2014 北京区域赛 dp

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.

InputThe 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.OutputFor 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

2
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.
题意：输入N和M，表示有N个数供你随机选择，求你选择的这些数（不能有重复的）的异或值大于等于M的方法有多少个，不选异或值就是零，选一个异或值就是本身。

分析：N最大为40，暴力法的复杂度为N!，必然不可行，考虑到N的范围比较小，而且这些数的异或值是有限的。所以想到是一道动态规划题。而异或的最大值为10^6约等于2^20，我们可以在这个范围内枚举异或，对于从第一个数取到第n个数的的结果就是d[n][∑j]，j为大于m的异或值，对于每一个阶段，我们可以选择异或a[i]或者不异或，对应的状态转移方程也就是a[i][j]=a[i-1][j]+a[i-1][j^a[i]]。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std; 
#define maxn 1000005
typedef long long ll;
ll a[maxn], dp[45][maxn];
ll T,cas,n,m,ans;
int main() {
    cin >> T;
    cas = 1;
    while( T -- ) {
        cin >> n >> m;
        ans = 0;
        for(ll i=1; i<=n; i++) {
            cin >> a[i];
        }
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for(ll i=1; i<=n; i++) {
            for(ll j=0; j<maxn; j++) {
                //dp[i][j]为用前面i个数（少于或等于）异或得到j的个数 
                dp[i][j] += dp[i-1][j];//加上用i-1个数（少于或等于）得到j的个数 
                dp[i][j^a[i]] += dp[i-1][j];
                //i-1个数与第i个数异或得到j的个数等于前i-1个数（少于或等于）得到j的个数
            }
        }
        for(ll j=m; j<maxn; j++) {
            ans += dp[n][j];
        }
        printf("Case #%lld: %lld
", cas++, ans);
    }
    return 0;
}