculatePrimorial(i));
}
}
}
// This Code is contributed by mits Python3
# Python3 program to find Primorial of given numbers
import math
MAX = 1000000;
# vector to store all prime less than and equal to 10^6
primes=[];
# Function for sieve of sundaram. This function stores all
# prime numbers less than MAX in primes
def sieveSundaram():
# In general Sieve of Sundaram, produces primes smaller
# than (2*x + 2) for a number given number x. Since
# we want primes smaller than MAX, we reduce MAX to half
# This array is used to separate numbers of the form
# i+j+2ij from others where 1 <= i <= j
marked=[False]*(int(MAX/2)+1);
# Main logic of Sundaram. Mark all numbers which
# do not generate prime number by doing 2*i+1
for i in range(1,int((math.sqrt(MAX)-1)/2)+1):
for j in range(((i*(i+1))<<1),(int(MAX/2)+1),(2*i+1)):
marked[j] = True;
# Since 2 is a prime number
primes.append(2);
# Print other primes. Remaining primes are of the
# form 2*i + 1 such that marked[i] is false.
for i in range(1,int(MAX/2)):
if (marked[i] == False):
primes.append(2*i + 1);
# Function to calculate primorial of n
def calculatePrimorial(n):
# Multiply first n primes
result = 1;
for i in range(n):
result = result * primes[i];
return result;
# Driver code
n = 5;
sieveSundaram();
for i in range(1,n+1):
print("Primorial(P#) of",i,"is",calculatePrimorial(i));
# This code is contributed by mitsC#
// C# program to find Primorial of given numbers
using System;
using System.Collections;
class GFG{
public static int MAX = 1000000;
// vector to store all prime less than and equal to 10^6
static ArrayList primes = new ArrayList();
// Function for sieve of sundaram. This function stores all
// prime numbers less than MAX in primes
static void sieveSundaram()
{
// In general Sieve of Sundaram, produces primes smaller
// than (2*x + 2) for a number given number x. Since
// we want primes smaller than MAX, we reduce MAX to half
// This array is used to separate numbers of the form
// i+j+2ij from others where 1 <= i <= j
bool[] marked = new bool[MAX];
// Main logic of Sundaram. Mark all numbers which
// do not generate prime number by doing 2*i+1
for (int i = 1; i <= (Math.Sqrt(MAX) - 1) / 2 ; i++)
{
for (int j = (i * (i + 1)) << 1 ; j <= MAX / 2 ; j += 2 * i + 1)
{
marked[j] = true;
}
}
// Since 2 is a prime number
primes.Add(2);
// Print other primes. Remaining primes are of the
// form 2*i + 1 such that marked[i] is false.
for (int i = 1; i <= MAX / 2; i++)
{
if (marked[i] == false)
{
primes.Add(2 * i + 1);
}
}
}
// Function to calculate primorial of n
static int calculatePrimorial(int n)
{
// Multiply first n primes
int result = 1;
for (int i = 0; i < n; i++)
{
result = result * (int)primes[i];
}
return result;
}
// Driver code
public static void Main()
{
int n = 5;
sieveSundaram();
for (int i = 1 ; i <= n; i++)
{
System.Console.WriteLine("Primorial(P#) of "+i+" is "+calculatePrimorial(i));
}
}
}
// This Code is contributed by mitsPHP
输出：
Primorial(P#) of 1 is 2
Primorial(P#) of 2 is 6
Primorial(P#) of 3 is 30
Primorial(P#) of 4 is 210
Primorial(P#) of 5 is 2310