2013成都网络赛 J A Bit Fun(水题)
A Bit Fun
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1148 Accepted Submission(s): 644
Problem Description
There are n numbers in a array, as a0, a1 ... , an-1, and another number m. We define a function f(i, j) = ai|ai+1|ai+2| ... | aj . 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 <= 230) Then n numbers come in the second line which is the array a, where 1 <= ai <= 230.
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 <= ai <= 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.
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
关键是理解位或运算
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include<stdio.h> int a[100010]; int main() { int t,cou=1,n,m,i,j,b; scanf("%d",&t); while(cou<=t) { int ans=0; scanf("%d%d",&n,&m); for(i=0;i<n;i++) scanf("%d",&a[i]); for(i=0;i<n;i++) { b=0; for(j=i;j<n;j++) { b|=a[j]; if(b>=m) break; ans++; } } printf("Case #%d: %d\n",cou,ans); cou++; } return 0; }