UVa12614 - Earn For Future

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本博客探讨了一种策略，通过理解位运算规则，玩家可以在特定的赌场游戏中选择卡片，以最大化其赢得的金额。文章详细解释了位运算的概念，并通过实例展示了如何在给定的卡片集合中找到最优解。

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In a lazy afternoon the great Froogrammer came to realize that, to make his future plans successful he needs a lot of money. To make some quick cash he decided to go to the casino to play a game. The rule of the game is the following:

The player is given N cards. Each card has a non-negative integer printed in it.
The player will choose some cards from the given cards.
The bitwise AND value of the chosen cards will be calculated and the player will be given the same amount of money. (i.e. equal to the bitwise AND value of the chosen cards).

After getting N cards Froogrammer was in a fix as usual. He could not decide which of the cards to choose. So he called you to help him. Please tell him the maximum amount he can win from these set of cards. If you are confused about bitwise AND operation see the notes section below.

Input

The first line of input will contain the number of test cases T (T < 101). Then there will be T test cases. Each of the test cases will start with an integer N ( 0 < N < 31) denoting the number of cards. Then the following line will contain N non-negative integers C_i ( 0 $\le$ C_i < 2³¹) separated by space, denoting the numbers printed on each of the cards.

Output

For each test case print one line of output denoting the case number and the maximum amount Froogrammer can win. See sample output for exact format.

Note:

A bitwise AND takes two binary representations of equal length and performs the logical AND operation on each pair of corresponding bits. The resulting bit of a position is 1 if the bit at that position of both numbers is 1; otherwise, that bit is 0.

For example:

        0101 (decimal 5)
    AND 0011 (decimal 3)
      = 0001 (decimal 1)

Sample Input

1
2
0 1

Sample Output

Case 1: 1

#include <cstdio>
#include <algorithm>

using namespace std;

const int N = 31;

int num[N];
int n;

void input()
{
	scanf("%d", &n);
	for (int i = 0; i < n; i++) {
		scanf("%d", &num[i]);
	}
}


void solve(int cas)
{
	int ans = *max_element(num, num + n);
	printf("Case %d: %d\n", cas, ans);
	
}

int main()
{
	#ifndef ONLINE_JUDGE
		freopen("d:\\OJ\\uva_in.txt", "r", stdin);
	#endif
	
	int t;
	scanf("%d", &t);
	for (int i = 1; i <= t; i++) {
		input();
		solve(i);
	}
	return 0;
}