hdu 6129 Just do it(递推)

Just do it

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

Total Submission(s): 155    Accepted Submission(s): 79


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

把每次变换的系数记录下来后可以发现是个杨辉三角，就拿第一个数字在每次变换后对后面的异或次数举例

1 0 0 0 0

1 1 1 1 1

1 2 3 4 5

1 3 6 10 15

斜着看，可以发现这是个杨辉三角 第x次变换第y项是C（x+y-2，y-1）;这是第一个数在第x次变换后对第y项的异或次数，后面的数可以依次类比

因为是杨辉三角，所以第一个数对第2个数如果是奇数次异或，那么第二个数对第三个数也是奇数次异或，后面的情况同理

利用 C(n,m)，如果n&m==m则C（n,m）来快速判断奇偶异或次数

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <map>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 2e6+10, inf = 0x3f3f3f3f;
int a
, b
;

int main()
{
int t;
scanf("%d", &t);
while(t--)
{
int n, m;
scanf("%d %d", &n, &m);
memset(b,0,sizeof(b));
for(int i=1; i<=n; i++)
{
scanf("%d", &a[i]);
//b[i]=a[i];
}
for(int i=1;i<=n;i++)
{
int nn=m+i-2,mm=i-1;
if((nn&mm)==mm)
{
for(int j=i;j<=n;j++)  b[j]^=a[j-i+1];
}
}
for(int i=1;i<=n;i++) printf("%d%c",b[i],i==n?'\n':' ');
}
return 0;
}