位运算的一些案例

C.Bits

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Let's denote as 

 the
number of bits set ('1' bits) in the binary representation of the non-negative integer x.

You are given multiple queries consisting of pairs of integers l and r.
For each query, find the x, such that l ≤ x ≤ r,
and 

is
maximum possible. If there are multiple such numbers find the smallest of them.

Input

The first line contains integer n — the number of queries (1 ≤ n ≤ 10000).

Each of the following n lines contain two integers li, ri —
the arguments for the corresponding query (0 ≤ li ≤ ri ≤ 1018).

Output

For each query print the answer in a separate line.

Sample test(s)

input

3
1 2
2 4
1 10

output

1
3
7

题目大意是找出二进制的l到r之间，含有1最多的数

由于n可能取到10000,且l和r都会取到10的18次方，所以不能一个一个的搜索

而且还要注重效率，这个时候用位运算就再巧妙不过了

#include<cstdio>
#include<cmath>
#include<cstring>
#include<cstdlib>
#include<vector>
#include<iostream>
#include<algorithm>
#include<windows.h>
using namespace std;
#define T cout<<"***"<<endl;
#define set0(arr) memset(arr,0,sizeof(arr));
#define fr(i,r) for(i=0;i<r;i++)
#define fsr(i,s,r) for(i=s;i<r;i++)
#define ll __int64
#define pf printf
#define sf scanf
int main()
{
int n,i,k;
cin>>n;
fr(k,n){
ll l,r;
sf("%I64d%I64d",&l,&r);
ll x=l;
fr(i,64){
if((x|(ll)1<<i)<=r)
x|=((ll)1<<i);
else{
pf("%I64d\n",x);
break;
}
}
}
}/**/