地址:http://codeforces.com/contest/768/problem/C

题目:

C. Jon Snow and his Favourite Number
time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Jon Snow now has to fight with White Walkers. He has n rangers, each of which has his own strength. Also Jon Snow has his favourite number x. Each ranger can fight with a white walker only if the strength of the white walker equals his strength. He however thinks that his rangers are weak and need to improve. Jon now thinks that if he takes the bitwise XOR of strengths of some of rangers with his favourite number x, he might get soldiers of high strength. So, he decided to do the following operation k times:

  1. Arrange all the rangers in a straight line in the order of increasing strengths.
  2. Take the bitwise XOR (is written as ) of the strength of each alternate ranger with x and update it's strength.

Suppose, Jon has 5 rangers with strengths [9, 7, 11, 15, 5] and he performs the operation 1 time with x = 2. He first arranges them in the order of their strengths, [5, 7, 9, 11, 15]. Then he does the following:

  1. The strength of first ranger is updated to , i.e. 7.
  2. The strength of second ranger remains the same, i.e. 7.
  3. The strength of third ranger is updated to , i.e. 11.
  4. The strength of fourth ranger remains the same, i.e. 11.
  5. The strength of fifth ranger is updated to , i.e. 13.

The new strengths of the 5 rangers are [7, 7, 11, 11, 13]

Now, Jon wants to know the maximum and minimum strength of the rangers after performing the above operations k times. He wants your help for this task. Can you help him?

Input

First line consists of three integers nkx (1 ≤ n ≤ 105, 0 ≤ k ≤ 105, 0 ≤ x ≤ 103) — number of rangers Jon has, the number of times Jon will carry out the operation and Jon's favourite number respectively.

Second line consists of n integers representing the strengths of the rangers a1, a2, ..., an (0 ≤ ai ≤ 103).

Output

Output two integers, the maximum and the minimum strength of the rangers after performing the operation k times.

Examples
input
5 1 2
9 7 11 15 5
output
13 7
input
2 100000 569
605 986
output
986 605

思路:因为每个数和异或后的数都很小,小于1024。所以可以用计数排序的思想来做。O(k*1024)。

  这题找循环节也可以做。

  

 #include <bits/stdc++.h>

 using namespace std;

 #define MP make_pair
#define PB push_back
typedef long long LL;
typedef pair<int,int> PII;
const double eps=1e-;
const double pi=acos(-1.0);
const int K=1e6+;
const int mod=1e9+; int n,k,x,sum[],tmp[]; int main(void)
{
cin>>n>>k>>x;
for(int i=,y;i<n;i++)
scanf("%d",&y),sum[y]++;
for(int i=;i<=k;i++)
{
for(int j=,t=;j<;t+=sum[j],j++)
if(t&)
{
tmp[j]+=(sum[j]+)/;
tmp[j^x]+=sum[j]/;
}
else
{
tmp[j^x]+=(sum[j]+)/;
tmp[j]+=sum[j]/;
}
memcpy(sum,tmp,sizeof(tmp));
memset(tmp,,sizeof(tmp));
}
for(int i=,ff=;i>=&&ff;i--)
if(sum[i]) printf("%d ",i),ff=;
for(int i=,ff=;i<=&&ff;i++)
if(sum[i]) printf("%d\n",i),ff=;
return ;
}

Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) C - Jon Snow and his Favourite Number的更多相关文章

  1. Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) D. Jon and Orbs

    地址:http://codeforces.com/contest/768/problem/D 题目: D. Jon and Orbs time limit per test 2 seconds mem ...

  2. Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) A. Oath of the Night's Watch

    地址:http://codeforces.com/problemset/problem/768/A 题目: A. Oath of the Night's Watch time limit per te ...

  3. Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) B. Code For 1

    地址:http://codeforces.com/contest/768/problem/B 题目: B. Code For 1 time limit per test 2 seconds memor ...

  4. Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined)

    C题卡了好久,A掉C题之后看到自己已经排在好后面说实话有点绝望,最后又过了两题,总算稳住了. AC:ABCDE Rank:191 Rating:2156+37->2193 A.Oath of t ...

