Little Girl and Maximum XOR
Time Limit: 2000ms
Memory Limit: 262144KB
This problem will be judged on CodeForces. Original ID: 276D
64-bit integer IO format: %I64d Java class name: Any
A little girl loves problems on bitwise operations very much. Here‘s one of them.
You are given two integers l and r. Let‘s consider the values of 
for all pairs of integers a and b (l ≤ a ≤ b ≤ r). Your task is to find the maximum value among all considered ones.
Expression means applying bitwise excluding or operation to integers x and y. The given operation exists in all modern programming languages, for example, in languages C++ and Java it is represented as "^", in Pascal — as «xor».
Input
The single line contains space-separated integers l and r (1 ≤ l ≤ r ≤ 1018).
Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.
Output
In a single line print a single integer — the maximum value of for all pairs of integers a, b (l ≤ a ≤ b ≤ r).
Sample Input
Input
1 2
Output
3
Input
8 16
Output
31
Input
1 1
Output
0
1 /*
2 2016年4月22日00:05:10
3 ^ 为异或运算
4 经过打表可得规律答案要么是0 要么是2的N次 - 1
5
6 要得到最大的XOR值，其值一定是2的N次 - 1
7
8 即在l 和 r的二进制中，从左到右遍历过去如果碰到 (2 ^ i)&l 为 1 , (2 ^ i)&r 为 0
9 即在 l 和 r 之间一定存在 形如 011111111 和 100000000 这样的数。
10
11 则可说明在[l , r]中存在 011111111111 和 10000000000 可得到最大XOR值为2的N次 - 1
12
13 PS：不会存在首先出现 l 为 0 r 为 1 的情况，因为 l < r
14 */
15
16 # include <iostream>
17 # include <cstdio>
18 # include <cstring>
19 # include <algorithm>
20 # include <queue>
21 # include <vector>
22 # include <cmath>
23 # define INF 0x3f3f3f3f
24 using namespace std;
25
26 int main(void)
27 {
28 int i;
29 long long l, r, ans;
30 while (cin>>l>>r){
31 // 1<<i 相当于 2的i次方
32 // 1000 0000 0 第9位 2的8次方 1<<8
33 // 所以第64位 2的63次方 即 1<<63
34 for (i = 63; i >= 0; i--){
35 if ((l&(1LL<<i)) ^ (r&(1LL<<i))){ // 1为int类型 要转为LL
36 break;
37 }
38 }
39 ans = (1LL<<(i+1)) - 1; // 要在原来位数上加一位 然后减1
40 cout<<ans<<endl;
41 }
42
43 return 0;
44 }