HDU 4737 A Bit Fun（2013成都网络赛）
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4737
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.
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.
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.
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：
求有多少对f(i, j) < m！
PS:
维护两个指针！
代码如下：
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long LL;
const int maxn = 100010;
int a[maxn];
int c[32];
int cal(int x, int v)
{
int ret = 0;
for(int i = 0; i <= 30; i++)
{
if(x&(1<<i))
{
c[i] += v;
}
if(c[i])
{
ret |= (1<<i);
}
}
return ret;
}
int main ()
{
int t;
int n, m;
int cas = 0;
scanf("%d",&t);
while(t--)
{
memset(c,0,sizeof(c));
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
{
scanf("%d",&a[i]);
}
int l = 0;
LL ans = 0;
for(int i = 0; i < n; i++)
{
int tt = cal(a[i], 1);
while(tt >= m)
{
tt = cal(a[l++], -1);
}
ans += (i-l+1);
}
printf("Case #%d: %I64d\n",++cas,ans);
}
return 0;
}