Given a range [L, R]. The task is to find the product of all the prime numbers in the given range from L to R both inclusive modulo 10^9 + 7.
Examples:
Input: L = 10, R = 20 Output: 46189 Prime numbers between [10, 20] are: 11, 13, 17, 19 Therefore, product = 11 * 13 * 17 * 19 = 46189 Input: L = 15, R = 25 Output: 7429
A Simple Solution is to traverse from L to R, check if the current number is prime. If yes, multiply it with product. Finally, print the product.
An Efficient Solution is to use Sieve of Eratosthenes to find all primes up to a given limit. Then, compute a prefix product array to store product till every value before the limit. Once we have prefix array, We just need to return (prefix[R] *modular_inverse( prefix[L-1]))%(10^9+7).
Note: prefix[i] will store the product of all prime numbers from 1 to i.
Below is the implementation of above approach:
// CPP program to find product of primes // in range L to R #include <bits/stdc++.h> using namespace std;
#define mod 1000000007 const int MAX = 10000;
// prefix[i] is going to store product of primes // till i (including i). int prefix[MAX + 1];
// Function to build the prefix product array void buildPrefix()
{ // Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
bool prime[MAX + 1];
memset (prime, true , sizeof (prime));
for ( int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true ) {
// Update all multiples of p
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
// Build prefix array
prefix[0] = prefix[1] = 1;
for ( int p = 2; p <= MAX; p++) {
prefix[p] = prefix[p - 1];
if (prime[p])
prefix[p] = (prefix[p] * p) % mod;
}
} /* Iterative Function to calculate (x^y)%p in O(log y) */ long long int power( long long int x, long long int y, int p)
{ // Initialize result
long long int res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
// y = y/2
y = y >> 1;
x = (x * x) % p;
}
return res;
} // Returns modular inverse long long int inverse( long long int n)
{ return power(n, mod - 2, mod);
} // Function to return product of prime in range long long int productPrimeRange( int L, int R)
{ return (prefix[R] * inverse(prefix[L - 1])) % mod;
} // Driver code int main()
{ buildPrefix();
int L = 10, R = 20;
cout << productPrimeRange(L, R) << endl;
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// C program to find product of primes // in range L to R #include <stdbool.h> #include <stdio.h> #include <string.h> #define mod 1000000007 const int MAX = 10000;
// Function to build the prefix product array void buildPrefix( int prefix[])
{ // Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
bool prime[MAX + 1];
memset (prime, true , sizeof (prime));
for ( int p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true ) {
// Update all multiples of p
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
// Build prefix array
prefix[0] = prefix[1] = 1;
for ( int p = 2; p <= MAX; p++) {
prefix[p] = prefix[p - 1];
if (prime[p])
prefix[p] = (prefix[p] * p) % mod;
}
} /* Iterative Function to calculate (x^y)%p in O(log y) */ long long int power( long long int x, long long int y, int p)
{ // Initialize result
long long int res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
// y = y/2
y = y >> 1;
x = (x * x) % p;
}
return res;
} // Returns modular inverse long long int inverse( long long int n)
{ return power(n, mod - 2, mod);
} // Function to return product of prime in range long long int productPrimeRange( int L, int R, int prefix[])
{ return (prefix[R] * inverse(prefix[L - 1])) % mod;
} // Driver code int main()
{ int prefix[MAX + 1];
buildPrefix(prefix);
int L = 10, R = 20;
printf ( "%lld" , productPrimeRange(L, R, prefix));
return 0;
} // This code is contributed by Aditya Kumar (adityakumar129) |
// Java program to find product of primes // in range L to R import java.io.*;
class GFG {
static int mod = 1000000007 ;
static int MAX = 10000 ;
// prefix[i] is going to store product of primes
// till i (including i).
static int [] prefix = new int [MAX + 1 ];
// Function to build the prefix product array
static void buildPrefix()
{
// Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
boolean prime[] = new boolean [MAX + 1 ];
for ( int i = 0 ; i < MAX + 1 ; i++)
prime[i] = true ;
// memset(prime, true, sizeof(prime));
for ( int p = 2 ; p * p <= MAX; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true ) {
// Update all multiples of p
for ( int i = p * 2 ; i <= MAX; i += p)
prime[i] = false ;
}
}
// Build prefix array
prefix[ 0 ] = prefix[ 1 ] = 1 ;
for ( int p = 2 ; p <= MAX; p++) {
prefix[p] = prefix[p - 1 ];
if (prime[p])
prefix[p] = (prefix[p] * p) % mod;
}
}
// Iterative Function to calculate (x^y)%p in O(log y)
static long power( long x, long y, int p)
{
// Initialize result
long res = 1 ;
// Update x if it is more than or equal to p
x = x % p;
while (y > 0 ) {
// If y is odd, multiply x with result
if ((y & 1 ) > 0 )
res = (res * x) % p;
// y must be even now y = y/2
y = y >> 1 ;
x = (x * x) % p;
}
return res;
}
// Returns modular inverse
static long inverse( long n)
{
return power(n, mod - 2 , mod);
}
// Function to return product of prime in range
static long productPrimeRange( int L, int R)
{
return (prefix[R] * inverse(prefix[L - 1 ])) % mod;
}
// Driver code
public static void main(String[] args)
{
buildPrefix();
int L = 10 , R = 20 ;
System.out.println(productPrimeRange(L, R));
