Happy Matt Friends HDU-5119

Happy Matt Friends

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

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.


14年北京站题~
题目意思：给出t组数据 n个数 求XOR出大于m的数的个数，每个数都有两种状态，取与不取。其实就是一个01背包。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e6+10;
const int MOD=1e9+7;
ll a[maxn];
int dp[45][maxn],n,m;
ll slove()
{
    dp[1][0]=dp[1][a[1]]=1;
    for(int i=2; i<=n; i++)
    {
        for(ll j=0; j<maxn; j++)
            {dp[i][j]+=dp[i-1][j];
            dp[i][j^a[i]]+=dp[i-1][j];}
    }
    ll ans=0;
    for(ll i=m; i<maxn; i++)
    {
        ans+=dp[n][i];
    }
    return ans;
}
int main()
{
    int t,k=0;
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        scanf("%d %d",&n,&m);
        for(int i=1; i<=n; i++)
            scanf("%lld",&a[i]);
        ll ans=slove();
        printf("Case #%d: %lld\n",++k,ans);
    }
    return 0;
}

PS：摸鱼怪的博客分享，欢迎感谢各路大牛的指点~

Happy Matt Friends HDU-5119

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