A Bit Fun

A Bit Fun

Time Limit : 5000/2500ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 43   Accepted Submission(s) : 13

Problem Description

There are n numbers in a array, as a[sub]0[/sub], a[sub]1[/sub] ... , a[sub]n-1[/sub], and another number m. We define a function f(i, j) = a[sub]i[/sub]|a[sub]i+1[/sub]|a[sub]i+2[/sub]| ... | a[sub]j[/sub] . 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[sup]30[/sup]) Then n numbers come in the second line which is the array a, where 1 <= a[sub]i[/sub] <= 2[sup]30[/sup].

 

Output

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

2013 ACM/ICPC Asia Regional Chengdu Online

 

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

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