【NEEPU OJ】1042--Xor-Sum

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描述

A xor-sum of a sequence of integers a_1, a_2,…a_m is defined as the bitwise XOR of all its elements: a_1⊕a_2⊕…⊕a_m, here ⊕ denotes the bitwise XOR operation which obeys the following rules:

0⊕0=0

0⊕1=1

1⊕0=1

1⊕1=0

Your task is to choose k integers from 1 to n to get the largest xor-sum.

输入

A single line contains two integers n and k (1≤k≤n≤10^18).

输出

Print one integer — the largest xor-sum you can obtain.

输入样例 1

4 3

输出样例 1

7

输入样例 2

6 6

输出样例 2

7

提示

In the first example,one optimal choice is 1, 2 and 4, giving the xor-sum of 7.

In the second example,one can, for example, take all six integers and obtain the xor-sum of 7.

来源

NEEPU 13th ACM

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这题是道找规律题:
k = 1时，结果一定是n。
假如n=4，k=3，用p来表示二进制数的有效位，那么

  
  

   
   

    
    

     
     
1 => 001 (p==1)
2 => 010 (p==2)
3 => 011 (p==2)
4 => 100 (p==3)
     
     
    011
XOR 100
      = 111 (7)
    
    
   
   
  
  
所以结果不会超过2p+1-1。
又因为2p和2p-1都不会超过n，而且它们的xor是2p+1-1。因此，当k>1时，总能得到最大值。

代码

#include <iostream>
using namespace std;

int main(){
    long long int n,k;
    long long int i;
    int c=0;
    cin>>n>>k;
    if(n>=1&&k>=1){
        if(k==1) cout<<n;
        else{
            for(i=1;c==0;i*=2){
                if(n>=i&&n<i*2){
                    cout<<i*2-1;
                    c=1;
                }
            }
        }
    }
    return 0;
}