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 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. Thet is the case number starting from 1.
Then follows the answer.

 

Sample Input

23 61 3 52 45 4

 

Sample Output

Case #1: 4Case #2: 0

 

Source

2013 ACM/ICPC Asia Regional Chengdu Online

 

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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;}