Bits - CodeForces 484 A 水题

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.

Sample test(s)

input

31 22 41 10

output

137

Note

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

题意：给一个l和r的区间，找出其中二进制中1的个数最多的数，如果有多解输出最小的数。

思路：首先l和r二进制化后从最高位往下看，如果存在不一样的时候，就让这位为0，低位全部为1，有一特殊情况就是r的这位后面全是1，这个时候就全变成1即可。

AC代码如下：

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<cmath>using namespace std;typedef long long ll;ll num1[100],num2[100],num3[100];int main(){    int T,t,n,m,i,j,k,len1,len2;    ll l,r,ans,ret;    bool flag;    scanf("%d",&T);    while(T--)    {        memset(num1,0,sizeof(num1));        memset(num2,0,sizeof(num2));        scanf("%I64d%I64d",&l,&r);        len1=0;        while(l)        {            num1[++len1]=l%2;            l/=2;        }        len2=0;        while(r)        {            num2[++len2]=r%2;            r/=2;        }        for(i=len2;i>=1;i--)        {            if(num1[i]==num2[i])              num3[i]=num2[i];            else            {                num3[i]=0;                break;            }        }        flag=true;        for(k=i;k>0;k--)           if(num2[k]==0)             flag=false;        for(k=i;k>0;k--)           num3[k]=1;        if(!flag)          num3[i]=0;        ans=0;        for(i=len2;i>=1;i--)           ans=ans*2+num3[i];        printf("%I64d\n",ans);    }}