D. Little Girl and Maximum XOR
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
题意: 在区间[l,r]中找出两个数a,b(a <= b),使得a ^ b达到最大;
题解:(1) 首先有这样一个现象,当2 ^ n <= r的n达到最大时,且2 ^ n - 1 >= l ,答案是(2 ^ n + 2 ^ n - 1),即其二进制数全部为一,且为异或的最大值,没法再大,因为2 ^ n已达到
极限; 另一种情况是当2^n < l 时就贪心处理. 从高位往低位枚举,对于当前为如果二进制数为1,比l还小的话,则置为1,如果当前值大于r的话,则置为0; 否则则后面的位符合
(1)说法;
AC代码:
#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-8
using 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;
}