cf-282e
“字典树”的变形,任意两数异或最大值,处理字典树的时候可以用递归,也可以用循环,下面有两个版本。
C - Sausage Maximization
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64uSubmit Status
Practice CodeForces 282E
Description
The Bitlandians are quite weird people. They have their own problems and their own solutions. They have their own thoughts and their own beliefs, they have their own values and their own merits. They have their own dishes and their own sausages! In Bitland a sausage is an array of integers! A sausage's deliciousness is equal to the bitwise excluding OR (the xor operation) of all integers in that sausage. One day, when Mr. Bitkoch (the local cook) was going to close his BitRestaurant, BitHaval and BitAryo, the most famous citizens of Bitland, entered the restaurant and each ordered a sausage. But Mr. Bitkoch had only one sausage left. So he decided to cut a prefix (several, may be zero, first array elements) of the sausage and give it to BitHaval and a postfix (several, may be zero, last array elements) of the sausage and give it to BitAryo. Note that one or both pieces of the sausage can be empty. Of course, the cut pieces mustn't intersect (no array element can occur in both pieces). The pleasure of BitHaval and BitAryo is equal to the bitwise XOR of their sausages' deliciousness. An empty sausage's deliciousness equals zero. Find a way to cut a piece of sausage for BitHaval and BitAryo that maximizes the pleasure of these worthy citizens.
Input
The first line contains an integer n (1 ≤ n ≤ 105). The next line contains n integers a1, a2, ..., an (0 ≤ ai ≤ 1012) — Mr. Bitkoch's sausage. Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64dspecifier.
Output
Print a single integer — the maximum pleasure BitHaval and BitAryo can get from the dinner.
Sample Input
Input
21 2
Output
3
Input
31 2 3
Output
3
Input
21000 1000
Output
1000
#include <stdio.h>
#include <algorithm>
using namespace std;
typedef __int64 LL;
const int maxn = +;
int ch[maxn*][], cnt, root, n;
LL num[maxn], ans;
void Insert(LL tar)
{
int cur = root;
for(int i = ; i >= ; i--)
{
int id = ((tar & (1LL<<(i-))) ? : );
if(ch[cur][id] == -)
{
ch[cnt][] = ch[cnt][] = -;
ch[cur][id] = cnt++;
}
cur = ch[cur][id];
}
}
void Find(LL tar)
{
int cur = root;
LL ret = ;
for(int i = ; i >= ; i--)
{
int id = ((tar & (1LL<<(i-))) ? : );
if(ch[cur][id^] != -)
{
ret |= (1LL << (i-));
cur = ch[cur][id^];
}
else
cur = ch[cur][id];
}
ans = max(ans, ret);
}
int main()
{
LL pre, suf;
while(scanf("%d", &n) != EOF)
{
pre = suf = cnt = ;
ch[cnt][] = ch[cnt][] = -;
root = cnt++;
Insert(0LL);
for(int i = ; i < n; i++)
{
scanf("%I64d", &num[i]);
suf ^= num[i];
}
ans = suf;
for(int i = ; i < n; i++)
{
pre ^= num[i];
suf ^= num[i];
Insert(pre);
Find(suf);
}
printf("%I64d\n", ans);
}
return ;
}
/*
#include <stdio.h>
#include <algorithm>
using namespace std;
typedef __int64 LL;
const int maxn = 100000+100;
int ch[maxn*40][2], cnt, root, n;
LL num[maxn], ans;
void Insert(LL tar)
{
int cur = root;
for(int i = 40; i >= 1; i--)
{
int id = ((tar & (1LL<<(i-1))) ? 1 : 0);
if(ch[cur][id] == -1)
{
ch[cnt][0] = ch[cnt][1] = -1;
ch[cur][id] = cnt++;
}
cur = ch[cur][id];
}
}
void Find(LL tar)
{
int cur = root;
LL ret = 0;
for(int i = 40; i >= 1; i--)
{
int id = ((tar & (1LL<<(i-1))) ? 1 : 0);
if(ch[cur][id^1] != -1)
{
ret |= (1LL << (i-1));
cur = ch[cur][id^1];
}
else
cur = ch[cur][id];
}
ans = max(ans, ret);
}
int main()
{
LL pre, suf;
while(scanf("%d", &n) != EOF)
{
pre = suf = cnt = 0;
ch[cnt][0] = ch[cnt][1] = -1;
root = cnt++;
Insert(0LL);
for(int i = 0; i < n; i++)
{
scanf("%I64d", &num[i]);
suf ^= num[i];
}
ans = suf;
for(int i = 0; i < n; i++)
{
pre ^= num[i];
suf ^= num[i];
Insert(pre);
Find(suf);
}
printf("%I64d\n", ans);
}
return 0;
}
#include <stdio.h> typedef __int64 LL;
const int maxn = + ;
int memory[maxn*][], allocp, root, n;
LL num[maxn], ans;
void Insert(LL tar)
{
LL cur = root;
for(int i = ; i >= ; i--) {
int k = (((1LL<<i) & tar)?:);
if(memory[cur][k] == -) {
memory[allocp][] = memory[allocp][] = -;
memory[cur][k] = allocp++;
}
cur = memory[cur][k];
}
}
void Find(LL x)
{
LL ret = ;
int cur = root;
for(int i = ; i >= ; i--) {
int k = (x & (1LL<<i))?:;
if(memory[cur][k^] != -) {
ret |= (1LL << i);
cur = memory[cur][k^];
} else {
cur = memory[cur][k];
}
}
ans = (ans>ret)?ans:ret;
}
int main()
{
LL pre, suf;
while(~scanf("%d", &n)) {
pre = suf = allocp = ;
memory[allocp][] = memory[allocp][] = -;
root = allocp++;
Insert(0LL);
for(LL i = ; i < n; i++) {
scanf("%I64d", &num[i]);
suf ^= num[i];
}
ans = suf;
for(int i = ; i < n; i++)
{
pre ^= num[i];
suf ^= num[i];
Insert(pre);
Find(suf);
}
printf("%I64d\n", ans);
}
return ;
}
cf-282e的更多相关文章
- ORA-00494: enqueue [CF] held for too long (more than 900 seconds) by 'inst 1, osid 5166'
凌晨收到同事电话,反馈应用程序访问Oracle数据库时报错,当时现场现象确认: 1. 应用程序访问不了数据库,使用SQL Developer测试发现访问不了数据库.报ORA-12570 TNS:pac ...
