D. Little Girl and Maximum XOR (greedy)

D. Little Girl and Maximum XOR

A little girl loves problems on bitwise operations very much. Here ' s one of them.

You are given the integers l and R. Let's consider the values of for all pairs of integersa 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 integersx and y. The given operation exists in all modern programming languages, for example, in languagesC+ + and Java it is represented as "^", inPascal -as? xor?.

Input

The single line contains space-separated integersl and R (1?≤? L? ≤? r≤?10).

%lld specifier to read or write 64-bit integers inс++. It is preferred to use thecin, cout streams or the%i64d specifier.

Output

In a single line print a single integer-the maximum value of pairs of integersa, b(l ? ≤? a? ≤? b? ≤? r).

Sample Test (s)Input

1 2

Output

3

Input

8 16

Output

31

Input

1 1

Output

0 Test instructions: In the interval [l,r] Find two number A, a (a <= b), so that a ^ B to reach the maximum; solution: (1) First there is such a phenomenon, when the 2 ^ n <= r of N reaches the maximum, and 2 ^ n-1 >= L, the answer is (2 ^ n + 2 ^ n-1), that is, the second binary number is all one, and the maximum value of XOR, can not be large, because 2 ^ n has reached the limit; Another case is greedy when 2^n < L. From high to low enumeration, for the current if the binary number is 1, smaller than L, then set to 1, if the current value is greater than R, then set to 0; Otherwise, the following bits meet (1); AC Code:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #include <bitset > #include <algorithm> #include <vector> #define Lson l,mid,u << 1#define rson mid + 1,r,u << 1 | 1#define ls u << 1#define rs u << 1 |    1#define Exp 1e-8using namespace Std;typedef long long ll;int main () {ll l,r;        while (Cin>>l>>r) {ll maxn = 0;            if (L = = r) {puts ("0");        Continue        } ll i = 0,ans;            while (true) {ans = 1LL << i;                if (Ans > R) {ans = 1LL << (i-1);            Break        } i++;        } i--;        if (Ans >= l && ans-1 >= l) {Cout<<ans + ans-1<<endl;                } else {while (true) {ll res = ans + (1LL << (i-1));                if (res <= l) ans + = 1LL << (i-1); else if (res <= r) {ans = 0;                    for (ll j = 0; J < i; J + +) {ans + = 1LL << J;                } break;            } i--;        } cout<<ans<<endl; }} return 0;}

D. Little Girl and Maximum XOR (greedy)