HDU4737：A Bit Fun（二分）

A Bit Fun

Time Limit: 5000/2500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3079    Accepted Submission(s): 1535

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

题意：求有多少个子区间满足区间内的数按位或结果小于m。

思路：a[i][j]表示前i个数二进制第j位有多少个1，然后二分即可。

# include <stdio.h>
# include <string.h>
# define MAXN 100000

int a[MAXN+1][36];
int judge(int i, int mid)
{
    int sum = 0;
    for(int j=0; j<36; ++j)
        if(a[mid][j]-a[i-1][j])
            sum |= (1<<j);
    return sum;
}

int main()
{
    int t, n, m, num,cas=1;
    scanf("%d",&t);
    while(t--)
    {
        long long ans = 0;
        memset(a, 0, sizeof(a));
        scanf("%d%d",&n,&m);
        for(int i=1; i<=n; ++i)
        {
            int len = 0;
            scanf("%d",&num);
            for(int j=0; j<36; ++j)
                a[i][j] += a[i-1][j];
            while(num)
            {
                a[i][len++] += num&1;
                num >>= 1;
            }
        }
        for(int i=1; i<=n; ++i)
        {
            int l = i;
            int r = n;
            while(l <= r)
            {
                int mid = (l+r)>>1;
                if(judge(i, mid) < m)
                    l = mid + 1;
                else
                    r = mid - 1;
            }
            ans += r - i + 1;
        }
        printf("Case #%d: %lld\n",cas++,ans);
    }
    return 0;
}