C++ Program to Count Primes in Ranges

Given a range [L, R], we need to find the count of total numbers of prime numbers in the range [L, R] where 0 <= L <= R < 10000. Consider that there are a large number of queries for different ranges.
Examples:

Input : Query 1 : L = 1, R = 10
        Query 2 : L = 5, R = 10
Output : 4
         2
Explanation
Primes in the range L = 1 to R = 10 are 
{2, 3, 5, 7}. Therefore for query, answer 
is 4 {2, 3, 5, 7}.
For the second query, answer is 2 {5, 7}.

A simple solution is to do the following for every query [L, R]. Traverse from L to R, check if current number is prime. If yes, increment the count. Finally, return the count.
An efficient solution is to use Sieve of Eratosthenes to find all primes up to the given limit. Then we compute a prefix array to store counts till every value before limit. Once we have a prefix array, we can answer queries in O(1) time. We just need to return prefix[R] – prefix[L-1]. 

C++

// CPP program to answer queries for count of
// primes in given range.
#include <bits/stdc++.h>
using namespace std;

const int MAX = 10000;

// prefix[i] is going to store count of primes
// till i (including i).
int prefix[MAX + 1];

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] = 0;
    for (int p = 2; p <= MAX; p++) {
        prefix[p] = prefix[p - 1];
        if (prime[p])
            prefix[p]++;
    }
}

// Returns count of primes in range from L to
// R (both inclusive).
int query(int L, int R)
{
    return prefix[R] - prefix[L - 1];
}

// Driver code
int main()
{
    buildPrefix();

    int L = 5, R = 10;
    cout << query(L, R) << endl;

    L = 1, R = 10;
    cout << query(L, R) << endl;

    return 0;
}

Output:

2
4

Time & Space Complexity will be same as Sieve of eratosthenes

Time Complexity: O(n*log(log(n)))

Auxiliary Space: O(n)

Please refer complete article on Count Primes in Ranges for more details!