HDU 6129 just do it（组合数奇偶性）

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Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1411    Accepted Submission(s): 823

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

 

Source

2017 Multi-University Training Contest - Team 7 

【分析】：

组合数C(n,m)的奇偶性：讲解：http://blog.csdn.net/millky/article/details/3206730

if((n&m)==m) 奇数

else     偶数

m次操作之后，只需计算每个元素对后面元素的贡献。

若贡献为偶数次，由于是亦或，可忽略

若贡献为奇数次，那么只需亦或一次。

试着写了写m=1,2,3,4,....的时候，得出规律：

第一个元素对第i个元素的贡献次数为C（i+m-2，m-1）我们只用到他的奇偶性。

【代码】：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long ll;int a[202020],b[202020];int e[202020];int T,n;ll h;int main(){    cin>>T;    while(T--)    {        cin>>n>>h;        for(int i=1;i<=n;i++)            scanf("%d",&a[i]);        int top=0;        ll x=h-1,y=h-1;        for(int i=1;i<=n;i++,x++)        {            if((x&y)==y)e[top++]=i;//i点组合数为奇数        }        memset(b,0,sizeof(b));        for(int i=1;i<=n;i++)        {            for(int j=0; i+e[j]-1<=n&&j<top; j++)                b[i+e[j]-1]^=a[i];            if(i<n)                printf("%d ",b[i]);            else printf("%d\n",b[i]);        }    }return 0;}