整整10个月后第二次搞这个问题才搞懂........第一次还是太随意了。
解题思路：
经过打表可得规律答案要么是0 要么是2的N次 - 1
要得到最大的XOR值，其值一定是2的N次 - 1
即在 l 和 r 的二进制中，从左到右遍历过去，如果碰到 (2 ^ i) & l 为 1 , (2 ^ i) & r 为 0
即在 l 和 r 之间一定存在 形如 10+ 和01+这样的数。
则可说明在[l , r]中存在 1000000000 和 0111111111 可得到最大XOR值为2的N次 - 1
PS：不会存在首先出现 l 为 0 r 为 1 的情况，因为 l < r
//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler #include <stdio.h> #include <iostream> #include <fstream> #include <cstring> #include <cmath> #include <stack> #include <string> #include <map> #include <set> #include <list> #include <queue> #include <vector> #include <algorithm> #define Max (a,b) (((a) > (b)) ? (a) : (b)) #define Min (a,b) (((a) < (b)) ? (a) : (b)) #define Abs (x) (((x) > 0) ? (x) : (-(x))) #define MOD 1000000007 #define pi acos(-1.0) using namespace std; typedef long long ll ; typedef unsigned long long ull ; typedef unsigned int uint ; typedef unsigned char uchar ; template <class T> inline void checkmin (T &a,T b) { if (a > b) a = b; } template <class T> inline void checkmax (T &a,T b) { if (a < b) a = b; } const double eps = 1e-7 ; const int N = 1 ; const int M = 200000 ; const ll P = 10000000097ll ; const int INF = 0x3f3f3f3f ; int main(){ int i, j, k, t, n, m, numCase = 0; ll l, r; while (cin >> l >> r){ for (i = 63; i >= 0; --i){ if ((l & (1LL << i)) ^ (r & (1LL << i))) break; } ll ans = pow (2, i + 1) - 1; cout << ans << endl; } return 0; }
题目描述
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».
输入
The input consists of multiple test cases.
Each test case contains space-separated integers l and r (1 ≤ l ≤ r ≤ 1018).
输出
In a single line print a single integer — the maximum value of for all pairs of integers a, b (l ≤ a ≤ b ≤ r).
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题目描述
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».
输入
The input consists of multiple test cases.
Each test case contains space-separated integers l and r (1 ≤ l ≤ r ≤ 1018).
输出
In a single line print a single integer — the maximum value of for all pairs of integers a, b (l ≤ a ≤ b ≤ r).