codeforces C. Bits 贪心

偏执的太偏执、 2022-07-24 09:16 201阅读 0赞

C. Bits

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Let's denote as ![c097ecd8556f4f8d61d8b143ef6afbc7928fce0b.png][] 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 ![c097ecd8556f4f8d61d8b143ef6afbc7928fce0b.png][]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 *l**i*, *r**i* — the arguments for the corresponding query (0 ≤ *l**i* ≤ *r**i* ≤ 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

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: 给定l和r, 求l和r之间的数中二进制中1的个数最多的数是哪一个, 如果有多个答案, 输出最小的那个数.

分析: 我们可以贪心的一步一步的将l的最低位0 变成1, 因为要使1的个数最多, 直到l再取大于r的时候就不能取了,这时候l的二进制中1的数目最多, 并且是最小的.

#include<bits/stdc++.h>
    
    #define inf 0x3f3f3f3f
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    const int N=100010,MOD=1e9+7;
    
    
    int main()
    {
        int n;
        cin>>n;
        while(n--)
        {
            ll l,r;
            cin>>l>>r;
            ll ans=l;
            while(l<=r)
            {
                ans = l;
                for(int i=0;i<61;i++){
                    if(!(l&(1LL<<i))){
                        l |= 1LL<<i;
                        break;
                    }
                }
            }
            cout<<ans<<endl;
        }
        return 0;
    }

#include<bits/stdc++.h>
    
    #define inf 0x3f3f3f3f
    using namespace std;
    
    typedef long long ll;
    typedef pair<int,int> pii;
    
    const int N=100010,MOD=1e9+7;
    
    
    int main()
    {
        int n;
        cin>>n;
        while(n--)
        {
            ll l,r;
            cin>>l>>r;
            ll ans=l;
            while(l<=r)
            {
                ans = l;
                ll m = (~l)&(-(~l));
                l += m;
            }
            cout<<ans<<endl;
        }
        return 0;
    }

[c097ecd8556f4f8d61d8b143ef6afbc7928fce0b.png]: /images/20220722/2d62bf3761d148baa974a2acc31b5249.png