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 = ;
 int a[MAXN];
 int num[];
 int bit[];

 int main()
 {
     //freopen("in.txt","r",stdin);
     //freopen("out.txt","w",stdout);

     int T;
     int n,m;
     bit[] = ;
     for(int i = ;i <= ;i++)
         bit[i] = *bit[i-];
     scanf("%d",&T);
     int iCase = ;
     while(T--)
     {
         iCase++;
         scanf("%d%d",&n,&m);
         for(int i = ;i < n;i++)
             scanf("%d",&a[i]);
         long long tot = (long long)n*(n+)/;
         int last = ;
         int i = ,j = ;
         memset(num,,sizeof(num));
         long long sum = ;
         while(j < n)
         {
             for(int k = ;k <=;k++)
                 if(a[j] & bit[k])
                     num[k]++;
             int s = ;
             for(int k = ;k <= ;k++)
                 if(num[k])
                     s += bit[k];
             if(s >= m)
             {
                 while(s >=m)
                 {
                     for(int k = ;k <= ;k++)
                         if(a[i] & bit[k])
                             num[k]--;
                     s = ;
                     for(int k = ;k <= ;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 ;
 }