HDU 6129 Just do it(找规律+杨辉三角)

Just do it

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 388    Accepted Submission(s): 218

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

 

题意：对一个非负整数序列a1-an进行异或操作获得新的序列，操作如下bi=a1^a2…^ai，然后ai=bi，求进行m次后的序列a

思路：我们可得到b[i]=b[i-1]^a[i]而a[i]就是上一次异或得到的b[i]，那么其实对于原序列中的第x元素在异或操作中被重复异或的次数所构成的系数为杨辉三角，打表如下

斜着看系数，系数构成的是一个杨辉三角，那么对于a1在第m次对bi的贡献就是i:C(m+i-2,i-1)，再根据异或的性质当系数为偶数时相互抵消异或结果为0，系数为奇数时才会有贡献，再根据组合数奇偶的判断方法对于C(n,k),若n&k == k 则c(n,k)为奇数，否则为偶数。对于a2的贡献，只是在a1的情况向后推一位，如a1在m=2，3号位的贡献是奇数那么a2就是在相同情况下的4号位贡献为奇数，a3就是在5号这样推下去

#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=200005;
int a[maxn],b[maxn];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
            memset(b,0,sizeof(b));
        for(int i=1;i<=n;i++)
        {
            int nn=m+i-2,k=i-1;
            if((nn&k)==k)
            for(int j=i;j<=n;j++) b[j]^=a[j-i+1];//a1有j有贡献，那么a2就在j+1有贡献
        }
        for(int i=1;i<=n;i++)
            i!=n?printf("%d ",b[i]):printf("%d\n",b[i]);
    }
    return 0;
}