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Product of all prime numbers in an Array in C++
Given an integer array arr[] with some elements, the task is to find the product of all the prime number of that numbers.
Prime number are those number which are either divided by 1 or the number itself, or a prime number is a number which is not divisible by any other number except 1 and the number itself. Like 1, 2, 3, 5, 7, 11, etc.
we have to find the solution for the given array −
Input −arr[] = { 11, 20, 31, 4, 5, 6, 70 }
Output − 1705
Explanation − Prime numbers in array are − 11, 31, 5 their product is 1705
Input − arr[] = { 1, 2, 3, 4, 5, 6, 7 }
Output − 210
Explanation − Prime numbers in array are − 1, 2, 3, 5, 7 their product is 210
Approach used below is as follows to solve the problem
Take input array arr[].
Loop every element and check if it is prime or not.
Product all the present primes in an array.
Return the product.
Algorithm
Start
In function int prodprimearr(int arr[], int n)
Step 1→ Declare and initialize max_val as max_val *max_element(arr, arr + n)
Step 2→ Declare vector<bool> isprime(max_val + 1, true)
Step 3→ Set isprime[0] and isprime[1] as false
Step 4→ Loop For p = 2 and p * p <= max_val and p++
If isprime[p] == true then,
Loop For i = p * 2 and i <= max_val and i += p
Set isprime[i] as false
Step 5→ Set prod as 1
Step 6→ For i = 0 and i < n and i++
If isprime[arr[i]]
Set prod = prod * arr[i]
Step 6→ Return prod
In function int main(int argc, char const *argv[])
Step 1→ Declare and initilalize arr[] = { 11, 20, 31, 4, 5, 6, 70 }
Step 2→ Declare and initialize n = sizeof(arr) / sizeof(arr[0])
Step 3→ Print the results of prodprimearr(arr, n)
StopExample
#include <bits/stdc++.h>
using namespace std;
int prodprimearr(int arr[], int n){
// To find the maximum value of an array
int max_val = *max_element(arr, arr + n);
// USE SIEVE TO FIND ALL PRIME NUMBERS LESS
// THAN OR EQUAL TO max_val
vector<bool> isprime(max_val + 1, true);
isprime[0] = false;
isprime[1] = false;
for (int p = 2; p * p <= max_val; p++) {
// If isprime[p] is not changed, then
// it is a prime
if (isprime[p] == true) {
// Update all multiples of p
for (int i = p * 2; i <= max_val; i += p)
isprime[i] = false;
}
}
// Product all primes in arr[]
int prod = 1;
for (int i = 0; i < n; i++)
if (isprime[arr[i]])
prod *= arr[i];
return prod;
}
int main(int argc, char const *argv[]){
int arr[] = { 11, 20, 31, 4, 5, 6, 70 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << prodprimearr(arr, n);
return 0;
}Output
If run the above code it will generate the following output −
1705
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