动态规划——Happy Matt Friends

最新推荐文章于 2021-03-11 22:03:18 发布

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本文介绍了一种通过动态规划算法解决特定游戏问题的方法，即在给定一系列整数的情况下，选择若干个数使得它们的异或和不低于指定值，进而计算出所有可能的选择方式。

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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 ≤ 10 ⁶).

In the second line, there are N integers ki (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.

题意：给n个数，任选若干个数，使其异或和不小于m，求有多少种选法。

思路：

由于异或有这样的性质：a^b^b=a，故设置dp[i][j]为从前i个数中选若干个数的异或和为j。

则有dp[i][j]=dp[i-1][j]+dp[i-1][j^a[i]]

同时根据题目数据1e6，可知异或和的最大值是1<<20-1。这个值不能大也不能小，否则会出错。

dp数组我用滚动数组优化了下，不用滚动数组应该也可以。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
const int maxn=(1<<20);
ll dp[45][maxn];
int a[45];
int n, m;
int main()
{
    int t;
    scanf("%d", &t);
    for(int cas=1; cas<=t; cas++)
    {
        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;
        int pos=1;
        for(int i=1; i<=n; i++)
        {
            for(int j=0; j<maxn; j++)
                dp[pos][j]=dp[pos^1][j]+dp[pos^1][j^a[i]];
            pos^=1;
        }
        ll ans=0;
        for(int i=m; i<maxn; i++)
            ans+=dp[pos^1][i];
        printf("Case #%d: %lld\n", cas, ans);
    }
    return 0;
}