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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.
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).
For each query print the answer in a separate line.
3
1 2
2 4
1 10
1
3
7
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; }
Codeforces Round #276 (Div. 1) A. Bits 贪心
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原文地址：http://www.cnblogs.com/zxhl/p/4965099.html
