HDU 5119 Happy Matt Friends (背包DP + 滚动数组)

题目链接：HDU 5119

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 \(N, M (1 \le N \le 40, 0 \le M \le 10^6)\).

In the second line, there are \(N\) integers \(k_i (0 ≤ k_i ≤ 10^6)\), 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

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）

Solution

题意

给定 \(n\) 个数 \(k[i]\)，从中取出一些数使得异或和大于等于 \(m\)，求有几种取法。

思路

背包DP 滚动数组

设 \(dp[i][j]\) 表示前 \(i\) 个数中异或和为 \(j\) 的所有取法。状态转移方程为 \(dp[i][j] = dp[i - 1][j] + dp[i - 1][j\ xor\ k[i]]\)。

由于当前状态只和前一个状态有关，因此可以用滚动数组优化。

Code

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int maxn = 1 << 20;

ll dp[10][maxn];

ll k[50];

int main() {

    ios::sync_with_stdio(false);

    cin.tie(0);

    int T;

    cin >> T;

    for(int _ = 1; _ <= T; ++_) {

        memset(dp, 0, sizeof(dp));

        dp[0][0] = 1;

        int n, m;

        cin >> n >> m;

        for(int i = 1; i <= n; ++i) {

            cin >> k[i];

        }

        for(int i = 1; i <= n; ++i) {

            for(int j = 0; j < maxn; ++j) {

                dp[i & 1][j] = dp[(i - 1) & 1][j] + dp[(i - 1) & 1][j ^ k[i]];

            }

        }

        ll ans = 0;

        for(int i = m; i < maxn; ++i) {

            ans += dp[n & 1][i];

        }

        cout << "Case #" << _ << ": " << ans << endl;

    }

    return 0;

}

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