修复我在C ++中实现Eratosthenes筛子的实现

Question

My algorithm runs correctly up to the first 100 primes but then something goes wrong. Please have a look at my code below, I tried to follow the pseudo-code given here https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes

#include <iostream>
#include <vector>
#include <cmath>
using namespace std;

int main()
{
  int n = 1000; //compute primes up to this number
  vector<bool> p(true,n); //all values set to true, from 0 to n

  for(int i = 2; i < sqrt(n)+1; i++){
    if( p[i-1] == true ){
      for(int j = i*i; j < n; j += i) //start looking for multiples of prime i at i*i (optimized)
    p[j-1] = false;
    }
  }

  for(int i = 2; i < n; i++){
    if( p[i-1] == true )
      cout << i << "\n";
  }
  return 0;
}

The output is:

2
3
5
7
11
13
17
19
23
29
31
37
41
43
47
53
59
61
193
199

Answer 1

I'm absolutely amazed the program runs at all. It has an absolute truckload of undefined behaviour !.

Unless I'm very much mistaken (in which case please reward the serenity of my Friday afternoon with downvotes), vector<bool> p(true, n) is creating a vector of size true , with the elements initialised to n .

You have the constructor arguments the wrong way round. This is effectively reversing the sieve for most of the values.

Have you absolutely clobbered your compiler's warning level?

Answer 2

First of all, you do not need to store a boolean value for each and every number. This way, you are wasting memory. You should store only the found primes instead, unless you have a very good reason not to do so.

I will not implement the code, since it would spoil the joy of learning. You should implement the following:

• Initialize p as a vector of integers.
• Store 2 as first value inside p
• Iterate all odd numbers starting from 3 and ending at the end number
• For each number, calculate its square root and store it into a variable
• Iterate all previous elements of the p until you reach a divisor or the value of the vector at the given index reaches the square root, starting from the second element, as pair numbers larger than 2 are ignored
• If you find a divisor inside your inner loop, store it into your vector and get out of the inner loop

At the end you will have a vector of primes, indexes will mean the prime index and the values will be actual primes. Each element will be a prime of its own.

Answer 3

You mess your numbering up. If p[k] is the primeness of number k+1 , your for loop is wrong

for(int j = i*i; j < n; j += i)

and should be

for(int j = (i-1)*(i-1); j < n; j += (i-1))

My advice would be to use more informative variable names and avoid those sources of confusion like p[k] giving information about integer k+1 .

Answer 4

Your vector construction is wrong. It must be

vector<bool> p(n, true); //all values set to true, from 0 to n

instead of

vector<bool> p(true, n); //all values set to true, from 0 to n