hdu 4737 A Bit Fun(思维&正解O(31*n))

A Bit Fun

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1104    Accepted Submission(s): 623

Problem Description

There are n numbers in a array, as a ₀, a ₁ ... , 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 <= 2 ³⁰) Then n numbers come in the second line which is the array a, where 1 <= a _i <= 2 ³⁰.

 

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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liuyiding

 

题意：

给你n个数字。问你有多少i,j.组合使得。a[i] | a[i+1] | a[i+2].......|a[j]的值小于m.i<=j。

思路：

对于|运算。很简单。但对于逆运算就不好想了。由于|运算只增不减。单调性可以利用。如果f(i,j)>=m了。那i和j以后的f值肯定>=m了。这样就可以用排除法做了。我们可以维护一个数组。存二进制中对应位1的个数。这样删除前面的数就好办了。

详细见代码：

#include<algorithm>
#include<iostream>
#include<string.h>
#include<sstream>
#include<stdio.h>
#include<math.h>
#include<vector>
#include<string>
#include<queue>
#include<set>
#include<map>
using namespace std;
const int INF=0x3f3f3f3f;
const int maxn=100010;
int bit[35],abit[35];
int a[maxn];
int main()
{
    int i,j,k,n,m,sum,pre,t,cas=1;
    __int64 ans;
    bit[0]=1;
    for(i=1;i<=31;i++)//预处理出二进制数组
        bit[i]=bit[i-1]<<1;
    scanf("%d",&t);
    while(t--)
    {
        sum=0;
        memset(abit,0,sizeof abit);//记录各二进制位1的个数
        scanf("%d%d",&n,&m);
        ans=(__int64)n*(n+1)/2;//用排除法。所以先算出总组合数。(n*(n-1))/2 + n。
        for(i=0;i<n;i++)
            scanf("%d",a+i);
        i=j=0;//i,j指针扫一遍
        while(j<n)
        {
            while(sum<m&&j<n)//如果小于m。可以向右扩展。
            {
                sum|=a[j];
                for(k=0;k<31;k++)
                    if(a[j]&bit[k])
                        abit[k]++;
                j++;
            }
            pre=i;
            while(sum>=m)
            {
                for(k=0;k<31;k++)
                    if(a[i]&bit[k])
                        abit[k]--;
                sum=0;
                for(k=0;k<31;k++)
                    if(abit[k])
                        sum|=bit[k];
                i++;
            }
            ans-=(i-pre)*(n-j+1);//(i-pre)的部分和(n-j+1)的部分肯定不能组合。
        }
        printf("Case #%d: %I64d\n",cas++,ans);
    }
    return 0;
}