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

21 113 31 2 3

 

Sample Output

11 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[N], b[N];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;}