Happy Matt Friends

时间：2023-03-09 00:43:44
Happy Matt Friends

Happy Matt Friends

Time Limit: 6000/6000 MS (Java/Others)    Memory Limit: 510000/510000 K (Java/Others)

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
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.

分析：由于只有40个数，答案上限为1e6，所以直接暴力枚举转移即可；
代码：
#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cmath>

#include <algorithm>

#include <climits>

#include <cstring>

#include <string>

#include <set>

#include <map>

#include <queue>

#include <stack>

#include <vector>

#include <list>

#define rep(i,m,n) for(i=m;i<=n;i++)

#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)

#define mod 1000000007

#define inf 0x3f3f3f3f

#define vi vector<int>

#define pb push_back

#define mp make_pair

#define fi first

#define se second

#define ll long long

#define pi acos(-1.0)

#define pii pair<int,int>

#define Lson L, mid, ls[rt]

#define Rson mid+1, R, rs[rt]

const int maxn=2e6+;

using namespace std;

ll gcd(ll p,ll q){return q==?p:gcd(q,p%q);}

ll qpow(ll p,ll q){ll f=;while(q){if(q&)f=f*p;p=p*p;q>>=;}return f;}

inline ll read()

{

    ll x=;int f=;char ch=getchar();

    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}

    return x*f;

}

int n,m,k,t,cas;

ll dp[][maxn],ans;

int main()

{

    int i,j;

    scanf("%d",&t);

    while(t--)

    {

        memset(dp,,sizeof(dp));

        ans=;

        dp[][]=;

        scanf("%d%d",&n,&m);

        rep(i,,n)

        {

            scanf("%d",&k);

            for(j=;j<=1e6;j++)dp[i%][j]=dp[(i-)%][j]+dp[(i-)%][j^k];

        }

        rep(i,m,1e6)ans+=dp[n%][i];

        printf("Case #%d: %lld\n",++cas,ans);

    }

    //system("Pause");

    return ;

}

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