PROBLEM LINK

Practice
Contest

DIFFICULTY

SIMPLE

PREREQUISITES

Sieve of Eratosthenes

PROBLEM

A number is called a K-prime if it has K distinct prime factors.

Find the number of K-prime numbers between some given A and B, inclusive.

QUICK EXPLANATION

Let us look at a simple implementation of the Sieve of Eratosthenes.

isPrime[N] = { YES }
for i = 2 to N
    if isPrime[i] = YES
        j = 2
        while i*j ≤ N
            isPrime[i*j] = NO
            j += i

Sieve of Eratosthenes has the wonderful property of iterating through all the multiples of a prime number, for each unique prime number below N. This means that the number of times the algorithm above marks a number not prime is equal to the number of unique prime factors of the number!

EXPLANATION

Let us maintain the number of times the Sieve of Erotosthenes algorithm would mark an item as not prime. This maintained in the array marked.

marked[N] = { 0 }
isPrime[N] = { YES }
for i = 2 to N
    if isPrime[i] = YES
        marked[i] = 1 // each prime is its own single factor
        j = 2
        while i*j ≤ N
            isPrime[i*j] = NO
            marked[i*j]++
            j += i

Now, for any given range [A,B] and value k, we can iterate through marked. If the value in marked is equal to k then we know that the number has exactly k distinct prime factors.

The complexity of Sieve of Eratosthenes is equivalent to the number of times the innermost statement is executed. This is equal to the sum of the number of distinct prime factors for each number below N. We know that a number cannot have more than log N prime factors. In fact, the complexity of Sieve of Eratosthenes is equal to O(N log log N).

There is one problem in our solution though. We are iterating between A and B for each input. It is given that there might be as many as 10000 inputs to be processed within a single second.

You will notice that the limit on possible values of k is very small. This can be exploited. Even if there were no limits on k, the maximum possible value in marked would be 7.

You can build a table to store the number of times some k occurs in marked before each position i, for each k.

table[5][N] = { 0 }
for i = 2 to N
    for j = 1 to 5
        table[j][i] = table[j][i-1]
    v = marked[i]
    if 1 ≤ v ≤ 5
        table[v][i]++

Now the answer for some range [A,B] and value k can be found in constant time form table.

SETTER'S SOLUTION

Can be found here.

TESTER'S SOLUTION

Can be found here.