题干：

Problem Description

Alice and Bob are playing a stone game. There are nnn piles of stones. In each turn, a player can remove some stones from a pile (the number must be positive and not greater than the number of remaining stones in the pile). One player wins if he or she remove the last stone and all piles are empty. Alice plays first.
To make this game even more interesting, they add a new rule: Bob can choose some piles and remove entire of them before the game starts. The number of removed piles is a nonnegative integer, and not greater than a given number ddd. Note ddd can be greater than nnn, and in that case you can remove all of the piles.
Let ansansans denote the different ways of removing piles such that Bob are able to win the game if both of the players play optimally. Bob wants you to calculate the remainder of ansansans divided by 109+710^9+7109+7.

Input

The first line contains an integer TTT, representing the number of test cases.
For each test cases, the first line are two integers nnn and ddd, which are described above.
The second line are nnn positive integers aia_iai​, representing the number of stones in each pile.
T≤5,n≤103,d≤10,ai≤103T \leq 5, n \leq 10^3, d \leq 10, a_i \leq 10^3T≤5,n≤103,d≤10,ai​≤103

Output

For each test case, output one integer (modulo 109+710^9 + 7109+7) in a single line, representing the number of different ways of removing piles that Bob can ensure his victory.

Sample Input

2
5 2
1 1 2 3 4
6 3
1 2 4 7 1 2

Sample Output

2
5

题目大意：

  剥去博弈的外皮，简化之后的题意是：给定n个数，一个x，让你最多可以选择d个数，求异或和为x的方案数。

解题报告：

  直接dp就行了，用滚动数组空间优化掉一维。（方法和01背包一样倒着遍历）

AC代码：

//C
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<vector>
#include<cctype>
using namespace std;
typedef long long ll;
const int MAX=1<<11;
const int mod=1e9+7;
//a=num[1]^num[2]^..^num[n];
//选i(0<=i<=d)个数使异或和等于a 
//dp[i][d][v]从前i个数中选d个数, 异或和为v的方案数 
//dp[i][d][v]=dp[i-1][d][v]+dp[i-1][d-1][v^val[i]];
int n,d,val[MAX+5];
ll dp[15][MAX+5]; //滚动减掉第一维 
int main()
{
	int t,q,i,j,k,sum;
	ll ans=0;
	cin>>t;
	for(;t;t--){
		scanf("%d%d",&n,&d); 
		d=min(d,n);
		ans=sum=0;
		for(i=1;i<=n;i++){
			scanf("%d",&val[i]);
			sum^=val[i];
		}
		//init
		for(i=0;i<=d;i++){
			for(j=0;j<=MAX;j++)
				dp[i][j]=0;
		}
		dp[0][0]=1;
		for(i=1;i<=n;i++){
			for(j=d;j>=1;j--){
				for(k=0;k<=MAX;k++){
					dp[j][k]=(dp[j][k]+dp[j-1][k^val[i]])%mod;
				}
			}	
		}
		for(i=0;i<=d;i++)
			ans=(ans+dp[i][sum])%mod;
		printf("%lld\n",ans);
	}
	return 0;
}