Uva 10311 - Goldbach and Euler

Problem Link :: Uva Goldbach and Euler

Code:: goes to below.....

/*
Verdict :: Accepted
Time :: 0.176

****** The best Rank (3rd) i have ever got in Uva online judge.
     to Solve that Problem..*******
*/

/// Header file begin
#include <iostream>
#include <cstdio>
#include <cstring>
#include <map>
#include <string>
#include <vector>
#include <cmath>
#include <cctype>
#include <sstream>
#include <set>
#include <list>
#include <stack>
#include<utility>
#include <queue>
#include <algorithm>
/// End
//..........
/// Macro
#define sf scanf
#define pf printf
#define sfint(a,b) scanf("%d %d",&a,&b)
#define sfl(a,b) scanf("%ld %ld",&a,&b)
#define sfll(a,b) scanf("%lld %lld",&a,&b)
#define sfd(a,b) scanf("%lf %lf",&a,&b)
#define sff(a,b) scanf("%f %f",&a,&b)
#define lp1(i,n) for(i=0;i<n;i++)
#define lp2(i,n) for(i=1;i<=n;i++)
#define mem(c,v) memset(c,v,sizeof(c))
#define cp(a) cout<<" "<<a<<" "<<endl
#define nl puts("")
#define sq(x) ((x)*(x))
#define all(x) x.begin(),x.end()
#define sz size()
#define gc getchar()
#define pb push_back
/// End.........

/// Size
#define mx7 20000100
#define mx6 1500000
#define mx5 100005
#define mx4 100001000
#define inf 1<<30                                           //infinity value
#define eps 1e-9
#define mx (65540)
#define mod 1000000007
#define pi acos(-1.0)

/// Macros for Graph

#define white 0
#define gray 1
#define black -1
#define nil -2

using namespace std;
//..................................................................................................................
typedef long long LL;
typedef long L;
typedef unsigned long long ull;
typedef unsigned long ul;
typedef unsigned int ui;
typedef pair<int, int> pii;

template<class T> T gcd(T a, T b ) {return b<=0?a:gcd(b,a%b);}
template<class T> T large(T a, T b ) {return a>b?a:b;}
template<class T> T small(T a, T b ) {return a<b?a:b;}
template<class T> T diffrnce(T a, T b) {return a-b<0?b-a:a-b;}

int prime[(mx4 >> 6)+1];

#define setbit(n) (prime[n>>6]|=(1<<((n>>1)&31)))
#define checkbit(n) (prime[n>>6]&(1<<((n>>1)&31)))
#define qrt 10000

vector<int>prm;
int plen;

void seieve()
{
    int i,j;

    for(i=3;i<=qrt;i+=2)
    {
        if(!checkbit(i))
        {
            for(j=i*i;j<=mx4;j+=i+i)
            {
                setbit(j);
            }
        }
    }

//    prm.push_back(2);
//
//    for(i=3;i<=mx4;i+=2)
//    {
//        if(!checkbit(i))
//        {
//            prm.push_back(i);
//        }
//    }
//
//    plen=prm.size();

    //lp1(i,plen)cp(prm[i]);
}

bool isPrime(int n)
{
    if(!(n&1) or n<2)
    {
        return false;
    }

    else
    {
        return (!checkbit(n));
    }

    return true;
}

int main()
{
    seieve();

    int n;

    while(~sf("%d",&n))
    {
        if(n&1)
        {
            if(isPrime(n-2))
            {
                pf("%d is the sum of 2 and %d.\n",n,n-2);
            }

            else
            {
                pf("%d is not the sum of two primes!\n",n);
            }
        }

        else
        {
            int div=n/2;
            bool f=false;

            for(int i=div;i<n;i++)
            {
                int b = n-i;

                if(isPrime(b) and isPrime(i) and b!=i)
                {
                    pf("%d is the sum of %d and %d.\n",n,b,i);
                    f=true;
                    break;
                }
            }

            if(!f)
            {
                pf("%d is not the sum of two primes!\n",n);
            }

            f=false;
        }
    }

    return 0;
}