A Bit Fun
A Bit Fun
Time Limit : 5000/2500ms (Java/Other) Memory Limit : 32768/32768K (Java/Other)
Total Submission(s) : 43 Accepted Submission(s) : 13
Problem Description
There are n numbers in a array, as a[sub]0[/sub], a[sub]1[/sub] ... , a[sub]n-1[/sub], and another number m. We define a function f(i, j) = a[sub]i[/sub]|a[sub]i+1[/sub]|a[sub]i+2[/sub]| ... | a[sub]j[/sub] . 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 <= 2[sup]30[/sup]) Then n numbers come in the second line which is the array a, where 1 <= a[sub]i[/sub] <= 2[sup]30[/sup].
Output
For every case, you should output "Case #t: " at first, without quotes. The [I]t[/I] 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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <string.h> 4 5 int main() 6 { 7 int T,A,B,sign; 8 long a[100000],m,n,sum,tmp,t,i,j; 9 scanf("%d",&T); 10 t=1; 11 while(T--) 12 { 13 sign=0; 14 scanf("%ld%ld",&n,&m); 15 for(i=0;i<n;i++) 16 scanf("%ld",&a[i]); 17 for(i=0;i<n;i++) 18 { 19 tmp=a[i]; 20 for(j=i;j<n;j++) 21 { 22 tmp=tmp|a[j]; 23 if(tmp<m) 24 sign++; 25 else 26 break; 27 } 28 } 29 printf("Case #%d: %d\n",t,sign); 30 t++; 31 } 32 return 0; 33 }
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* 作者: Wurq
* 博客: https://www.cnblogs.com/Wurq/
* Gitee: https://gitee.com/wurq
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**************************************
* 作者: Wurq
* 博客: https://www.cnblogs.com/Wurq/
* Gitee: https://gitee.com/wurq
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