http://codeforces.com/contest/484/problem/A
题意：
询问[a,b]中二进制位1最多且最小的数
贪心，假设开始每一位都是1
从高位i开始枚举，
如果当前数>b，且减去1<<i后仍>=a,就减1<<i
当当前数在[a,b]之间时，输出
因为从高位开始减，所以保证当前数是最小的
#include<cstdio>
#include<iostream>
using namespace std;
typedef long long LL;
LL bit[61];
void read(LL &x)
{
x=0; char c=getchar();
while(!isdigit(c)) c=getchar();
while(isdigit(c)) { x=x*10+c-'0'; c=getchar(); }
}
int main()
{
bit[0]=1;
for(int i=1;i<=60;++i) bit[i]=bit[i-1]<<1;
int n;
scanf("%d",&n);
LL a,b;
LL ans;
while(n--)
{
read(a); read(b);
ans=(1LL<<61)-1;
for(int i=60;i>=0;--i)
{
if(ans>b && ans-bit[i]>=a) ans-=bit[i];
if(ans>=a && ans<=b) break;
}
cout<<ans<<'\n';
}
}
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.
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).
For each query print the answer in a separate line.
3
1 2
2 4
1 10
1
3
7
The binary representations of numbers from 1 to 10 are listed below:
110 = 12
210 = 102
310 = 112
410 = 1002
510 = 1012
610 = 1102
710 = 1112
810 = 10002
910 = 10012
1010 = 10102
