链接:http://acm.hdu.edu.cn/showproblem.php?pid=5536

题面;

Chip Factory

Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 6277    Accepted Submission(s): 2847

Problem Description
John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage large amounts of products, every processor has a serial number. More specifically, the factory produces n

chips today, the i

-th chip produced this day has a serial number si

.

At the end of the day, he packages all the chips produced this day, and send it to wholesalers. More specially, he writes a checksum number on the package, this checksum is defined as below:

maxi,j,k(si+sj)⊕sk

which i,j,k

are three different integers between 1

and n

. And ⊕

is symbol of bitwise XOR.

Can you help John calculate the checksum number of today?

 
Input
The first line of input contains an integer T

indicating the total number of test cases.

The first line of each test case is an integer n

, indicating the number of chips produced today. The next line has n

integers s1,s2,..,sn

, separated with single space, indicating serial number of each chip.

1≤T≤1000

3≤n≤1000

0≤si≤109

There are at most 10

testcases with n>100

 
Output
For each test case, please output an integer indicating the checksum number in a line.
 
Sample Input
2
3
1 2 3
3
100 200 300
 
Sample Output
6
400
 

模板题

Source
 
实现代码;
#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int M = 1e3+;
int tot;
int ch[*M][],vis[*M];
ll val[*M],a[M]; void init(){
memset(vis,,sizeof(vis));
tot = ;
ch[][] = ch[][] = ;
} void ins(ll x){
int u = ;
for(int i = ;i >= ;i --){
int v = (x>>i)&;
if(!ch[u][v]){
ch[tot][] = ch[tot][] = ;
val[tot] = ;
vis[tot] = ;
ch[u][v] = tot++;
}
u = ch[u][v];
vis[u]++;
}
val[u] = x;
} void update(ll x,int c){
int u = ;
for(int i = ;i >= ;i --){
int v = (x>>i)&;
u = ch[u][v];
vis[u] += c;
}
} ll query(ll x){
int u = ;
for(int i = ;i >= ;i --){
int v = (x>>i)&;
if(ch[u][v^]&&vis[ch[u][v^]]) u = ch[u][v^];
else u = ch[u][v];
}
return x^val[u];
} int main()
{
ios::sync_with_stdio();
cin.tie(); cout.tie();
int t,n,m;
cin>>t;
while(t--){
cin>>n;
ll mx = ;
init();
for(int i = ;i <= n;i ++)
cin>>a[i],ins(a[i]);
for(int i = ;i <= n;i ++){
for(int j = i+;j <= n;j ++){
update(a[i],-); update(a[j],-);
mx = max(mx,query(a[i]+a[j]));
update(a[i],); update(a[j],);
}
}
cout<<mx<<endl;
}
}

hdu 5536 Chip Factory (01 Trie)的更多相关文章

  1. HDU 5536 Chip Factory 【01字典树删除】

    题目传送门:http://acm.hdu.edu.cn/showproblem.php?pid=5536 Chip Factory Time Limit: 18000/9000 MS (Java/Ot ...

  2. HDU 5536 Chip Factory 字典树+贪心

    给你n个数,a1....an,求(ai+aj)^ak最大的值,i不等于j不等于k 思路:先建字典树,暴力i,j每次删除他们,然后贪心找k,再恢复i,j,每次和答案取较大的,就是答案,有关异或的貌似很多 ...

  3. HDU 5536 Chip Factory 字典树

    Chip Factory Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid= ...

  4. 2015ACM/ICPC亚洲区长春站 J hdu 5536 Chip Factory

    Chip Factory Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)T ...

  5. HDU 5536 Chip Factory

    Chip Factory Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)T ...

  6. hdu 5536 Chip Factory 字典树+bitset 铜牌题

    Chip Factory Time Limit: 18000/9000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)T ...

  7. HDU 5536 Chip Factory Trie

    题意: 给出\(n(3 \leq n \leq 1000)\)个数字,求\(max(s_i+s_j) \bigoplus s_k\),而且\(i,j,k\)互不相等. 分析: 把每个数字看成一个\(0 ...

  8. HDU 5536 Chip Factory (暴力+01字典树)

    <题目链接> 题目大意: 给定一个数字序列,让你从中找出三个不同的数,从而求出:$\max_{i,j,k} (s_i+s_j) \oplus s_k$的值. 解题分析:先建好01字典树,然 ...

  9. ACM学习历程—HDU 5536 Chip Factory(xor && 字典树)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5536 题目大意是给了一个序列,求(si+sj)^sk的最大值. 首先n有1000,暴力理论上是不行的. ...

随机推荐

  1. Vysor Pro破解助手

    Vysor更新到1.7.7版本后,原来提供的VysorPro助手无法正常破解了. 针对新版本的改动,更新了一下Vysor破解助手,支持破解Vysor 1.6.6和Vysor1.7.7之间的版本. Vy ...

  2. <4>Python切片功能剖析

    引用文章:https://mp.weixin.qq.com/s/NZ371nKs_WXdYPCPiryocw 切片基础法则: (1)公式,禁止0. (2)i, n同号:从序列的第i位索引起,向右取n- ...

  3. vulnhub writeup - 持续更新

    目录 wakanda: 1 0. Description 1. flag1.txt 2. flag2.txt 3. flag3.txt Finished Tips Basic Pentesting: ...

  4. mybatis使用oracle的nulls first/nulls last

    nulls first/nulls last 顾名思义,就是在检索结果集里,有null值的时候,把null值认为是最大值,还是最小值. nulls first 放置在结果集最前面 nulls last ...

  5. 【PAT】B1018 锤子剪刀布

    抄的柳婼小姐姐的,感觉三个数求最大那里用的真棒 #include <stdio.h> int main() { int N; scanf("%d", &N); ...

  6. .NET 术语

    .NET 术语 1. AOT 预编译器.与 JIT 类似,此编译器还可将 IL 转换为机器代码. 与 JIT 编译相比,AOT 编译在应用程序执行前进行并且通常在不同计算机上执行. 由于在运行时 AO ...

  7. WIN提权总结【本地存档-转载】

    [ web提权 ] 1.能不能执行cmd就看这个命令:net user,net不行就用net1,再不行就上传一个net到可写可读目录,执行/c c:\windows\temp\cookies\net1 ...

  8. log4j控制指定包下的日志

    最近观察日志发现如下两个问题: 1.项目用的是springboot项目,整合了rabbitmq,项目启动后,会自动监控rabbitmq谅解是否正常,导致控制台一直输出监控日志,此时就想阻止该类日志输出 ...

  9. 爬虫系列---selenium详解

    一 安装 pip install Selenium 二 安装驱动 chrome驱动文件:点击下载chromedriver (yueyu下载) 三 配置chromedrive的路径(仅添加环境变量即可) ...

  10. 第五节 matplotlib库

    一.Matplotlib基础知识 1.1Matplotlib中的基本图表包括的元素 x轴和y轴 axis水平和垂直的轴线 x轴和y轴刻度 tick刻度标示坐标轴的分隔,包括最小刻度和最大刻度 x轴和y ...