Given three integers a, b, n .Your task is to print number of numbers between a and b including them also which have n-divisors. A number is called n-divisor if it has total n divisors including 1 and itself.
Input : a = 1, b = 7, n = 2 Output : 4 There are four numbers with 2 divisors in range [1, 7]. The numbers are 2, 3, 5, and 7.
The naive approach is to check all the numbers between a and b how many of them are n-divisor number for doing this find out the number of each divisors for each number . If it is equal to n then it is a n-divisor number
Any number can be written in the form of its prime factorization let the number be x and p1, p2..pm are the prime numbers which divide x so x = p1e1 * p2e2….pmem where e1, e2…em are the exponents of prime numbers p1, p2….pm. So the number of divisors of x will be (e1+1)*(e2+1)…*(em+1).
Now the second observation is for prime numbers greater than sqrt(x) their exponent cannot exceed 1. Let’s prove this by contradiction suppose there is a prime number P greater than sqrt(x) and its exponent E in prime factorization of x is greater than one (E >= 2) so P^E sqrt(x) so P^E > (sqrt(x))E and E >= 2 so PE will always be greater than x
Third observation is that number of prime numbers greater than sqrt(x) in the prime factorization of x will always be less than equal to 1. This can also be proved similarly by contradiction as above.
Now to solve this problem
Step 1: Apply sieve of erastothenes and calculate prime numbers upto sqrt(b).
Step 2: Traverse through each number from a to b and calculate exponents of each prime number in that number by repeatedly dividing that number by prime number and use the formula numberofdivisors(x) = (e1+1)*(e2+1)….(em+1).
Step 3: If after dividing by all the prime numbers less than equal to square root of that number if number > 1 this means there is a prime number greater than its square root which divides and its exponent will always be one as proved above.
# Python3 program to count numbers
# with n divisors
# applying sieve of erastothenes
def sieve(primes, x):
primes = False;
# if a number is prime mark all
# its multiples as non prime
i = 2;
while (i * i <= x): if (primes[i] == True): j = 2; while (j * i <= x): primes[i * j] = False; j += 1; i += 1; # function that returns numbers of number # that have n divisors in range from a to b. # x is sqrt(b) + 1. def nDivisors(primes, x, a, b, n): # result holds number of numbers # having n divisors result = 0; # vector to hold all the prime # numbers between 1 and sqrt(b) v = ; for i in range(2, x + 1): if (primes[i]): v.append(i); # Traversing all numbers in given range for i in range(a, b + 1): # initialising temp as i temp = i; # total holds the number of # divisors of i total = 1; j = 0; # we need to use that prime numbers that # are less than equal to sqrt(temp) k = v[j]; while (k * k <= temp): # holds the exponent of k in prime # factorization of i count = 0; # repeatedly divide temp by k till it is # divisble and accordingly increase count while (temp % k == 0): count += 1; temp = int(temp / k); # using the formula no.of divisors = # (e1+1)*(e2+1).... total = total * (count + 1); j += 1; k = v[j]; # if temp is not equal to 1 then there is # prime number in prime factorization of i # greater than sqrt(i) if (temp != 1): total = total * 2; # if i is a ndvisor number # increase result if (total == n): result += 1; return result; # Returns count of numbers in [a..b] # having n divisors. def countNDivisors(a, b, n): x = int(math.sqrt(b) + 1); # primes[i] = true if i is a prime number # initialising each number as prime primes = [True] * (x + 1); sieve(primes, x); return nDivisors(primes, x, a, b, n); # Driver code a = 1; b = 7; n = 2; print(countNDivisors(a, b, n)); # This code is contributed by mits [tabby title="C#"]
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