Codeforces Round #456 B. New Year's Eve

最新推荐文章于 2024-05-07 09:13:24 发布

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本文探讨了一个有趣的问题：如何从编号1到n的不同甜度糖果中选取不超过k个糖果，使得这些糖果甜度的异或和最大。文章给出了具体的解决思路与实现代码。

Since Grisha behaved well last year, at New Year's Eve he was visited by Ded Moroz who brought an enormous bag of gifts with him! The bag contains n sweet candies from the good ol' bakery, each labeled from 1 to n corresponding to its tastiness. No two candies have the same tastiness.

The choice of candies has a direct effect on Grisha's happiness. One can assume that he should take the tastiest ones — but no, the holiday magic turns things upside down. It is the xor-sum of tastinesses that matters, not the ordinary sum!

A xor-sum of a sequence of integers a1, a2, ..., am is defined as the bitwise XOR of all its elements: $\text{[math]}$ , here $\text{[math]}$ denotes the bitwise XOR operation; more about bitwise XOR can be found here.

Ded Moroz warned Grisha he has more houses to visit, so Grisha can take no more than k candies from the bag. Help Grisha determine the largest xor-sum (largest xor-sum means maximum happiness!) he can obtain.

Input

The sole string contains two integers n and k (1 ≤ k ≤ n ≤ 1018).

Output

Output one number — the largest possible xor-sum.

   Examples
 
     input
   
4 3

     output
   
7

     input
   
6 6

     output
   
7

Note

In the first sample case, one optimal answer is 1, 2 and 4, giving the xor-sum of 7.

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

这题真是醉了，我无语，把不超过k个翻译成了一定要k个，之前一直做不出来，看清题目后立马会

如果只呢去一家的话，肯定是n最大，因为Max=0^Max

如果大于等于2那么就是它的所有位都为1，这个也是很好证明的

因为最大那个最高位肯定是1，那么它其它是0的只要异或1就ok，那么你一定能在1-n这范围里找到这样一个数，因为二进制进位

#include<cstdio>
using namespace std;
long long f[61]={1};
int main(){
    for(int i=1;i<=60;i++)
        f[i]=f[i-1]*2;
    long long n,k;
    scanf("%I64d%I64d",&n,&k);
    if(k==1)
        printf("%I64d\n",n);
    else{
        int i;
        for(i=0;i<60;i++)
            if(f[i]<=n&&n<f[i+1])
                break;
        printf("%I64d\n",f[i+1]-1);
    }
    return 0;
}

在这里我想问一下各位大佬，如果一定要去k家，那么要怎么办？