2013成都网络赛 J A Bit Fun（水题）

A Bit Fun

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1148    Accepted Submission(s): 644

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

Source

2013 ACM/ICPC Asia Regional Chengdu Online

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关键是理解位或运算

#include<stdio.h>
int a[];
int main()
{
    int t,cou=,n,m,i,j,b;
    scanf("%d",&t);
    while(cou<=t)
    {
        int ans=;
        scanf("%d%d",&n,&m);
        for(i=;i<n;i++)
        scanf("%d",&a[i]);
        for(i=;i<n;i++)
        {
            b=;
            for(j=i;j<n;j++)
            {
                b|=a[j];
                if(b>=m)
                break;
                ans++;
            }
        }
        printf("Case #%d: %d\n",cou,ans);
        cou++;
    }
    return ;
}