J - A Bit Fun-----（2015 summer training #9）

J - A Bit Fun
时限:2500MS 内存:32768KB 64位IO格式:%I64d
& %I64u

问题描述

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.

输入

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.

输出

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.

样例输入

2
3 6
1 3 5
2 4
5 4

样例输出

Case #1: 4
Case #2: 0

分析：或运算只会越计算越大，值是一个递增的，所以中间必有一个分界点，找到就ok了。

CODE：

#include <iostream>
using namespace std;

int main()
{
int t,cases=0,arr[100005];
cin>>t;
while(t--){
int n,m;
cin>>n>>m;
cases++;
for(int i=0;i<n;i++)
cin>>arr[i];
cout<<"Case #"<<cases<<": ";
int cnt=0;
long long ans;
for(int i=0;i<n;i++){
ans=0;
for(int j=i;j<n;j++){
ans=ans|arr[j];
cnt++;
if(ans>=m){
cnt--;
break;
}
}
}
cout<<cnt<<endl;
}
return 0;
}