A. A Bit Fun

A. A Bit Fun

Time Limit: 2500ms

Memory Limit: 32768KB

64-bit integer IO format: %I64d      Java class name: Main

Submit Status PID: 33859

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

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

int main()
{
    int N,i,j,sign,Times;
    long  a,b,tmp,sum[100000];
    scanf("%d",&N);
    Times=1;
    while(N--)
    {
        sign=0;
        scanf("%ld%ld",&a,&b);
        for(i=0;i<a;i++)
            scanf("%ld",&sum[i]);
        for(i=0;i<a;i++)
        {
            for(j=i,tmp=sum[i];j<a;j++)
             {
                tmp=tmp|sum[j];
                if(tmp<b)
                    sign++;
                else 
                    break;
            }
        }
        printf("Case #%d: %d\n",Times++,sign);
    }
    return 0;
}