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?.
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.
In a single line print a single integer — the maximum value of for all pairs of integers a, b (l?≤?a?≤?b?≤?r).
1 2
3
8 16
31
1 1
0
【题解】
刚开始一看到数据范围惊呆了，但是最后仔细思考以后发现是有技巧可言的，首先，异或值要为1，那么两个数的对应2位进制位要不同，而异或值要最大，则二进制的高位要尽可能的为1，所以这就是切入点，从给定的区间上下限入手（l和R），从l和r二进制的最高位开始比较，如果出现对应位异或值位1，就从该处开始，低位都置为1，此时其表示的数就是最大异或值了。
【AC代码】
#include<cstdio>
#include<algorithm>
#include<cstdio>
#include<math.h>
#include<string.h>
#include<iostream>
using namespace std;
typedef __int64 ll;
ll l,r;int main()
{while(~scanf("%I64d%I64d",&l,&r)){ll max_xor = 0;int i = 61;for(;i>=0;--i){if(l&(1LL<<i) ^ r&(1LL<<i)) break;}for(;i>=0;--i){max_xor|= 1LL<<i;}printf("%I64d\n",max_xor);}return 0;
}