I wrote this solution for project Euler #7, to find the 10001th prime number using sieve of eratosthenes. It's really slow though. Any suggestions to improve performance (without using more complex sieves like sieve of atkin)?

#define max 500000
using namespace  std;

int main()
 int ctr = 0;
 int i=2,j,n=10001,p=2;
 int arr[max];


  if(arr[p] == 0)
        arr[j] = 1;





  • \$\begingroup\$ It causes stack overflow on windows, VC++14. \$\endgroup\$ Commented Jul 29, 2016 at 16:45
  • \$\begingroup\$ I used code blocks ,windows and it worked all right. \$\endgroup\$
    – Syed Saad
    Commented Jul 29, 2016 at 18:25
  • \$\begingroup\$ I'm not saying the code is broken, just saying that on VC++14 it might fail. \$\endgroup\$ Commented Jul 29, 2016 at 18:31

2 Answers 2

  • Why is everybody using #include<bits/stdc++.h>? It is so hard to include <cstring> to get memset working?
  • Why do you need <climits>?
  • Don't use using namespace std;. It's a bad practice.
  • Static array of ints with size 500000 is likely too big to fit into stack. Use int * arr = new int[max] and delete [] arr; instead.
  • Using whole integers to store boolean value is just a waste of memory.
  • It takes 42 seconds to find result. My version (without sieve) takes about real: 0.015s, user: 0.013s.

    #include <iostream>
    bool isPrime(int num) {
      for (int i = 2; i*i <= num; ++i) {
        if ((num % i) == 0) return false;
      return num > 1;
    int main() {
      int i = 2;
      for (int primes = 0; ; ++i) {
        if (isPrime(i)) {
          if (++primes == 10001) {
            std::cout << i << "\n";
  • \$\begingroup\$ Thanks , i tried it without sieve and got similar results \$\endgroup\$
    – Syed Saad
    Commented Jul 29, 2016 at 19:13

If the prime number to be found is always going to be the 10001st prime number, then you can greatly improve the performance by hard-coding the result:

#include <iostream>
#define PRIME_10001 104743

int main(void)
    std::cout << PRIME_10001 << '\n';
    return 0;

If 104743 is banned for some reason, here are some alternatives:

  • 0x19927 in base 16,
  • 0314447 in octal,
  • 0b11001100100100111 in binary with C++14 (or GNU C),
  • 104'743 with a C++14 digit separator

If you do use the for-loop approach mentioned in the previous answer, you should note that the condition of this loop may overflow:

for (int i = 2; i*i <= num /* for i as little as 256: kaboom! */; ++i) {

Instead, using division eliminates several possibilities:

for (int i = 2; i <= num/i; ++i) {

You should probably use an unsigned integer to remove the rest of the possibilities for overflow. Alternatively, check if num is negative and return false if it is.


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