Counting Positive Integers with a Given Number of Divisors

问题

basically what i was trying to do is insert an integer k that represents the number of divisors and then finding all the numbers that have k divisors from 1-100000

#include <stdio.h>

int main(void)
{
    int k, x = 1, y = 100000, divisor, count;

    printf("Enter the target number of divisors:\n");
    scanf("%d", &k);



    for (divisor = 0; divisor <= 1; divisor++)
        if (x % divisor == 0 && y % divisor == 0)
           count++;

    printf("There are %d numbers between 1 and 100000 inclusive which have exactly %d   divisors\n", k, divisor);

    return 0;
}

However I can't seem to be able to do it, please do help me as I'm fairly new to the programming scene and haven't found an answer elsewhere.

回答1:

There is a theorem that states if you have the canonical representation of an integer being a₁^b₁ * a₂^b₂ ... a_n^b_n then the number of divisors of this integer is (b₁ + 1) * (b₂ + 1) ... (b_n + 1).

Now that you have this theorem, you can modify slightly Eratosthenes's sieve to get all integers up to 100 000 in canonical form.

Here is some code that does what I mean by modified erathosthenes's sieve.

const int size = 100000;
int devs[size + 1];
void compute_devs() {
  for (int i = 0; i < size + 1; ++i) {
    devs[i] = (i%2 == 0) ? 2 : 1;
  }

  int o = sqrt(size);    
  for (int i = 3; i <= size; i += 2) {
    if (devs[i] != 1) {
      continue;
    }
    devs[i] = i;
    if (i <= o) {
      for (int j = i * i; j < size; j += 2 * i) {
        devs[j] = i;
      }
    }
  }
}

After calling compute_devs the value of devs will store the value of the greatest prime divisor of each number up to size. I will leave the rest of the task to you, but having this array it becomes pretty straight forward.

来源：https://stackoverflow.com/questions/26587512/counting-positive-integers-with-a-given-number-of-divisors