题目

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 ≤ 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

2
3 2
1 2 3
3 3
1 2 3

Sample Output

Case #1: 4
Case #2: 2

Hint

In the rst 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.

题解

动态规划问题
选取任意个人,使所有人的数异或后大于 m 求有多少种选法
用 dp[i][j] 表示前 i 个人中选取任一个异或和为 j 的选法个数

• 如果不选择第 i 个人,则有 dp[i][j] = dp[i-1][j]
• 如果选择第 i 个人,则有 dp[i][j^a[i]] = dp[i-1][j]

计算后从 m 开始计算总共有多少即可

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
using namespace std;

const int maxn = 45;
const int maxm = 1000005;

int a[maxn];
int dp[maxn][maxm];

void Do() {
    int n,m;
    scanf("%d%d",&n,&m);
    memset(dp,0,sizeof(dp));

    for(int i = 1;i <= n;i++){
        scanf("%d",&a[i]);
    }
    dp[0][0] = 1;

    for(int i = 1;i <= n;i++) {
        for(int j = 0;j < maxm;j++) {
            dp[i][j] += dp[i - 1][j];
            dp[i][j ^ a[i]] += dp[i - 1][j];
        }
    }

    long long ans = 0;
    for(int i = m;i <= maxm;i++)
        ans += dp[n][i];

    printf("%lld\n",ans);
}

int main(){
    int T;
    scanf("%d",&T);
    for(int i = 1;i <= T;i++) {
        printf("Case #%d: ",i);
        Do();
    }
    return 0;
}