hdu5269 ZYB loves Xor I【字典树】

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(AiAi xor AjAj)) (i,j∈[1,n])(i,j∈[1,n]) 
We define that lowbit(x)=2k2k,k is the smallest integer satisfied ((xx and 2k2k)>0) 
Specially,lowbit(0)=0 
Because the ans may be too big.You just need to output ansans 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 nn 
The second line has nn integers A1A1,A2A2….AnAn 
n∈[1,5∗104]n∈[1,5∗104]，Ai∈[0,229]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

思路：首先要明白lowbit(x)的意思，lowbit(x)就相当于x转换成二进制后1的最低位转换成十

进制的数；因为是这是求两个值的异或，所以当前两个数转换成二进制后前缀必须相同的，那

才能使这一位如果不相同才是1的最低位，所以可以把数字转换成二进制倒着插入字典数中，

然后遍历整棵树，每个节点都记录了当前相同前缀串的个数，所以左右相乘；

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