Given an even number ( greater than 2 ), return two prime numbers whose sum will be equal to given number

Question

vector<int> Solution::primesum(int n) {
  int prime[n+1];
  for(int i=0;i<n+1;i++){
    prime[i]=1;
  }
   prime[0]=0;
   prime[1]=0;
   for(int i=2;i<=sqrt(n);i++){

        if(prime[i]==1){
            for(int j=2;i*j<=n;j++)
               prime[i*j]=0;
        }

  }
  vector <int> sum;
  for(int i=2;i<=(n)/2;i++){
      if(prime[i]==1&&prime[n-i]==1){
       sum.push_back(i);
       sum.push_back(n-i);
       return sum;
      }
      }
    return sum;
     }

The above code is working fine for smaller inputs but when I try to provide it with numbers something like >100000000, it is showing segmentation fault.Can someone please help me? Thanks!

Try to use `unsigned long` or `unsigned long long` instead of `int` in your program. — Michael, Jul 04 '17 at 06:59
This is not standard C++; it makes use of VLA extensions that a couple compilers support. — Justin, Jul 04 '17 at 07:01

MikeMB · Answer 1 · 2017-07-04T07:16:28.263

0

Most likely int prime[n+1] leads to a stackoverflow (100'000'000 requires ~400 MB on the stack). I'd recommend using std::vector<int> prime(n+1) instead.

edited Jul 04 '17 at 07:16

answered Jul 04 '17 at 07:03

MikeMB

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Thanks but that was also showing the same error. – Namrata Tanwar Jul 05 '17 at 02:17
@NamrataTanwar: Can't reproduce your problem in this case. Can you open a new question with your new code, including information about your platform, compiler, flags a.s.o. – MikeMB Jul 05 '17 at 09:36

Given an even number ( greater than 2 ), return two prime numbers whose sum will be equal to given number

1 Answers1