1269 - Consecutive Sum
Time Limit: 3 second(s) | Memory Limit: 64 MB |
Little Jimmy is learning how to add integers. As in decimal the digits are 0 to 9, it makes a bit hard for him to understand the summation of all pair of digits. Since addition of numbers requires the knowledge of adding digits. So, his mother gave him a software that can convert a decimal integer to its binary and a binary to its corresponding decimal. So, Jimmy's idea is to convert the numbers into binaries, and then he adds them and turns the result back to decimal using the software. It's easy to add in binary, since you only need to know how to add (0, 0), (0, 1), (1, 0), (1, 1). Jimmy doesn't have the idea of carry operation, so he thinks that
1 + 1 = 0
1 + 0 = 1
0 + 1 = 1
0 + 0 = 0
Using these operations, he adds the numbers in binary. So, according to his calculations,
3 (011) + 7 (111) = 4 (100)
Now you are given an array of n integers, indexed from 0 to n-1, you have to find two indices i j in the array (0 ≤ i ≤ j < n), such that the summation (according to Jimmy) of all integers between indices i and jin the array, is maximum. And you also have to find two indices, p q in the array (0 ≤ p ≤ q < n), such that the summation (according to Jimmy) of all integers between indices p and q in the array, is minimum. You only have to report the maximum and minimum integers.
Input
Input starts with an integer T (≤ 10), denoting the number of test cases.
Each case starts with a line containing an integer n (1 ≤ n ≤ 50000). The next line contains n space separated non-negative integers, denoting the integers of the given array. Each integer fits into a 32 bit signed integer.
Output
For each case, print the case number, the maximum and minimum summation that can be made using Jimmy's addition.
Sample Input |
Output for Sample Input |
2 5 6 8 2 4 2 5 3 8 2 6 5 |
Case 1: 14 2 Case 2: 15 1 |
Note
Dataset is huge, use faster I/O methods.
题意:求连续区间的异或的最大和最小值;
思路:trie树,贪心;
先求一遍前缀异或和;然后先将0加入tire树,0异或任何值为那个值本身,然后我们知道pre[i]^pre[j];j<i;
ans[j+1]^ans[j+2]...^ans[i];也就是区间[i,j],那么我们把前面的前缀加入tire,然后我们在树中贪心选取,如果是找最大的,就贪心选与当前位不同的,因为是按最高位
依次往下,所以我们保证高位数要尽量大,如果不存在与当前位相同,那么只能走相同的。知道咋贪心最大的,最小的和这差不多。
复杂度为(n*log(1<<31-1));
1 #include<stdio.h>
2 #include<algorithm>
3 #include<string.h>
4 #include<stdlib.h>
5 #include<iostream>
6 using namespace std;
7 typedef long long LL;
8 LL ans[500005];
9 int id[32];
10 struct node
11 {
12 node *p[2];
13 node()
14 {
15 memset(p,0,sizeof(p));
16 }
17 };
18 node *root;
19 void in(int *v)
20 {
21 int i,j,k;
22 node *c=root;
23 for(i=30; i>=0; i--)
24 {
25 int pre=v[i];
26 if(c->p[v[i]]==NULL)
27 {
28 c->p[v[i]]=new node();
29 }
30 c=c->p[v[i]];
31 }
32 }
33 LL ma(int *v)
34 {
35 LL sum=0;
36 LL l=1;
37 node*c=root;
38 int i;
39 for(i=30; i>=0; i--)
40 {
41 int pre=v[i];
42 if(c->p[(pre+1)%2]==NULL)
43 {
44 c=c->p[pre];
45 sum*=2;
46 sum+=0;
47 }
48 else
49 {
50 c=c->p[(pre+1)%2];
51 sum*=2;
52 sum+=1;
53 }
54 }
55 return sum;
56 }
57 LL mi(int *v)
58 {
59 LL sum=0;
60 LL l=2;
61 node*c=root;
62 int i;
63 for(i=30; i>=0; i--)
64 {
65 int pre=v[i];
66 if(c->p[pre]==NULL)
67 {
68 c=c->p[(pre+1)%2];
69 sum*=2;
70 sum+=1;
71 }
72 else
73 {
74 c=c->p[pre];
75 sum*=2;
76 sum+=0;
77 }
78 }
79 return sum;
80 }
81 void del(node *c)
82 {
83 for(int i=0; i<2; i++)
84 {
85 if(c->p[i]!=NULL)
86 del(c->p[i]);
87 }
88 free(c);
89 }
90 int main(void)
91 {
92 int i,j,k;
93 scanf("%d",&k);
94 int s;
95 int n,m;
96 for(s=1; s<=k; s++)
97 {
98 scanf("%d",&n);
99 for(i=1; i<=n; i++)
100 {
101 scanf("%lld",&ans[i]);
102 ans[i]^=ans[i-1];
103 }
104 root=new node();
105 LL maxx=0;
106 LL minn=1e12;
107 memset(id,0,sizeof(id));
108 in(id);
109 for(i=1; i<=n; i++)
110 {
111 memset(id,0,sizeof(id));
112 LL kk=ans[i];
113 int cnt=0;
114 while(kk)
115 {
116 id[cnt++]=kk%2;
117 kk/=2;
118 }
119 maxx=max(maxx,ma(id));
120 minn=min(minn,mi(id));
121 in(id);
122 }del(root);
123 printf("Case %d: ",s);
124 printf("%lld %lld\n",maxx,minn);
125 }return 0;
126 }
1269 - Consecutive Sum的更多相关文章
- LightOJ 1269 Consecutive Sum (Trie树)
Jan's LightOJ :: Problem 1269 - Consecutive Sum 题意是,求给定序列的中,子序列最大最小的抑或和. 做法就是用一棵Trie树,记录数的每一位是0还是1.查 ...
