Happy Matt Friends

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 89 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/dyzll/p/5943866.html

文章标签：

#java

本文介绍了一种通过暴力枚举转移的方法来解决特定问题的算法实现。问题中，Matt和他的朋友们进行一场游戏，通过计算被选中朋友的魔法数字的异或和是否大于等于给定值M来判断Matt是否能赢得游戏。文章给出了具体的输入输出样例，并提供了一份详细的代码实现。

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 ≤ 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 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+10;
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p;p=p*p;q>>=1;}return f;}
inline ll read()
{
    ll x=0;int f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,k,t,cas;
ll dp[2][maxn],ans;
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        ans=0;
        dp[0][0]=1;
        scanf("%d%d",&n,&m);
        rep(i,1,n)
        {
            scanf("%d",&k);
            for(j=0;j<=1e6;j++)dp[i%2][j]=dp[(i-1)%2][j]+dp[(i-1)%2][j^k];
        }
        rep(i,m,1e6)ans+=dp[n%2][i];
        printf("Case #%d: %lld\n",++cas,ans);
    }
    //system("Pause");
    return 0;
}