HDU6129-Just do it

Just do it

Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1205 Accepted Submission(s): 704

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

Source
2017 Multi-University Training Contest - Team 7

题目大意：给出n个数，进行m次前缀异或和操作，问最后的结果是多少？
解题思路：

1112113611410201

斜地看就是一个杨辉三角，从1开始标号的话，第i行第j列的数是Cj−1i+j−2,第k个数对第m行第j列的贡献就是第k+1个数对m行第j+1列的贡献，贡献为奇数时才记，组合数Cmn为奇数的条件为n&m==m。

#include<iostream>#include<cstdio>#include<string>#include<cmath>using namespace std;const int INF=0x3f3f3f3f;const int MAXN=2e5+5;const int MOD=1e9+7;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] );      b[i]=0;    }    for(int i=1;i<=n;i++)    {      if(((m+i-2)&(i-1))==(i-1))      {        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;}/*21 113 31 2 3*/