Problem:
Pair of Primes

We arrange the numbers between 1 and N (1 <= N <= 10000) in increasing
order and decreasing order like this:

1 2 3 4 5 6 7 8 9 . . . N
N . . . 9 8 7 6 5 4 3 2 1

Two numbers faced each other form a pair. Your task is to compute the number of pairs P such that both
numbers in the pairs are prime.

Input

Input contains several test cases. Each test case consists of an integer N in
one line.

Output

For each line of input, output P .

Sample Input

1
4
7
51

Output for the Sample Input

0
2
2
6

Solution in C++11 (judge verdict: Accepted)

//include
//------------------------------------------
#include <bits/stdc++.h>

using namespace std;

//typedef
//------------------------------------------
typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<string> vs;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef pair<int, int> pii;
typedef long long ll;
typedef unsigned long long ull;

//container util
//------------------------------------------
#define all(a)  (a).begin(),(a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define pb push_back
#define mp make_pair
#define iter(i,c) for(auto i=(c).begin(); i!=(c).end(); ++i)

//repetition
//------------------------------------------
#define loop(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)  loop(i,0,n)

//constant
//--------------------------------------------
const double eps = 1e-10;
const double pi  = acos(-1.0);
const double inf = (int)1e8;

// output vector
// ------------------------------------------
namespace std {
    template<typename T> ostream & operator<<(ostream& os, vector<T> const& v){
        for(auto it=v.begin(); it!=v.end(); ++it){
            os << *it << (it+1==v.end() ? "" : ",");
        }
        return os;
    }
}

//clear memory
#define clr(a) memset((a), 0 ,sizeof(a))

int gcd(int a, int b){
   int tmp;
   while(a%b!=0){
      tmp=a%b;
      a=b;
      b=tmp;
   }
}
 
int lcm(int a,int b){
   return a*b/gcd(a,b);
}
   vb pn((int)1e4+10); 
void prime_number(int n){
   if(n==0||n==1) //return false;
   fill(all(pn),true);
   pn[0]=false;pn[1]=false;
   loop(i,2,n/2+1){
      if(!pn[i]) continue;
      for(int j=2;i*j<=n;j++){
         pn[i*j]=false;
      }
   }
   //return pn[n];
}

int main(){
   int a,ans=0;
   while(cin>>a){
      ans=0;
      prime_number(a);
      rep(i,a){
         if(pn[i+1]&&pn[a-i]) ans++;
      }
      cout << ans << endl;
   }
}