Codeforces Round #456 (Div. 2) B题
B. New Year's Eve
time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
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: , here 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.
Input
Output
思路:
题意:从1~n中选出k个数,使得它们的异或值最大
解题思路:可以发现从了k等于1的时候答案为n,否则一定有使n的所有位为1的数。
AC代码:
#include<bits/stdc++.h> using namespace std; #define int unsigned long long signed main(){
int n,k;
cin>>n>>k;
if(k==){
cout<<n;return ;
}
int cnt=;
while(n){
cnt++;
n/=;
}
cout<<(int)(pow(,cnt))-;
return ;
}
/*
1 0
2 1
4 2
8 3
16 4 */
最新文章
- Linux编程 ---- dup函数
- Android多线程文件下载
- Final-阶段站立会议1
- hdu 3635 Dragon Balls(并查集)
- M1分数分配
- [Jquery] js验证手机号
- Delphi在win7/vista下写注册表等需要管理员权限的解决方案
- HDU 1787 GCD Again
- sql server 的约束的作用 于 理解 [转]
- Wannafly模拟赛5 A	Split 暴力 二分加速
- 设计模式之 - 策略模式(Strategy Pattern)
- MongoDB- 简单操作命令
- sqlite比较时间秒
- _ZNote_Qt_按钮增加图片
- dispatchEvent(AWTEvent) 分派事件
- Android 心跳动画
- 在简历中使用STAR法则
- 【TP3.2.*】解决session过期不失效 和 设置不成功问题
- 扩展Spring切面
- isKindOfClass isMemeberOfClass 的区分