A Bit Fun
Time Limit: 5000/2500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3079 Accepted Submission(s): 1535 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.
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.
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
Source
题意:求有多少个子区间满足区间内的数按位或结果小于m。
思路:a[i][j]表示前i个数二进制第j位有多少个1,然后二分即可。
# include# include # define MAXN 100000int a[MAXN+1][36];int judge(int i, int mid){ int sum = 0; for(int j=0; j<36; ++j) if(a[mid][j]-a[i-1][j]) sum |= (1< >= 1; } } for(int i=1; i<=n; ++i) { int l = i; int r = n; while(l <= r) { int mid = (l+r)>>1; if(judge(i, mid) < m) l = mid + 1; else r = mid - 1; } ans += r - i + 1; } printf("Case #%d: %lld\n",cas++,ans); } return 0;}