Just do it

Just do it

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)

Total Submission(s): 1420    Accepted Submission(s): 831

Problem Description

There is a nonnegative integer sequence a1...n of length n . HazelFan wants to do a type of transformation called prefix-XOR, which means a1...n changes into b1...n , where bi equals to the XOR value of a1,...,ai . He will repeat it for m times, please tell him the final sequence.

 

Input

The first line contains a positive integer T(1≤T≤5) , denoting the number of test cases.
For each test case:
The first line contains two positive integers n,m(1≤n≤2×105,1≤m≤109) .
The second line contains n nonnegative integers a1...n(0≤ai≤230−1) .

 

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
  
  

 

Source

2017 Multi-University Training Contest - Team 7

 

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打表找规律，发现他们的系数如下：

1 1 1 1 1 1 1

2 1 2 3 4 5 6

3 1 3 6 10 15 21

4 1 4 10 20 35 56

5 1 5 15 35 70 126

..............................................

..............................................

i .......c(m+i-2,i-1).................

发现他们的系数是杨辉三角

我们只需要系数的奇偶，c(m,n) ：if((m&n)==n) 则c(m,n)是一个奇数，一个组合数的性质

代码如下：

#include<stdio.h>
#include<string.h>
const int N=2*1e5+5;
int a[N],b[N];
int main()
{
	int t,n,m;
	scanf("%d",&t);
	while(t--)
	{
		scanf("%d%d",&n,&m);
		memset(b,0,sizeof(b));
		for(int i=1;i<=n;i++)
			scanf("%d",&a[i]);
		for(int i=1;i<=n;i++)
		{
			int mm=m+i-2,nn=i-1;
			if((mm&nn)==nn)
			{
				for(int j=i;j<=n;j++)
				{
					b[j]^=a[j-i+1];
				}
			}
		}
		for(int i=1;i<=n;i++)
		{
			if(i==1)
				printf("%d",b[i]);
			else
				printf(" %d",b[i]);
		}
		puts("");
	}
	return 0;
}