- LightOJ 1269 - Consecutive Sum Trie树
题意:给出一串序列,求区间连续异或值的最大和最小. 思路:如果不是出在专题里,想不到可以用字典树做.先求前缀异或值,转为二进制,加入Trie树中,如果要求最大,就是尽可能走和当前位数字相反的,这样异或 ...
- Consecutive Sum LightOJ - 1269(区间异或和)
Consecutive Sum 又来水一发blog... 本来是昨天补codechef的题,最后一道题是可持久化字典树,然后去黄学长博客看了看 觉得字典树写法有点不太一样,就想着用黄学长的板子写码几道 ...
- hdoj 1977 Consecutive sum II
Consecutive sum II Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Other ...
- HDOJ(HDU) 1977 Consecutive sum II(推导、、)
Problem Description Consecutive sum come again. Are you ready? Go ~~ 1 = 0 + 1 2+3+4 = 1 + 8 5+6+7+8 ...
- hdu 1231, dp ,maximum consecutive sum of integers, find the boundaries, possibly all negative, C++ 分类: hdoj 2015-07-12 03:24 87人阅读 评论(0) 收藏
the algorithm of three version below is essentially the same, namely, Kadane's algorithm, which is o ...
- BZOJ 2157: 旅游( 树链剖分 )
树链剖分.. 样例太大了根本没法调...顺便把数据生成器放上来 -------------------------------------------------------------------- ...
- Codeforces Round #747 (Div. 2) Editorial
Codeforces Round #747 (Div. 2) A. Consecutive Sum Riddle 思路分析: 一开始想起了那个公式\(l + (l + 1) + - + (r − 1) ...
- Codeforces Round #821(Div.2) (A-C) 题解
Codeforces Round #821(Div.2) (A-C) A.Consecutive Sum 大致题意 给定一组共 n 个数据 ,如果俩个数的下标在 mod k 意义下同余,则可以交换a[ ...
随机推荐
- 【leetcode】986. Interval List Intersections (双指针)
You are given two lists of closed intervals, firstList and secondList, where firstList[i] = [starti, ...
- final&static
final 1.final修饰类,那么该类不能有子类,那么也就没有子类重写父类的方法,也就没有多态 2.final修饰成员变量,那么成员变量要么显式赋值(用第一种),要么在构造方法中赋值 无论哪一种, ...
- 通信协议 HTTP TCP UDP
TCP HTTP UDP: 都是通信协议,也就是通信时所遵守的规则,只有双方按照这个规则"说话",对方才能理解或为之服务. TCP HTTP UDP三者的关系: T ...
- zabbix之邮件报警
创建媒介类型 如果用QQ邮箱的话,先设置一下授权码 为用户设置报警 创建一个用户 配置动作 测试
- 小程序中使用less(最优方式)
写惯了less/sass,但是现在开发小程序缺还是css,很不习惯. 在网上搜的教程,要么是gulp,要么就是vscode的Easy-less的插件. 传统方式 我们来对比,这两种方式的优劣. Gul ...
- 【Service】【Database】【MySQL】基础
1. 概念 1.1. 作者:Unireg 1.2. MySQL AB --> MySQL Solaris:二进制版本: 1.3. 官方网站: MySQL: www.mysql.com Maria ...
- 【Linux】【Basis】网络
Linux网络属性配置 计算机网络: TCP/IP:协议栈(使用) ISO,OSI:协议栈(学习) ...
- 如何优雅正确地通过interrupt方法中断线程
为什么废弃Thread的stop函数? 简单来说就是stop方法中断线程太过暴力随意,且会是否线程持有的锁,会导致线程安全问题.还有可能导致存在需要被释放的资源得不到释放,引发内存泄露.所以用stop ...
- TSN 时间敏感网络:缘起 (TSN历史与现状)
前言 随着工业物联网(IIoT)的兴起和工业4.0的提出,越来越多的设计师.工程师和最终用户关注时间敏感网络(Time-Sensitive Networking,下简称为TSN).TSN为以太网提供确 ...
- 成本计算?(Project)
<Project2016 企业项目管理实践>张会斌 董方好 编著 成本各种输入以后就该计算了是吗? 其实,计算有我什么事啊,不都是些四则运算吗?Project要是连这都搞不定,他还在地球上 ...