A Bit Fun
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 199 Accepted Submission(s): 123
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
import java.io.*;
import java.util.*;
public class Main {
BufferedReader bu = new BufferedReader(new InputStreamReader(System.in));;
PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out), true);
StringTokenizer st;
int t, n;
long m;
long[] a;
public static void main(String[] args) throws Exception {
new Main().work();
}
void work() throws Exception {
t = nextInt();
for (int p = 1; p <= t; p++) {
pw.print("Case #" + p + ": ");
n = next().intValue();
m = next();
a = new long[n + 1];
for (int i = 1; i <= n; i++) {
a[i] = next();
}
long count = 0,x;
for (int i = 1; i <=n; i++) {
long kp=a[i]|a[i];
if(kp>=m)
continue;
else{
x=a[i];
count++;
}
for (int j = i+1; j <= n; j++) {
if(a[j]>=m)
break;
x = x | a[j];
if (x>=m)
break;
count++;
}
}
pw.println(count);
}
}
public int nextInt() throws Exception {
return Integer.parseInt(bu.readLine());
}
public Long next() throws Exception {
while (st == null || !st.hasMoreTokens())
st = new StringTokenizer(bu.readLine());
return Long.parseLong(st.nextToken());
}
}