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++14 (judge verdict: Accepted)

#define _CRT_SECURE_NO_WARNINGS
#include<bits/stdc++.h>
using namespace std;
#ifndef FAST_IO
 #define IOS ios_base::sync_with_stdio(false);cin.tie(nullptr);
#endif

#define int long long
#define FILE_IO freopen("input.txt","r",stdin)/#define int long long
#define f first
#define s second
#define endl '\n'
#define all(c) c.begin(),c.end()
#define bp __builtin_popcount
#define debug(x) cerr << #x << " = " << x << endl;

const int MOD = 1e9 + 7;
const int INF = 1e18;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>void print(T t){for (auto &it : t) cout << it << " ";}
inline int powerexp(int a, int b, int m = MOD){ int res = 1;while(b){if(b&1)res = res * a % m;a = a * a % m,b >>= 1;}return res % m;}
inline int modinv(int a){ return powerexp(a,MOD-2,MOD);}
inline int lcm(int a, int b){ int g = __gcd(a,b);return a/g*b;}
inline int add(int x, int y){ x += y; if (x >= MOD) x -=  MOD; return x;}
inline int mul(int x, int y){ return (x * 1ll * y) % MOD;}
inline int sub(int x, int y){ x -= y; if (x<0) x += MOD; return x ;}

//-------------------< start here >-------------------------// 

const int N = 1e4 + 11;
bool isprime[N];
vector<int> primes;

void sieve(){
  fill(isprime + 2, isprime + N, true);
  
  for (int i = 2; i < N; i++){
	  if (isprime[i]){
		primes.push_back(i);
		for (int j = 2; i * j < N; j++){
		   isprime[i * j] = false;	
	    }  
	  }  
	}	
}

void solve(){
  sieve();
  int n;
  while (cin >> n){
	  vector<int> v1, v2;
	  int ret = 0;
	  for (int i = 1; i <= n; i++) v1.push_back(i);
	  for (int i = n; i >= 1; i--) v2.push_back(i);
      
	  for (int i = 1; i <= n; i++){
		 ret += isprime[i] && isprime[n - i + 1]; 
	  }
	 cout << ret << endl;  
	}	
}

//-----------------------------
//comment from here for TOPCODER 
int32_t main(){
   IOS;
   solve();
   //find();
   #ifdef LOCAL_DEFINE
   cerr << "time elapsed " << (double)clock()/CLOCKS_PER_SEC << endl;
   #endif
}