hdu 4737 A Bit Fun 尺取法

A Bit Fun

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

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

#include<bits/stdc++.h>
using namespace std;
#define ll __int64
#define esp 0.00000000001
const int N=1e5+,M=1e6+,inf=1e9+,mod=;
int a[N];
int flag[];
void init()
{
    memset(flag,,sizeof(flag));
}
void update(int x,int hh)
{
    int sum=;
    while(x)
    {
        flag[sum++]+=x%*hh;
        x>>=;
    }
}
int getnum()
{
    int ans=;
    for(int i=;i<=;i++)
    {
        if(flag[i])
        ans+=<<i;
    }
    return ans;
}
int main()
{
    int x,y,z,i,t;
    int T,cas=;
    scanf("%d",&T);
    while(T--)
    {
        init();
        scanf("%d%d",&x,&y);
        for(i=;i<=x;i++)
        scanf("%d",&a[i]);
        int st=;
        int en=;
        int cnt=;
        ll ans=;
        int qu=;
        while()
        {
            while(getnum()<y&&en<=x)
            update(a[en++],);
            if(getnum()<y)
            {
                ans+=(ll)(en-st)*(en-st+)/;
                break;
            }
            ans+=max(,en-st-);
            update(a[st++],-);
        }
        printf("Case #%d: %I64d\n",cas++,ans);
    }
    return ;
}