Uva 10168 Summation of Four Primes 素数

/**   Status: 1054090310168Summation of Four PrimesAcceptedC++0.8642012-08-30 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  10000000int 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 ;}*/