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 sequencea1...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 integerT(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

21 113 31 2 3

 

Sample Output

11 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：

11 1 1 11 1

21 2 3 45 6

31 3 6 1015 21

41 4 10 2035 56

51 5 15 3570 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]);elseprintf(" %d",b[i]);}puts("");}return 0;}