Codeforces Round #169 (Div. 2) D. Little Girl and Maximum XOR（贪心，中等）

D. Little Girl and Maximum XOR

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 test(s)

Input

1 2

Output

3

Input

8 16

Output

31

Input

1 1

Output

0

思路简述：对于l和r，先化为对应的二进制数，自高位向下逐位比较，不同不需变化，相同则若同为1，r的此位变为0（如果是l的此位变为0会
使得得到的数小于l），同为0则l的此位变为1（若r的此位变为1，则得到的数大于r）。

#include <iostream>
#include <string>
#include <set>
#include <map>
#include <vector>
#include <stack>
#include <queue>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
#define LL long long
#define cti const int
#define ctll const long long
#define dg(i) cout << "*" << i << endl;

LL l, r;
char sl[65], sr[65];

int NumToStr(char *s, LL x)
{
    int i;
    for(i = 0; x != 1; i++)
    {
        s[i] = (x & 1) + '0';
        x >>= 1;
    }
    s[i] = '1';
    return i;
}

LL StrToNum(char *s)
{
    LL x = 0;
    int i, j;
    for(i = 63; s[i] == '0'; i--){}
    for(j = 0; j <= i; j++)
        if(s[j] == '1') x += ((LL)1 << j);
    return x;
}

int main()
{
    int i, j;
    while(scanf("%I64d %I64d", &l, &r) != EOF)
    {
        if(l == r)
        {
            puts("0");
            continue ;
        }
        memset(sl, '0', sizeof(sl));
        memset(sr, '0', sizeof(sr));
        int lenl = NumToStr(sl, l);
        int lenr = NumToStr(sr, r);
        int tag;
        if(lenr > lenl) tag = 1;
        else tag = 0;
        for(i = lenr - 1; tag != 2 && i >= 0; i--)
        {
            if(sr[i] > sl[i]) tag = 1;
            if(tag && sr[i] == sl[i]) tag = 2;
        }
        if(tag != 2) printf("%I64d\n", l^r);
        else
        {
            for(j = i + 1; j >= 0; j--)
            {
                if(sl[j] == sr[j])
                {
                    if(sl[j] == '1') sr[j] = '0';
                    else sl[j] = '1';
                }
            }
            printf("%I64d\n", StrToNum(sl)^StrToNum(sr));
        }
    }
    return 0;
}