UVA

//例题3-4 猜数字游戏的提示
//收获：处理思路：在猜测和答案中都出现某数字，且在猜测和答案位置不同的次数 = 都出现的次数 -  数字位置正确的次数； 都出现的次数 = min(答案中出现次数，猜测串中出现次数) 

#include <iostream>
using namespace std;
int main()
{
	int n, *a, *b;
	int sum1, sum2, c1, c2, kase = 0;
	while (cin >> n)
	{
		if (!n) break;
		a = new int [n];
		b = new int [n];
		
		cout << "Game " << ++kase << ":" << endl;
		for (int i = 0; i < n; i++) cin >> a[i];
		while(1)
		{
			sum1 = sum2 = 0;
			for (int i = 0; i < n; i++)
			{
				cin >> b[i];
				if (a[i] == b[i]) sum1++;
			}
			if (b[0] == 0) break;
			
			for (int j = 1; j <= 9; j++)
			{
				c1 = c2 = 0;
				for (int i = 0; i < n; i++)
				{
					if (a[i] == j) c1++;
					if (b[i] == j) c2++;
				}
				sum2 += min(c1, c2);
			}
			cout << "    (" << sum1 << "," << sum2 - sum1 << ")" << endl;
		}
	}
	delete[]a;
	delete[]b;
	return 0;
}