Codeforces 484A. Bits 贪心

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最新推荐文章于 2020-05-15 11:31:42 发布

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本文探讨了一种高效的方法来解决多个查询中找到非负整数x的问题，该整数满足特定的位数条件，并在给定的范围内达到最大可能的位数。通过实例分析，展示了如何通过迭代和位运算技巧，实现快速且准确地找到符合条件的最小x值。

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Let's denote as $\text{[math]}$ 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 $\text{[math]}$ 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

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

/**
 * Created by ckboss on 14-11-7.
 */
import java.util.*;

public class CF484A
{
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int T_T = in.nextInt();
        while (T_T-- > 0) {

            long l = in.nextLong(), r = in.nextLong();

            long low=0;
            while(true) {

                int mid = -1;
                long low_leve=-1;
                for (int i = 0; i <= 60; i++) {
                    long temp=(1L << i) - 1L;
                    if(temp>r) break;
                    if ( temp >= l && temp <= r) {
                        mid = i;
                    }
                    if(temp < l){
                        low_leve=temp;
                    }
                }

                if (mid != -1) {
                    System.out.println((1L << mid) - 1L+low);
                    break;
                }
                else{
                    low_leve++;
                    low+=low_leve;
                    l-=low_leve; r-=low_leve;
                }
            }
        }
    }
}