#include <iostream> #include <cstdio> #include <sstream> #include <cstring> #include <map> #include <set> #include <vector> #include <stack> #include <queue> #include <algorithm> #include <cma…

一.字典树描述:Trie树,即字典树,又称单词查找树或键树,是一种树形结构,是一种哈希树的变种.典型应用是用于统计和排序大量的字符串(但不仅限于字符串),所以经常被搜索引擎系统用于文本词频统计.它的优点是:最大限度地减少无谓的字符串比较,查询效率比哈希表高. Trie的核心思想是空间换时间.利用字符串的公共前缀来降低查询时间的开销以达到提高效率的目的.它有3个基本性质: 1.根节点不包含字符,除根节点外每一个节点都只包含一个字符.2.从根节点到某一节点,路径上经过的字符连接起来,为该节点对应的字…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 1555    Accepted Submission(s): 657 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Ze…

Oil Skimming Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3426    Accepted Submission(s): 1432 Problem Description Thanks to a certain "green" resources company, there is a new profitabl…

/*H E A D*/ struct Trie{ int son[maxn<<2][2]; int b[67],tot; void init(){ // memset(son,0,sizeof son); tot=0; son[0][0]=son[0][1]=0; } void insert(ll x){ int now=0; rep(i,0,32) b[i]=(x>>i)&1; rrep(i,32,0){ if(!son[now][b[i]]){ son[now][b[i…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1203 I NEED A OFFER! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33303    Accepted Submission(s): 13470 Problem Description Speakless很早就想出国,现在…

在客人能够拿到的伞与客人之间建边  跑hc就好了.... 看看别人的:https://blog.csdn.net/wall_f/article/details/8248350 #include <iostream> #include <cstdlib> #include <cstdio> #include <cstring> #include <queue> #include <cmath> using namespace std;…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=4825 题面: Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 6430    Accepted Submission(s): 2783 Problem Description Zeus 和 Prometheus 做了一个游戏,…

hdu 4825 Xor Sum(trie+贪心) 刚刚补了前天的CF的D题再做这题感觉轻松了许多.简直一个模子啊...跑树上异或x最大值.贪心地让某位的值与x对应位的值不同即可. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #define CLR(a,b) memset((a),(b),sizeof(…

点击打开链接 Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 291    Accepted Submission(s): 151 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包括了N个正整数,随后 Prometheus…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Zeus 发起M次询问,每次询问中包含一个正整数 S ,之后 Zeus 需要在集合当中找出一个正整数 K ,使得 K…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 3647    Accepted Submission(s): 1595 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Ze…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 4144    Accepted Submission(s): 1810 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Ze…

Bone Collector Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 60469    Accepted Submission(s): 25209 Problem Description Many years ago , in Teddy’s hometown there was a man who was called “Bo…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 Many years ago , in Teddy’s hometown there was a man who was called “Bone Collector”. This man like to collect varies of bones , such as dog’s , cow’s , also he went to the grave …  The bone collect…

Xor Sum 一遍A了之后大呼一声好(keng)题!debug了两小时~~~~百度之星资格赛,可以. 题意:给你一个n个元素的数组,m次查询,每次输入一个数k要求从数组中找到一个数与k异或值最大,输出这个数. 思路:因为拉的字典树专题,所以自然想到用字典树去想思路,手推了一下样例果然发现规律了,把这些数的二进制全部竖着列出来,不足高位补0,然后每次比较比较一个数,从高位开始比较,取其相反即可.于是可以用字典树把输入的数的二进制建树,查找只需转化成二进制然后再转成补码(字典树匹配),特别特别注意…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 0    Accepted Submission(s): 0 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包括了N个正整数,随后 Prometheus 将向 Zeus 发起…

Robberies Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 31769    Accepted Submission(s): 11527 Problem Description The aspiring Roy the Robber has seen a lot of American movies, and knows that…

Jim has a balance and N weights. (1≤N≤20) The balance can only tell whether things on different side are the same weight. Weights can be put on left side or right side arbitrarily. Please tell whether the balance can measure an object of weight M. In…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others) Total Submission(s): 4445    Accepted Submission(s): 652 Problem Description Zeus 和 Prometheus 做了一个游戏.Prometheus 给 Zeus 一个集合,集合中包括了N个正整数.随后 Prometheus 将向 Ze…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 4119    Accepted Submission(s): 1796 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Ze…

Xor Sum Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包括了N个正整数,随后 Prometheus 将向 Zeus 发起M次询问,每次询问中包括一个正整数 S ,之后 Zeus 须要在集合其中找出一个正整数 K ,使得 K 与 S 的异或结果最大.Prometheus 为了让 Zeus 看到人类的伟大,随即允许 Zeus 能够向人类求助.你能证明人类的智慧么? Input 输入包括若干组測试数…

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38909   Accepted: 16862 Description Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible fro…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 2403    Accepted Submission(s): 1041 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Ze…

最大报销额 Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 21113    Accepted Submission(s): 6326 Problem Description 现 有一笔经费可以报销一定额度的发票.允许报销的发票类型包括买图书(A类).文具(B类).差旅(C类),要求每张发票的总额不得超过1000元,每张发 票上,单项物品…

Xor Sum Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 132768/132768 K (Java/Others)Total Submission(s): 1786    Accepted Submission(s): 758 Problem Description Zeus 和 Prometheus 做了一个游戏,Prometheus 给 Zeus 一个集合,集合中包含了N个正整数,随后 Prometheus 将向 Zeu…

省选前刷道LCT板题(话说之前没做这道题-) CODE #include<bits/stdc++.h> using namespace std; inline void read(int &num) { char ch; int flg = 1; while(!isdigit(ch=getchar()))if(ch=='-')flg = -flg; for(num=0; isdigit(ch); num=num*10+ch-'0', ch=getchar()); num*=flg; }…

emmm-标题卡着长度上限- LCT板题-(ε=ε=ε=┏(゜ロ゜;)┛) CODE #include <cctype> #include <cmath> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; typedef long long LL; char cb[1<<15],*cs=cb,*ct=cb; #define ge…

题意:给定两个长为n的数组a和b:重新排列a和b,生成数组c,c[i]=a[i] xor b[i]:输出字典序最小的c数组. 分析:将a中的数插入一颗01字典树a中:将b中的数插入一颗01字典树b中:在trie树上查找n次,每次同时在a和b中下移一层:if 能同时走0,则同时走0:else if 能同时走1,则同时走1:else if 树a能走0&&树b能走1,则a走0.b走1:else if 树a能走1&&树b能走0,则a走1.b走0:else 向c中插入一个新数为这两个…

Description     约翰遭受了重大的损失:蟑螂吃掉了他所有的干草,留下一群饥饿的牛．他乘着容量为C(1≤C≤50000)个单位的马车,去顿因家买一些干草．  顿因有H(1≤H≤5000)包干草,每一包都有它的体积Vi(l≤Vi≤C).约翰只能整包购买, 他最多可以运回多少体积的干草呢? Input     第1行输入C和H,之后H行一行输入一个Vi． Output       最多的可买干草体积． Sample Input 7 3  //总体积为7,用3个物品来背包 2 6 5 Th…