HDU 4737 A Bit Fun （2013成都网络赛）

A Bit Fun

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 423    Accepted Submission(s): 270

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

 

Recommend

liuyiding

 

 用位数num维护l,r两个指针。

扫描一遍。

复杂度31*n

/* ***********************************************
Author        :kuangbin
Created Time  :2013/9/14 星期六 11:59:04
File Name     :2013成都网络赛1010.cpp
************************************************ */

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>
using namespace std;
const int MAXN = 100010;
int a[MAXN];
int num[40];
int bit[40];

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    
    int T;
    int n,m;
    bit[0] = 1;
    for(int i = 1;i <= 31;i++)
        bit[i] = 2*bit[i-1];
    scanf("%d",&T);
    int iCase = 0;
    while(T--)
    {
        iCase++;
        scanf("%d%d",&n,&m);
        for(int i = 0;i < n;i++)
            scanf("%d",&a[i]);
        long long tot = (long long)n*(n+1)/2;
        int last = 0;
        int i = 0,j = 0;
        memset(num,0,sizeof(num));
        long long sum = 0;
        while(j < n)
        {
            for(int k = 0;k <=31;k++)
                if(a[j] & bit[k])
                    num[k]++;
            int s = 0;
            for(int k = 0;k <= 31;k++)
                if(num[k])
                    s += bit[k];
            if(s >= m)
            {
                while(s >=m)
                {
                    for(int k = 0;k <= 31;k++)
                        if(a[i] & bit[k])
                            num[k]--;
                    s = 0;
                    for(int k = 0;k <= 31;k++)
                        if(num[k])
                            s += bit[k];
                    i++;
                    //cout<<i<<endl;
                }
                sum += (long long)(n-j)*(i-last);
                last = i;
            }
            j++;
        }
        printf("Case #%d: ",iCase);
        //cout<<tot<<" "<<sum<<endl;
        cout<<tot-sum<<endl;
    }
    return 0;
}