A Bit Fun

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

There are n numbers in a array, as a

₀, a

₁ ... , a

_n-1, and another number m. We define a function f(i, j) = a

_i|a

_i+1|a

_i+2| ... | a

_j . Where "|" is the bit-OR operation. (i <= j)

The problem is really simple: please count the number of different pairs of (i, j) where f(i, j) < m.

 

 

Input

The first line has a number T (T <= 50) , indicating the number of test cases.

For each test case, first line contains two numbers n and m.(1 <= n <= 100000, 1 <= m <= 2

³⁰) Then n numbers come in the second line which is the array a, where 1 <= a

_i <= 2

³⁰.

 

 

Output

For every case, you should output "Case #t: " at first, without quotes. The

t is the case number starting from 1.

Then follows the answer.

 

 

Sample Input

2 3 6 1 3 5 2 4 5 4

 

 

Sample Output

Case #1: 4 Case #2: 0

 

 

Source

#include
     
      using namespace std;#define ll __int64#define esp 0.00000000001const int N=1e5+10,M=1e6+10,inf=1e9+10,mod=1000000007;int a[N];int flag[40];void init(){    memset(flag,0,sizeof(flag));}void update(int x,int hh){    int sum=0;    while(x)    {        flag[sum++]+=x%2*hh;        x>>=1;    }}int getnum(){    int ans=0;    for(int i=0;i<=35;i++)    {        if(flag[i])        ans+=1<