2013 ACM/ICPC Asia Regional Chengdu Online A Bit Fun

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main()
{
    int T,k=0;
    scanf("%d",&T);
    while(T--)
    {
      int m,n,i,j,a[110000],s,cnt=0;
      scanf("%d%d",&m,&n);
      for(i = 0;i < m; i++)
      scanf("%d",&a[i]);
      for(i = 0;i < m; i++)
      {
            s = a[i];
            for(j = i;j < m; j++)
                {
                    s |= a[j];
                    if(s < n) cnt++;
                    else break;
                }
        }    
        printf("Case #%d: %d\n",++k,cnt);
      
      
    }
    //system("pause");
    return 0;
}

A Bit Fun

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

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

2 3 6 1 3 5 2 4 5 4

 

 

Sample Output

Case #1: 4 Case #2: 0