[DP] 取若干数异或>=m的方案数 HDU5119

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在一场由Matt和他的朋友们参与的游戏中，每个朋友都有一个魔法数字。Matt的目标是选择一些朋友，使得所选朋友的魔法数字的异或和不小于给定阈值M，从而赢得比赛。本文探讨了如何计算所有可能的胜利方式的数量，通过动态规划算法解决这个问题。

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others) Memory Limit: 510000/510000 K (Java/Others)
Total Submission(s): 5399 Accepted Submission(s): 2065

Problem Description

Matt has N friends. They are playing a game together.

Each of Matt’s friends has a magic number. In the game, Matt selects some (could be zero) of his friends. If the xor (exclusive-or) sum of the selected friends’magic numbers is no less than M , Matt wins.

Matt wants to know the number of ways to win.

Input

The first line contains only one integer T , which indicates the number of test cases.

For each test case, the first line contains two integers N, M (1 ≤ N ≤ 40, 0 ≤ M ≤ 106).

In the second line, there are N integers ki (0 ≤ ki ≤ 106), indicating the i-th friend’s magic number.

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1) and y indicates the number of ways where Matt can win.

Sample Input

3 2

1 2 3

3 3

1 2 3

Sample Output

Case #1: 4

Case #2: 2

Hint

In the ﬁrst sample, Matt can win by selecting: friend with number 1 and friend with number 2. The xor sum is 3. friend with number 1 and friend with number 3. The xor sum is 2. friend with number 2. The xor sum is 2. friend with number 3. The xor sum is 3. Hence, the answer is 4.

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

前 i 项异或成 j 的方案数DP
dp[i][j] = dp[i - 1][j] + dp[i - 1][j ^ a[i]];
<= 1e6 的数最大异或得20位的数

#include <bits/stdc++.h>
#define ll long long
using namespace std;

const int mm = 1050000;

int a[45];
ll dp[45][mm + 10];
int main()
{
	int T;
	scanf("%d", &T);
	for (int cas = 1; cas <= T; cas++)
	{
		int n, m;
		scanf("%d %d", &n, &m);
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		
		memset(dp, 0, sizeof dp);
		dp[0][0] = 1;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 0; j < (1 << 20); j++)
				dp[i][j] = dp[i - 1][j ^ a[i]] + dp[i - 1][j];
		}
		
		ll ans = 0;
		for (int i = m; i < (1 << 20); i++)
			ans += dp[n][i];
		printf("Case #%d: %lld\n", cas, ans);
	}
	return 0;
}