}
} // This code is contributed by Aditya Kumar (adityakumar129) |
# Python 3 program to find product of primes # in range L to R mod = 1000000007
MAX = 10000
# prefix[i] is going to store product of primes # till i (including i). prefix = [ 0 ] * ( MAX + 1 )
# Function to build the prefix product array def buildPrefix():
# Create a boolean array "prime[0..n]". A
# value in prime[i] will finally be false
# if i is Not a prime, else true.
prime = [ True ] * ( MAX + 1 )
p = 2
while p * p < = MAX :
# If prime[p] is not changed, then
# it is a prime
if (prime[p] = = True ) :
# Update all multiples of p
for i in range ( p * 2 , MAX + 1 , p):
prime[i] = False
p + = 1
# Build prefix array
prefix[ 0 ] = prefix[ 1 ] = 1
for p in range ( 2 , MAX + 1 ) :
prefix[p] = prefix[p - 1 ]
if (prime[p]):
prefix[p] = (prefix[p] * p) % mod
# Iterative Function to calculate # (x^y)%p in O(log y) def power(x, y,p):
# Initialize result
res = 1
# Update x if it is more than or
# equal to p
x = x % p
while (y > 0 ) :
# If y is odd, multiply x with result
if (y & 1 ):
res = (res * x) % p
# y must be even now
# y = y//2
y = y >> 1
x = (x * x) % p
return res
# Returns modular inverse def inverse( n):
return power(n, mod - 2 , mod)
# Function to return product of prime in range def productPrimeRange(L, R):
return (prefix[R] * inverse(prefix[L - 1 ])) % mod
# Driver code if __name__ = = "__main__" :
buildPrefix()
L = 10
R = 20
print (productPrimeRange(L, R))
# this code is contributed by # ChitraNayal |
// C# program to find product of // primes in range L to R using System;
class GFG
{ static int mod = 1000000007;
static int MAX = 10000;
// prefix[i] is going to store product // of primes till i (including i). static int []prefix = new int [MAX + 1];
// Function to build the prefix // product array static void buildPrefix()
{ // Create a boolean array "prime[0..n]".
// A value in prime[i] will finally be
// false if i is Not a prime, else true.
bool []prime = new bool [MAX + 1];
for ( int i = 0; i < MAX + 1; i++)
prime[i] = true ;
//memset(prime, true, sizeof(prime));
for ( int p = 2; p * p <= MAX; p++)
{
// If prime[p] is not changed,
// then it is a prime
if (prime[p] == true )
{
// Update all multiples of p
for ( int i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
// Build prefix array
prefix[0] = prefix[1] = 1;
for ( int p = 2; p <= MAX; p++)
{
prefix[p] = prefix[p - 1];
if (prime[p])
prefix[p] = (prefix[p] * p) % mod;
}
} /* Iterative Function to calculate (x^y)%p in O(log y) */
static long power( long x, long y, int p)
{ // Initialize result
long res = 1;
// Update x if it is more
// than or equal to p
x = x % p;
while (y > 0)
{
// If y is odd, multiply x
// with result
if ((y & 1) > 0)
res = (res * x) % p;
// y must be even now
// y = y/2
y = y >> 1;
x = (x * x) % p;
}
return res;
} // Returns modular inverse static long inverse( long n)
{ return power(n, mod - 2, mod);
} // Function to return product // of prime in range static long productPrimeRange( int L, int R)
{ return (prefix[R] *
inverse(prefix[L - 1])) % mod;
} // Driver code public static void Main ()
{ buildPrefix();
int L = 10, R = 20;
Console.WriteLine(productPrimeRange(L, R));
} } // This code is contributed by anuj_67 |
<script> // Javascript program to find product of primes // in range L to R var mod = 100000007
var MAX = 10000;
// prefix[i] is going to store product of primes // till i (including i). var prefix = Array(MAX+1);
// Function to build the prefix product array function buildPrefix()
{ // Create a boolean array "prime[0..n]". A
// value in prime[i] will finally be false
// if i is Not a prime, else true.
var prime = Array(MAX+1).fill( true );
for ( var p = 2; p * p <= MAX; p++) {
// If prime[p] is not changed, then
// it is a prime
if (prime[p] == true ) {
// Update all multiples of p
for ( var i = p * 2; i <= MAX; i += p)
prime[i] = false ;
}
}
// Build prefix array
prefix[0] = prefix[1] = 1;
for ( var p = 2; p <= MAX; p++) {
prefix[p] = prefix[p - 1];
if (prime[p])
prefix[p] = (prefix[p] * p) % mod;
}
} /* Iterative Function to calculate (x^y)%p in O(log y) */ function power(x, y, p)
{ // Initialize result
var res = 1;
// Update x if it is more than or
// equal to p
x = x % p;
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
// y = y/2
y = y >> 1;
x = (x * x) % p;
}
return res;
} // Returns modular inverse function inverse( n)
{ return power(n, mod - 2, mod);
} // Function to return product of prime in range function productPrimeRange(L, R)
{ return (prefix[R] * inverse(prefix[L - 1])) % mod;
} // Driver code buildPrefix(); var L = 10, R = 20;
document.write( productPrimeRange(L, R)); </script> |
46189