求素数的爱沙托散筛法

//author: smileonce
#include 
using namespace std;

static const int N = 1000;
int main(int argc, char *argv[])
{
  int i, a[N];
  for (i=2; i<N; i++) a[i] = 1;
  for (i=2; i<N; i++)
    if (a[i])
        for(int j=i; j*i<N; j++) a[i*j] = 0;
  for (i=2; i<N; i++)
    if (a[i]) cout << " " << i;
  cout << endl;
 
  return 0;
}

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