这是队友的代码，只是转载一下。

A Bit Fun

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 

今年ACM成都赛区网赛的题最后一题，自己花了不少时间做的。题目数据较大，本题用追逐法解决。表示一次AC，有点开心。令l[i]表示a[i]之前到a[i]的异或和小于m的最左边的下标值，则最后答案为ans+=i-l[i]+1 (1<=i<=n) 代码如下

#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
#include <cstring>
#define maxn 100010
using namespace std;
int a[maxn];
int l[maxn];
int f(int x,int y)
{
    int temp=a[x];
    for(int i=x+1;i<=y;i++)
    {
        temp|=a[i];
    }
    return temp;
}
int main()
{
    int T;
    scanf("%d",&T);
    getchar();
    for(int kase=1;kase<=T;kase++)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        memset(l,0,sizeof(l));
        long long ans=0;
        for(int i=1;i<=n;i++)
            scanf("%d",a+i);
        int L,R;
        for(R=1;R<=n;R++)
        {
            L=R;
            while(L>=1&&f(L,R)<m) L--;//找最左边的下标值
            l[R]=++L;
        }
        for(int i=1;i<=n;i++)
        {
            //printf("l[%d]=%d\n",i,l[i]);
            if(l[i]) ans+=i-l[i]+1;
        }
        printf("Case #%d: %lld\n",kase,ans);
    }
    return 0;
}