A. Bits
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n个询问，每次询问你l,r之间的数，在二进制下1位数最多的是哪个数
题解：我们从l向上构造，在0位补齐，贪心从小位到大位的补齐，一定是最优
///1085422276
#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std ;
typedef long long ll;
typedef unsigned long long ull;
#define mem(a) memset(a,0,sizeof(a))
#define pb push_back
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<‘0‘||ch>‘9‘){
if(ch==‘-‘)f=-1;ch=getchar();
}
while(ch>=‘0‘&&ch<=‘9‘){
x=x*10+ch-‘0‘;ch=getchar();
}return x*f;
}
//****************************************
const int N=6000+50;
#define mod 10000007
#define inf 1000000001
#define maxn 10000
int d[5000];
ll test(ll x,ll y) {
for(int i=0;i<63;i++) {
if(!(x&(1ll<<i))&&x+(1ll<<i)<=y)
x+=(1ll<<i);
}
return x;
}
int main() {
int n=read();ll l,r;
for(int i=1;i<=n;i++) {
cin>>l>>r;
cout<<test(l,r)<<endl;
}
return 0;
}
代码