华为OJ_2129_素数伴侣

输入:
有一个正偶数N（N≤100），表示待挑选的自然数的个数。后面给出具体的数字，范围为[2,30000]。
输出:
输出一个整数K，表示你求得的“最佳方案”组成“素数伴侣”的对数。
输入: 输入说明
1 输入一个正偶数n
2 输入n个整数

输出:求得的“最佳方案”组成“素数伴侣”的对数。
 
样例输入:
4
2 5 6 13          
样例输出:

2

#include <iostream>
#include <cmath>

using namespace std;

#define N 100
#define M 30000

int isPrime( int nNum );
int getBestSol( int *nNums, char *bUsed, int nLen, int nDoubles = 0 );
char bPrimes[M + 1];

int main( void )
{
	int n;
	int nNums[N];
	char bUsed[N];
	

	bPrimes[0] = 0;
	for( int i = 1; i  <= M; ++i ){
		bPrimes[i] = isPrime( i );
	}

	cin >> n;
	for( int i = 0; i < n; ++i ){
		cin >> nNums[i];
		bUsed[i] = 0;
	}

//	int nDoubles = 0;
	cout << getBestSol( nNums, bUsed, n ) << endl;

	return 0;
}

int isPrime( int nNum )
{
	if( nNum == 1 )
		return 0;
	if( nNum == 2 )
		return 1;
	
	int nSqrt = sqrt( static_cast<double>( nNum ) );
	for( int i = 2; i <= nSqrt; ++i )
		if( nNum % i == 0 )
			return 0;
	return 1;
}

int getBestSol( int *nNums, char *bUsed, int nLen, int nDoubles )
{
	int i = 0;
	while( bUsed[i] && i < nLen )
		++i;
	if( i == nLen )
		return nDoubles;

	bUsed[i] = 1;
	int maxDoubles = 0, nDoublesOut = 0;
	for( int j = i + 1; j < nLen; ++j ){
		if( bUsed[j] == 0 ){
			bUsed[j] = 1;
			int nDoublesIn = ( bPrimes[nNums[i] + nNums[j]] ? ( nDoubles + 1 ) : nDoubles );
			nDoublesOut = getBestSol( nNums, bUsed, nLen, nDoublesIn );
			bUsed[j] = 0;

			if( maxDoubles < nDoublesOut )
				maxDoubles = nDoublesOut;
			if( maxDoubles == nLen / 2 )
				return maxDoubles;
		}
	}
	bUsed[i] = 0;
	return maxDoubles;
}