Given a positive integer N, calculate the product of the first N prime numbers.
Examples:
Input : N = 3 Output : 30 Explanation : First 3 prime numbers are 2, 3, 5. Input : N = 5 Output : 2310
Approach:
- Create a sieve which will help us to identify if the number is prime or not in O(1) time.
- Run a loop starting from 1 until and unless we find n prime numbers.
- Multiply all the prime numbers and neglect those which are not prime.
- Then, display the product of 1st N prime numbers.
Time Complexity – O( Nlog(logN) )
Below is the implementation of above approach:
C++
// C++ implementation of above solution #include "cstring" #include <iostream> using namespace std;
#define MAX 10000 // Create a boolean array "prime[0..n]" and initialize // all entries it as true. A value in prime[i] will // finally be false if i is Not a prime, else true. bool prime[MAX + 1];
void SieveOfEratosthenes()
{ memset(prime, true, sizeof(prime));
prime[1] = false;
for (int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == true) {
// Set all multiples of p to non-prime
for (int i = p * 2; i <= MAX; i += p)
prime[i] = false;
}
}
} // find the product of 1st N prime numbers int solve(int n)
{ // count of prime numbers
int count = 0, num = 1;
// product of prime numbers
long long int prod = 1;
while (count < n) {
// if the number is prime add it
if (prime[num]) {
prod *= num;
// increase the count
count++;
}
// get to next number
num++;
}
return prod;
} // Driver code int main()
{ // create the sieve
SieveOfEratosthenes();
int n = 5;
// find the value of 1st n prime numbers
cout << solve(n);
return 0;
} |
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Java
// Java implementation of above solution class GFG
{ static int MAX=10000;
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
static boolean[] prime=new boolean[MAX + 1];
static void SieveOfEratosthenes()
{ prime[1] = true;
for (int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == false) {
// Set all multiples of p to non-prime
for (int i = p * 2; i <= MAX; i += p)
prime[i] = true;
}
}
} // find the product of 1st N prime numbers static int solve(int n)
{ // count of prime numbers
int count = 0, num = 1;
// product of prime numbers
int prod = 1;
while (count < n) {
// if the number is prime add it
if (!prime[num]) {
prod *= num;
// increase the count
count++;
}
// get to next number
num++;
}
return prod;
} // Driver code public static void main(String[] args)
{ // create the sieve
SieveOfEratosthenes();
int n = 5;
// find the value of 1st n prime numbers
System.out.println(solve(n));
} } // This code is contributed by mits |
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C#
// C# implementation of above solution class GFG
{ static int MAX=10000;
// Create a boolean array "prime[0..n]" and initialize
// all entries it as true. A value in prime[i] will
// finally be false if i is Not a prime, else true.
static bool[] prime=new bool[MAX + 1];
static void SieveOfEratosthenes()
{ prime[1] = true;
for (int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then it is a prime
if (prime[p] == false) {
// Set all multiples of p to non-prime
for (int i = p * 2; i <= MAX; i += p)
prime[i] = true;
}
}
} // find the product of 1st N prime numbers static int solve(int n)
{ // count of prime numbers
int count = 0, num = 1;
// product of prime numbers
int prod = 1;
while (count < n) {
// if the number is prime add it
if (!prime[num]) {
prod *= num;
// increase the count
count++;
}
// get to next number
num++;
}
return prod;
} // Driver code public static void Main()
{ // create the sieve
SieveOfEratosthenes();
int n = 5;
// find the value of 1st n prime numbers
System.Console.WriteLine(solve(n));
} } // This code is contributed by mits |
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Python
''' python3 implementation of above solution'''import math as mt
MAX=10000
''' Create a boolean array "prime[0..n]" and initialize all entries it as true. A value in prime[i] will finally be false if i is Not a prime, else true.''' prime=[True for i in range(MAX+1)]
def SieveOfErastosthenes():
prime[1]=False
for p in range(2,mt.ceil(mt.sqrt(MAX))):
#if prime[p] is not changes, then it is a prime
if prime[p]:
#set all multiples of p to non-prime
for i in range(2*p,MAX+1,p):
prime[i]=False
#find the product of 1st N prime numbers def solve(n):
#count of prime numbers
count,num=0,1
#product of prime numbers
prod=1
while count<n:
#if the number is prime add it
if prime[num]:
prod*=num
#increase the count
count+=1
num+=1
return prod
#Driver code #create the sieve SieveOfErastosthenes() n=5
#find the valu of 1st n prime numbers print(solve(n))
#this code is contributed by Mohit Kumar 29 |
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PHP
<?php // PHP implementation of above solution $MAX = 10000;
// Create a boolean array "$prime[0..$n]" and // initialize all entries it as true. A value // in $prime[i] will finally be false if i is // Not a $prime, else true. $prime = array_fill(0, $MAX + 1, true);
function SieveOfEratosthenes()
{ global $MAX;
global $prime;
$prime = array_fill(0, $MAX + 1, true);
$prime[1] = false;
for ($p = 2; $p * $p <= $MAX; $p++)
{
// If $prime[$p] is not changed,
// then it is a $prime
if ($prime[$p] == true)
{
// Set all multiples of $$p to non-$prime
for ($i = $p * 2; $i <= $MAX; $i += $p)
$prime[$i] = false;
}
}
} // find the product of 1st N $prime numbers function solve($n)
{ global $prime;
// $count of $prime numbers
$count = 0;
$num = 1;
// product of $prime numbers
$prod = 1;
while ($count < $n)
{
// if the number is $prime add it
if ($prime[$num]== true)
{
$prod *= $num;
// increase the $count
$count++;
}
// get to next number
$num++;
}
return $prod;
} // Driver code // create the sieve SieveOfEratosthenes(); $n = 5;
// find the value of 1st $n $prime numbers echo solve($n);
// This code is contributed by ihritik ?> |
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Output:
2310
NOTE: For larger values of N, the product may be give integer overflow errors.
Also for multiple queries, prefix array technique can be used which will give output of each query in O(1) after making the prefix array first which will take O(N) time.
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