- cf之路,1,Codeforces Round #345 (Div. 2)
cf之路,1,Codeforces Round #345 (Div. 2) ps:昨天第一次参加cf比赛,比赛之前为了熟悉下cf比赛题目的难度.所以做了round#345连试试水的深浅..... ...
- cf Round 613
A.Peter and Snow Blower(计算几何) 给定一个点和一个多边形,求出这个多边形绕这个点旋转一圈后形成的面积.保证这个点不在多边形内. 画个图能明白 这个图形是一个圆环,那么就是这个 ...
- ARC下OC对象和CF对象之间的桥接(bridge)
在开发iOS应用程序时我们有时会用到Core Foundation对象简称CF,例如Core Graphics.Core Text,并且我们可能需要将CF对象和OC对象进行互相转化,我们知道,ARC环 ...
- [Recommendation System] 推荐系统之协同过滤(CF)算法详解和实现
1 集体智慧和协同过滤 1.1 什么是集体智慧(社会计算)? 集体智慧 (Collective Intelligence) 并不是 Web2.0 时代特有的,只是在 Web2.0 时代,大家在 Web ...
- CF memsql Start[c]UP 2.0 A
CF memsql Start[c]UP 2.0 A A. Golden System time limit per test 1 second memory limit per test 256 m ...
- CF memsql Start[c]UP 2.0 B
CF memsql Start[c]UP 2.0 B B. Distributed Join time limit per test 1 second memory limit per test 25 ...
- CF #376 (Div. 2) C. dfs
1.CF #376 (Div. 2) C. Socks dfs 2.题意:给袜子上色,使n天左右脚袜子都同样颜色. 3.总结:一开始用链表存图,一直TLE test 6 (1)如果需 ...
- CF #375 (Div. 2) D. bfs
1.CF #375 (Div. 2) D. Lakes in Berland 2.总结:麻烦的bfs,但其实很水.. 3.题意:n*m的陆地与水泽,水泽在边界表示连通海洋.最后要剩k个湖,总要填掉多 ...
- CF #374 (Div. 2) D. 贪心,优先队列或set
1.CF #374 (Div. 2) D. Maxim and Array 2.总结:按绝对值最小贪心下去即可 3.题意:对n个数进行+x或-x的k次操作,要使操作之后的n个数乘积最小. (1)优 ...
随机推荐
- 【C++】函数指针宏定义
看耗子叔文章学习虚函数表(http://blog.csdn.net/haoel/article/details/1948051)的时候被例子的第一句惊到了 typedef void(*Fun)(voi ...
- c#中DropDownList控件绑定枚举数据
c# asp.net 中DropDownList控件绑定枚举数据 1.枚举(enum)代码: private enum heros { 德玛 = , 皇子 = , 大头 = , 剑圣 = , } 如果 ...
- logback详细配置(三)
转自:http://blog.csdn.net/haidage/article/details/6794540 <filter>: 过滤器,执行一个过滤器会有返回个枚举值,即DENY,NE ...
- Failed to load PDF in chrome/Firefox/IE
笔者的公司搭建了一个Nexus服务器,用来管理我们自己的项目Release构件和Site文档. 今天的问题是当用户访问一个Site里的PDF文件的时候,报错说“detected that the ne ...
- 源码搭建SVN+Apache+Setpass
1.安装配置apache2.2.18 http://download.csdn.net/download/YH555/3299526tar xf httpd-2.2.18.tar.bz2cd http ...
- 清除Outlook 2013中缓存的邮件地址
1.删除相关文件(可能会没有访问权限): 路径:C:\Documents and Settings\user\用户名\Application Data\Microsoft\Outlook 文件名:ou ...
- axis2 WebService的发布与调用
1:准备: JDK:http://www.oracle.com/technetwork/java/javase/downloads/jdk6downloads-1902814.html e ...
- AWK只打印某个域后的所有域
如转载请指明(博客http: yangzhigang cublog cn).前言:有时我们需要将某个域之后的所有域打印出来,而且每个记录(行)的域的个数也不一定,所以用$4,$5,… $n,… $(N ...
- TRUNCATE引起CPU异常上涨
13:05 2015/9/11 午睡醒来收到几封CPU使用率预警邮件.登录对应服务器,打开资源监视器,看到sqlservr.exe进程的CPU达到40%(平常服务器CPU消耗在10%以内).查看CPU ...
- sqlplus登陆
cd \sqlplus sys@test_id as sysdba 切换用户SQL> connect system@test_id