CodeForces 276D – Little Girl and Maximum XOR 贪心

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 $\text{[math]}$ 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 $\text{[math]}$ 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 $\text{[math]}$ for all pairs of integers a, b (l ≤ a ≤ b ≤ r).

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