HDU6129-Just do it

Just do it

                                                                         Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
                                                                                                   Total Submission(s): 719    Accepted Submission(s): 415

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

 

题意：有n个数，每经过一轮，就对前缀做一次异或，求出最终的序列

解题思路：我们可以找出经过每一轮后的每个数字是由哪些数字异或得到的，比如对于题目中给出的a[1]，我们可以找出经过每一轮后哪些数字是由它异或得到的，后面的数字就是将这个表往右移一位

#include <iostream>#include <cstdio>#include <cstring>#include <map>#include <set>#include <string>#include <cmath>#include <algorithm>#include <vector>#include <bitset>#include <stack>#include <queue>#include <unordered_map>#include <functional>using namespace std;const int INF=0x3f3f3f3f;#define LL long longint ans[200009],a[200009];int dfs(int x,int y,int len){    if(x>len/2&&y>len/2) return 0;    if(x==1||y==1) return 1;    if(x>len/2) return dfs(x-len/2,y,len/2);    else if(y>len/2) return dfs(x,y-len/2,len/2);    else return dfs(x,y,len/2);}int main(){    int t,n,m;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&n,&m);        for(int i=1;i<=n;i++)scanf("%d",&a[i]);        memset(ans,0,sizeof ans);        int k=2;        while(k<n) k*=2;        m%=k;        if(!m) m=k;        for(int i=1;i<=n;i++)        {            if(dfs(m,i,k))            {                for(int j=i;j<=n;j++)                    ans[j]^=a[j-i+1];            }        }        int flag=0;        for(int i=1;i<=n;i++)        {            if(flag) printf(" ");            else flag=1;            printf("%d",ans[i]);        }        printf("\n");    }    return 0;}