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：
看起来很难其实想清楚了就很水的一题，就是求数组x，y之间的数异或的最大值
思路：
最后找规律可以发现只要把x，y都转化为二进制，从高位往低位找到第一个“1 0”（顺序不能反）情况，记录下位置，然后输出pow（2，tag+1）-1就行了
原理是可以把该位后面的数字改成每一位是0 1或者1 0，也就是异或一定为1,
代码：
#include<iostream>
#include<cstring>
#include<stdio.h>
#include<math.h>
#include<string>
#include<stdio.h>
#include<queue>
#include<stack>
#include<map>
#include<vector>
#include<deque>
#include<algorithm>
#define ll long long
#define INF 1008611111
#define M (t[k].l+t[k].r)/2
#define lson k*2
#define rson k*2+1
using namespace std;
int main()
{int a[105],b[105],c[105],len1=0,len2=0;memset(a,0,sizeof(a));memset(b,0,sizeof(b));ll l,r,i,j;scanf("%lld%lld",&l,&r);if(l==r){printf("0\n");return 0;}while(l!=0){a[len1]=l%2;l/=2;len1++;}while(r!=0){b[len2]=r%2;r/=2;len2++;}int tag=0;for(i=max(len1,len2)-1;i>=0;i--){if(b[i]==1&&a[i]==0){tag=i;break;}}printf("%lld\n",(ll)pow(2,tag+1)-1);return 0;
}