codeforces 484A A. Bits(贪心)

题目链接：

A. 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.

Examples

input

3
1 2
2 4
1 10

output

1
3
7

题意：

在[l,r]中找一个数字使得这个数的二进制的1数目最多,如果有多种答案就输出最小的那个;

思路：

跟上次一道hdu的题目相似,按二进制位贪心,并更新l和r,如果当前为一个为0一个为1,那么就知道后面为都可以取1,还要判断这一位是否可取1,如果都是0那么就下一位,如果都是1那么就更新答案和l,r再进行下一位;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>
 
using namespace std;
 
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
 
typedef  long long LL;
 
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}
 
const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e5+20;
const int maxn=4e3+220;
const double eps=1e-12;
int n;
LL ans=0;
LL solve(LL l,LL r,int pos)
{

    int tl=((l>>pos)&1),tr=((r>>pos)&1);//cout<<l<<" "<<r<<" "<<pos<<" "<<tl<<" "<<tr<<endl;
    if(tl==0&&tr==1)
    {
            if(pos==0)ans+=1;
            else 
            {
                if(r!=(1LL<<(pos+1))-1)ans+=(1LL<<pos)-1;
                else ans+=r;
            }
    }
    else if(tl==0&&tr==0&&pos)solve(l,r,pos-1);
    else if(tl==1&&tr==1)
    {
        ans+=(1LL<<pos);
        if(pos)solve(l-(1LL<<pos),r-(1LL<<pos),pos-1);
    }
}
int main()
{
    read(n);
    LL l,r;
    For(i,1,n)
    {
        ans=0;
        read(l);read(r);
        solve(l,r,63);
        print(ans);
    }

    
    return 0;
}