class ComputeNumberOfDivisors {

// Method to find prime numbers

static void SieveOfEratosthenes(int n, boolean prime[]) {

for (int p = 2; p * p <= n; p++) {

if (prime[p] == true) {

for (int i = p * 2; i <= n; i += p)

prime[i] = false;

}

}

}

// Method to count divisors

static int countDivisors(int n) {

if (n == 1) {

return 1;

}

boolean prime[] = new boolean[n + 1];

boolean primeSquare[] = new boolean[(n * n) + 1];

int a[] = new int[n];

prime[1] = false;

for (int i = 2; i <= n; i++) {

prime[i] = true;

}

SieveOfEratosthenes(n, prime);

int j = 0;

for (int p = 2; p <= n; p++) {

if (prime[p]) {

a[j] = p;

primeSquare[p * p] = true;

j++;

}

}

int ans = 1;

for (int i = 0;; i++) {

if (n < a[i] * a[i] * a[i]) {

break;

}

int cnt = 1;

while (n % a[i] == 0) {

n = n / a[i];

cnt = cnt + 1;

}

ans = ans * cnt;

}

if (prime[n]) {

ans = ans * 2;

} else if (primeSquare[n]) {

ans = ans * 3;

} else if (n != 1) {

ans = ans * 4;

}

return ans;

}

public static void main(String[] args) {

int x = 100;

System.out.println("Total number of divisors of " + x + " are " + countDivisors(x));

}

}

// Time Complexity : O( n ^ (1/3) )