hdu5269ZYB loves Xor I

这个题还挺有意思的，虽然做法还是很显然的，但是还是值得写篇博客

标程是字典树，看起来比较蛋疼

直接分治就好了，

排序之后，使得ai不被排在它后面的管控，也就是lowbit(a[i])<=lowbit(a[j]),i

然后接下来分治，每次把区间分成两半，使得前面一半管控后面一半，一边拆一边算ans

这里，我直接写了一发dfs，很显然递归深度最多也就logA，放心跑

最后打出ans即可

排序的复杂度是nlognlogA

分治就是nlogA

时间主要花在排序上，事实上，排序的实际时间应该更好，所以跟标程的nlogA并没有太大区别

这里因为hd服务器的原因，交g++会蜜汁re，c++很快就过了

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 188    Accepted Submission(s): 103

Problem Description

Memphis loves xor very musch.Now he gets an array A.The length of A is n.Now he wants to know the sum of all (lowbit( Ai xor  Aj))  (i,j∈[1,n])
We define that lowbit(x)= 2k,k is the smallest integer satisfied (( x and  2k)>0)
Specially,lowbit(0)=0
Because the ans may be too big.You just need to output  ans mod 998244353

 

Input

Multiple test cases, the first line contains an integer T(no more than 10), indicating the number of cases. Each test case contains two lines
The first line has an integer  n
The second line has  n integers  A1, A2.... An
n∈[1,5∗104]， Ai∈[0,229]

 

Output

For each case, the output should occupies exactly one line. The output format is Case #x: ans, here x is the data number begins at 1.

 

Sample Input

2 5 4 0 2 7 0 5 2 6 5 4 0

 

Sample Output

Case #1: 36 Case #2: 40

 

Source

BestCoder Round #44

#include