2013年成都赛区网络赛之水题A Bit Fun

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

题意：给定n个数字，相邻数字进行异或操作，求满足操作的和小于某个数m的数量

题解：随着异或数字的增多，操作值必然越来越大。暴力求解

#include<stdio.h>
int p[100005];
int main()
{
    int test,n,i,ca=0,j,m;
 scanf("%d",&test);
 while(test--)
 {
  ca++;
  int cnt=0;
  scanf("%d %d",&n,&m);
  for(i=0;i<n;i++)
  {
   scanf("%d",&p[i]);
  }
  printf("Case #%d: ",ca);
        for(i=0;i<n;i++)
  {
   int t=p[i];
   for(j=i;j<n;j++)
   {
               t=t|p[j];
      if(t<m) cnt++;
      else break;
   }
  }
  printf("%d\n",cnt);
 }
 return 0;
}