  5. Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) A B 水 搜索

    A. Oath of the Night's Watch time limit per test 2 seconds memory limit per test 256 megabytes input ...

  6. 【博弈论】【SG函数】【找规律】Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) E. Game of Stones

    打表找规律即可. 1,1,2,2,2,3,3,3,3,4,4,4,4,4... 注意打表的时候,sg值不只与剩下的石子数有关,也和之前取走的方案有关. //#include<cstdio> ...

  7. 【概率dp】Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) D. Jon and Orbs

    直接暴力dp就行……f(i,j)表示前i天集齐j种类的可能性.不超过10000天就能满足要求. #include<cstdio> using namespace std; #define ...

  8. 【基数排序】Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) C. Jon Snow and his Favourite Number

    发现值域很小,而且怎么异或都不会超过1023……然后可以使用类似基数排序的思想,每次扫一遍就行了. 复杂度O(k*1024). #include<cstdio> #include<c ...

  9. 【找规律】Divide by Zero 2017 and Codeforces Round #399 (Div. 1 + Div. 2, combined) B. Code For 1

    观察一下,将整个过程写出来,会发现形成一棵满二叉树,每一层要么全是0,要么全是1. 输出的顺序是其中序遍历. 每一层的序号形成等差数列,就计算一下就可以出来每一层覆盖到的区间的左右端点. 复杂度O(l ...

随机推荐

  1. jQuery实现高亮显示网页关键词的方法

    本文实例讲述了jQuery实现高亮显示网页关键词的方法.分享给大家供大家参考.具体如下: 这是一款基于jquery实现的高亮显示网页上搜索关键词的代码,当你在文本框中输入的时候,如果下面的正文中包括你 ...

  2. 20 个常用的 CSS 技巧

    1. 黑白图像 这段代码会让你的彩色照片显示为黑白照片,是不是很酷? img.desaturate {    filter: grayscale(100%);    -webkit-filter: g ...

  3. 系统之间通讯方式—SOAP(web service)

    [web service 简介] 是一种新的web应用程序分支,是自包含.自描述.模块化的应用,可以发布.定位.通过web调用.web service是一个应用组件,它逻辑性的为其他应用程序提供数据与 ...

  4. box-sizing在什么情况下会失效

    众所周知,box-sizing是将w3c盒模型与IE盒模型相互转换的利器,但是有时候也会失效,那么在什么情况下回失效呢,MD在没设置高度的时候回失效的透透的,所以一定记得需要转换的时候设置个高度!!!

  5. Http协议原理解析第一篇

    一:http的由来: OSI模型把网络通信分成七层:物理层.数据链路层.网络层.传输层.会话层.表示层和应用层,对于开发网络应用人员来说,一般把网络分成五层,这样比较容易理解.这五层为:物理层.数据链 ...

  6. 【BZOJ2287】【POJ Challenge】消失之物 背包动规

    [BZOJ2287][POJ Challenge]消失之物 Description ftiasch 有 N 个物品, 体积分别是 W1, W2, ..., WN. 由于她的疏忽, 第 i 个物品丢失了 ...

  7. 【BZOJ3442】学习小组 费用流

    [BZOJ3442]学习小组 Description [背景] 坑校准备鼓励学生参加学习小组. [描述] 共有n个学生,m个学习小组,每个学生有一定的喜好,只愿意参加其中的一些学习小组,但是校领导为学 ...

  8. nvm-windows 安装后,node 命令报错

    下载地址: https://github.com/coreybutler/nvm-windows/releases 我是点下面这个直接安装的: 配置文件,我都是习惯在软件目录下建个 files 文件夹 ...

  9. 160317(二)、按sort排序,sort为空的在后面

    按sort排序,sort为空的在后面 select * from 表名 order by (case when sort is null or sort='' then 1 else 0 end),s ...

  10. Java代码规范、基本类型和实例演练

    1.代码分段 当一个方法内部的代码超过7行时,就要考虑分成段落. 使用空行分隔代码 按照代码的功能进行分段 最终效果是一眼就能在宏观上把握代码的结构 (1)举例 Card 分成2段 第一段:定义变量保 ...