686 - Goldbach's Conjecture (II)

Problem Link

For any integer n (n ≥ 4) there exist two prime numbers p1 and p2 such that p1 + p2 = n. In a problem we might need to find the number of essentially different pairs (p1, p2), satisfying the condition in the conjecture for a given even number n (4 ≤ n ≤ 2 15). (The word ‘essentially’ means that for each pair (p1, p2) we have p1 ≤p2.)
For example, for n = 10 we have two such pairs: 10 = 5 + 5 and 10 = 3 + 7.

Thus my solution in C++ :

1.  #include<bits/stdc++.h>
2. using namespace std;
3. #define MAX 10000000  ///~10^7
4. bool prime[MAX];
5. vector <int> p;
6. void sieve()
7. {
8.     memset(prime,0,sizeof(prime));
9.     prime[0]=prime[1]=1;
10.     for(int i = 4; i < MAX; i += 2)
11.     {
12.         prime[i] = 1;
13.     }
14.     for(int i = 3; i*i < MAX; i += 2)
15.     {
16.         if(!prime[i])
17.         {
18.             for(int j = i*i; j < MAX; j += i)
19.                 prime[j] = 1; /// not prime
20.         }
21.     }
22.     for(int i = 2; i < MAX; i++)
23.         if(!prime[i])
24.             p.push_back(i);
25. }
26. int FindSol(int n)
27. {
28.     int res = 0;
29.     for(int i = 2; i <= n/2; i++)
30.         if(!prime[i] && !prime[n-i])
31.             res++;
32.     return res;
33. }
34. int main()
35. {
36.     int n;
37.     sieve();
38.     while(cin>>n && n)
39.         cout<<FindSol(n)<<endl;
40.     return 0;
41. }