Codeforces 276D. Little Girl and Maximum XOR（模拟）-蒲公英云

Codeforces 276D. Little Girl and Maximum XOR（模拟）

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 4044ca87d9d7298f84b83138e9d1049c_v_1508041262 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 b6c25253add7f19654760f9ea40fe9c0_v_1508041262 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 4044ca87d9d7298f84b83138e9d1049c_v_1508041262 for all pairs of integers a, b (l ≤ a ≤ b ≤ r).

Example

Input

1 2

Output

Input

8 16

Output

Input

1 1

Output

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;
}