素数查找程序的效率（c++）

Question

I'm wondering if there is something inefficient in this code I've made, or if there is a faster way to find prime numbers.

#include <stdio.h>

int main(void)
{
    int count;

    for(int i=3; i<1000; i+=2){//search number in range of 3~999
        count=0;//init count
        for(int j=3; j*j<=i; j+=2){
            if(count==1){//if i has aliquot already, break the loop
                break;
            }
            if(i%j==0){
                count=1;//if i has aliquot, change count to 1
            }
        }
        if(count==0){
            printf("%d ", i);//if there are no aliquot, print i
        }
    }

    return 0;
}

Answer 1

Seems that you are using trial division , which takes O (√ n ) time to determine the primality of a single number, thus inefficient for finding all prime numbers in a range. To find all prime numbers in a range efficiently, consider using the sieve of Eratosthenes (with time complexity O ( n log n loglog n )) or Euler's sieve (with time complexity O ( n )). Here are simple implementations for these two algorithms.

Implementation for the sieve of Eratosthenes

bool isPrime[N + 5];

void eratosthenes(int n) {
    for (int i = 2; i <= n; ++i) {
        isPrime[i] = true;
    }
    isPrime[1] = false;
    for (int i = 2; i * i <= n; ++i) {
        if (isPrime[i]) {
            for (int j = i * i; j <= n; j += i) {
                isPrime[j] = false;
            }
        }
    }
}

Implementation for Euler's sieve

bool isPrime[N + 5];
std::vector<int> primes;

void euler(int n) {
    for (int i = 2; i <= n; ++i) {
        isPrime[i] = true;
    }
    isPrime[1] = false;
    for (int i = 2; i <= n; ++i) {
        if (isPrime[i]) primes.push_back(i);
        for (size_t j = 0; j < primes.size() && i * primes[j] <= n; ++j) {
            isPrime[i * primes[j]] = false;
            if (i % primes[j] == 0) break;
        }
    }
}

Answer 2

Eratosthenes Sieve is cool, but deprecated ? Why not use my Prime class ? it has an incrementation method, does not not use division. Prime class describes a number through its congruences to lower primes. Incrementing a prime is to increase all congruences, a new congruence is created if the integer is prime (all congruences -modulos_- differ from 0).

#include <iostream>
#include <vector>
#include <algorithm>
#include <utility>

class Prime {
public :
  Prime () : n_ (2), modulos_ (std::vector<std::pair<int, int> > ())
  {
    if (!modulos_.capacity ()) modulos_.reserve (100000000);
    std::pair<int, int> p (2, 0);
    modulos_.push_back (p);
  }
  ~Prime () {}
  Prime (const Prime& i) : n_ (i.n_), modulos_ (i.modulos_)
   {}
  bool operator == (const Prime& n) const {
    return (n_ == n.n_);
  }
  bool operator != (const Prime& n) const {
    return !operator == (n);
  }
  Prime& operator = (const Prime& i) {
    n_ = i.n_,
    modulos_ = i.modulos_;
    return *this;
  }
  void write (std::ostream& os) const {
    os << n_;
  }
  void operator ++ () {
    int prime (1);
    do {
      ++n_;
      prime = 1;
      std::for_each (modulos_.begin (), modulos_.end (), [&prime] (std::pair<int, int>& p) {
        ++p.second;
        if (p.first == p.second) {
          p.second = 0;
          prime  = 0;
        }
      });
    }
    while (!prime);
    std::pair<int, int> p (n_, 0);
    modulos_.push_back (p);
  }
  bool operator < (const int& s) const {
    return n_ < s;
  }
private :
  int n_;
  std::vector<std::pair<int, int> > modulos_; 
};

Usage :

int main (int, char**) {
  Prime p;
  do {
    p.write (std::cout);
    std::cout << std::endl;
    ++p;
  }
  while (p < 20);
}

Results : 2 3 5 7 11 13 17 19

Answer 3

For the primes up to 1000, an efficient method is

cout << "2  3   5   7   11  13  17  19  23
29  31  37  41  43  47  53  59  61  67
71  73  79  83  89  97  101 103 107 109
113 127 131 137 139 149 151 157 163 167
173 179 181 191 193 197 199 211 223 227
229 233 239 241 251 257 263 269 271 277
281 283 293 307 311 313 317 331 337 347
349 353 359 367 373 379 383 389 397 401
409 419 421 431 433 439 443 449 457 461
463 467 479 487 491 499 503 509 521 523
541 547 557 563 569 571 577 587 593 599
601 607 613 617 619 631 641 643 647 653
659 661 673 677 683 691 701 709 719 727
733 739 743 751 757 761 769 773 787 797
809 811 821 823 827 829 839 853 857 859
863 877 881 883 887 907 911 919 929 937
941 947 953 967 971 977 983 991 997" << endl;