Uva 10168 Summation of Four Primes 素数

/*
*   Status: 10540903	10168	Summation of Four Primes	Accepted	C++	0.864	2012-08-30 02:49:48
*   url:  http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1109
*   Time: 2012.8.30 10:30 around
*   Stratege: 通过打表，发现奇数是2，3开始的，偶数是2，2开始的，所以只要一层循环，就可以了
*
*/
#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <queue>
using namespace std ;

#define LL long long
#define MAXN  10000000
int p[MAXN], prilen ;
bool NotPrime[MAXN] ;
int n ;

void getPrime ()
{
	int i, j ;
	for (i = 4; i < MAXN; i += 2)
		NotPrime[i] = true ;
	NotPrime[0] = NotPrime[1] = true ;
	prilen = 1 ;
	p[0] = 2 ;
	for (i = 3; i < MAXN; i += 2)
	{
		if (!NotPrime[i])
		{
			p[prilen++] = i ;
			for (j = 2*i; j < MAXN; j += i)
				NotPrime[j] = true ;
		}
	}
}

int main ()
{
	getPrime () ;
	int i ;
	while (scanf ("%d", &n) != EOF)
	{
		if (n < 8)
		{
			printf ("Impossible.\n") ;
			continue ;
		}
		if (n % 2 == 0)
		{
			printf ("2 2 ") ;
			n -= 4 ;
			for (i = 0; i < prilen; i ++)
			{
				if (!NotPrime[p[i]] && !NotPrime[n-p[i]])
				{
					printf ("%d %d\n", p[i], n-p[i]) ;
					break ;
				}
			}
		}
		else if (n % 2)
		{
			printf ("2 3 ") ;
			n -= 5 ;
			for (i = 0; i < prilen; i ++)
			{
				if (!NotPrime[p[i]] && !NotPrime[n-p[i]])
				{
					printf ("%d %d\n", p[i], n-p[i]) ;
					break ;
				}
			}
		}
	}
	return 0 ;
}
//打表代码
/*
int main ()
{
	getPrime () ;
	int i, j, k, l, q ;
	for (i = 100; i < 200; i ++)
	{
		for (j = 0; p[j] < i; j ++)
			for (k = 0; p[k] < i; k ++)
				for (q = 0; q < i; q ++)
				{
					if (!NotPrime[i-p[j]-p[k]-p[q]] && i-p[j]-p[k]-p[q] >= 2)
					{
						cout << i << " " << p[j] << " " << p[k] << " " << p[q] << " " << i-p[j]-p
							[k]-p[q] << endl ; 
						goto L ;
					}
				}
		cout << i << " " << "impossible" << endl ;
L:		continue ;
	}
	cin >> q; 
	return 0 ;
}
*/