[HDU6129]Just do it

Description
There is a nonnegative integer sequence \(a_{1...n}\) of length \(n\). HazelFan wants to do a type of transformation called prefix-XOR, which means \(a_{1...n}\) changes into \(b_{1...n}\), where \(b_i\) equals to the XOR value of \(a_1,...,a_i\). He will repeat it for m times, please tell him the final sequence.

Input
The first line contains a positive integer \(T(1\leqslant T\leqslant 5)\), denoting the number of test cases.
For each test case:
The first line contains two positive integers \(n,m(1\leqslant n\leqslant 2×10^5,1\leqslant m\leqslant 10^9)\).
The second line contains \(n\) nonnegative integers \(a_{1...n}(0\leqslant a_i\leqslant 2^{30}−1)\).

Output
For each test case:
A single line contains \(n\) nonnegative integers, denoting the final sequence.

Sample Input

2
1 1
1
3 3
1 2 3

Sample Output

1
1 3 1

---

显然，直接暴力做m次前缀异或和是不可行的，我们先用变量列几个数出来找一下规律

（记\(\otimes\)为异或操作，\(a^k\)表示有\(k\)个\(a\)进行异或操作）

次数	\(a\)	\(b\)	\(c\)	\(d\)
1	\(a\)	\(a\otimes b\)	\(a\otimes b\otimes c\)	\(a\otimes b\otimes c\otimes d\)
2	\(a\)	\(a^2\otimes b\)	\(a^3\otimes b^2\otimes c\)	\(a^4\otimes b^3\otimes c^2\otimes d\)
3	\(a\)	\(a^3\otimes b\)	\(a^6\otimes b^3\otimes c\)	\(a^{10}\otimes b^6\otimes c^3\otimes d\)

可以发现，对于表格中的某个位置\(A[i][j]\)而言，其可以写成\(A[i][j]=A[i-1][j]\otimes A[i][j-1]\)，这个形式十分类似于杨辉三角，而将整个表格顺时针旋转45°后，单独查看某个变量的系数，其就是杨辉三角

稍加推理可得，对于数列中第\(i\)个数，其在做完\(m\)次操作后，对第\(j\)位数\((i<j)\)的贡献是\(\binom{(m-1)+(j-i)}{m-1}\)

但直接计算的话肯定会超时。我们注意到这是异或操作，故只有奇数次的贡献才是有效的，根据组合数奇偶性定理可得，当\(n\ \&\ k=k\)时，\(\binom{n}{k}\)才为奇数

将该定理代入之前的式子，很容易得到\((m-1)\ \&\ (j-i)=0\)，即\((j-i)_2\)中1的位置必定是\((m-1)_2\)中0的位置

我们再按照01背包的思路，每次枚举出一个\((m-1)_2\)的0的位置，可得到\(2^k((m-1)\ \&\ 2^k=0)\)，再将每个数的贡献累加到其之后\(2^k\)的位置上的数即可

/*program from Wolfycz*/
#include<map>
#include<cmath>
#include<cstdio>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define Fi first
#define Se second
#define ll_inf 1e18
#define MK make_pair
#define sqr(x) ((x)*(x))
#define pii pair<int,int>
#define int_inf 0x7f7f7f7f
using namespace std;
typedef long long ll;
typedef unsigned int ui;
typedef unsigned long long ull;
inline char gc(){
	static char buf[1000000],*p1=buf,*p2=buf;
	return p1==p2&&(p2=(p1=buf)+fread(buf,1,1000000,stdin),p1==p2)?EOF:*p1++;
}
template<typename T>inline T frd(T x){
	int f=1; char ch=gc();
	for (;ch<'0'||ch>'9';ch=gc())	if (ch=='-')    f=-1;
	for (;ch>='0'&&ch<='9';ch=gc())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
template<typename T>inline T read(T x){
	int f=1; char ch=getchar();
	for (;ch<'0'||ch>'9';ch=getchar())	if (ch=='-')	f=-1;
	for (;ch>='0'&&ch<='9';ch=getchar())	x=(x<<1)+(x<<3)+ch-'0';
	return x*f;
}
inline void print(int x){
	if (x<0)	putchar('-'),x=-x;
	if (x>9)	print(x/10);
	putchar(x%10+'0');
}
const int N=2e5;
int A[N+10];
int main(){
//	freopen(".in","r",stdin);
//	freopen(".out","w",stdout);
	int T=read(0);
	while (T--){
		int n=read(0),m=read(0)-1;
		for (int i=1;i<=n;i++)	A[i]=read(0);
		for (int j=30;~j;j--){
			if (m&(1<<j))	continue;
			for (int i=n;i>1<<j;i--)
				A[i]^=A[i-(1<<j)];
		}
		for (int i=1;i<=n;i++){
			printf("%d",A[i]);
			putchar(i==n?'\n':' ');
		}
	}
	return 0;
}