Problem Description

There are n numbers in a array, as a₀, a₁ ... , 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 <= 2³⁰) Then n numbers come in the second line which is the array a, where 1 <= a_i <= 2³⁰.

 

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

 

Sample Output

 

Source

 

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 /* ***********************************************

 Author        :kuangbin

 Created Time  :2013/9/14 星期六 11:59:04

 File Name     :2013成都网络赛\1010.cpp

 ************************************************ */

 #pragma comment(linker, "/STACK:1024000000,1024000000")

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <set>

 #include <map>

 #include <string>

 #include <math.h>

 #include <stdlib.h>

 #include <time.h>

 using namespace std;

 const int MAXN = ;

 int a[MAXN];

 int num[];

 int bit[];

 int main()

 {

     //freopen("in.txt","r",stdin);

     //freopen("out.txt","w",stdout);

     int T;

     int n,m;

     bit[] = ;

     for(int i = ;i <= ;i++)

         bit[i] = *bit[i-];

     scanf("%d",&T);

     int iCase = ;

     while(T--)

     {

         iCase++;

         scanf("%d%d",&n,&m);

         for(int i = ;i < n;i++)

             scanf("%d",&a[i]);

         long long tot = (long long)n*(n+)/;

         int last = ;

         int i = ,j = ;

         memset(num,,sizeof(num));

         long long sum = ;

         while(j < n)

         {

             for(int k = ;k <=;k++)

                 if(a[j] & bit[k])

                     num[k]++;

             int s = ;

             for(int k = ;k <= ;k++)

                 if(num[k])

                     s += bit[k];

             if(s >= m)

             {

                 while(s >=m)

                 {

                     for(int k = ;k <= ;k++)

                         if(a[i] & bit[k])

                             num[k]--;

                     s = ;

                     for(int k = ;k <= ;k++)

                         if(num[k])

                             s += bit[k];

                     i++;

                     //cout<<i<<endl;

                 }

                 sum += (long long)(n-j)*(i-last);

                 last = i;

             }

             j++;

         }

         printf("Case #%d: ",iCase);

         //cout<<tot<<" "<<sum<<endl;

         cout<<tot-sum<<endl;

     }

     return ;

 }