题意:给出n个数,找出两个整数a[i],a[j](i < j),使得a[i] - a[j]尽量大 从小到大枚举j,在这个过程中维护a[i]的最大值 maxai晚于ans更新, 可以看这个例子 1 8 9 10 11 正确的应该是-1 如果更早更新的话,算出来就是0 用数组来存的 #include<iostream> #include<cstdio> #include<cstring> #include <cmath> #include<stack…

Open Credit System In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are more senior than other students. The professor of such a course has found quite a number of such students who…

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=102419#problem/A In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are more senior than other students. The professor of such a c…

11078 - Open Credit System Time limit: 3.000 seconds Problem E Open Credit System Input: Standard Input Output: Standard Output In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are m…

Open Credit SystemInput: Standard Input Output: Standard Output In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are more senior than other students. The professor of such a course h…

Open Credit System Input:Standard Input Output: Standard Output In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are more senior than other students. The professor of such a course h…

题意:给定一个数k,每次计算k的平方,然后截取最高的n位,然后不断重复这两个步骤,问这样可以得到的最大的数是多少? Floyd判圈算法:这个算法用在循环问题中,例如这个题目中,在不断重复中,一定有一个不断重复的过程,就是说出现的数字一定有一个周期, 这就是一个典型的循环问题.Floyd判圈算法的思想就是,例如两个小孩在一个圆形的操场上跑步,两个人以恒定的速度从同一个点出发,然后一个小孩跑步的 速度比另一个小孩的速度要快,很显然,一定的时间后,快一点的小孩一定会追赶一个循环之后追上跑的慢的那个小孩…

题目大意:给长度N的A1.....An 求(Ai-Aj)MAX 枚举n^2 其实动态维护最大值就好了 #include<iostream> #include<cstdio> using namespace std; ],n,ans,MAX; int main() { scanf("%d",&t); while (t--) { scanf("%d",&n); ;i<=n;i++) scanf("%d",…

CALCULATOR CONUNDRUM Alice got a hold of an old calculator that can display n digits. She was bored enough to come up with the following time waster. She enters a number k then repeatedly squares it until the result overflows. When the result overflo…

题目链接:http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2544解题报告:小明有一个老式计算器,这个计算器只能显示n位数,现在小明输入一个数k,并且将这个数k一直做平方运算,当得到的结果大于n位的时候 ,这个计算器会自动取最高的n位,问这样一直进行下去能得到的最大的数是多少? 很显然,屏幕上显示的数会有一个周期,例如,当输入1 6 的时候,…

题文: You are given the task to design a lighting system for a huge conference hall. After doing a lot of calculation and sketching, you have figured out the requirements for an energy-efficient design that can properly illuminate the entire hall. Acco…

CALCULATOR CONUNDRUM   Alice got a hold of an old calculator that can display n digits. She was bored enough to come up with the following time waster. She enters a number k then repeatedly squares it until the result overflows. When the result overf…

题意:有个老式计算器,每次只能记住一个数字的前n位.现在输入一个整数k,然后反复平方,一直做下去,能得到的最大数是多少.例如,n=1,k=6,那么一次显示:6,3,9,1... 思路:这个题一定会出现循环,所以一个个模拟,遇到相同的就再之前所有数中找最大的输出即可. 怎么判断遇到相同的呢?如果装在数组里一一比较显然很慢,如果用v[k]判断k是否出现过,k范围太大空间不够用. 第一种方法是使用STL里的set来判重. #include<cstdio> #include<set> us…

https://vjudge.net/problem/UVA-11549 题意: 有一个老式计算器,只能显示n位数字,输入一个整数k,然后反复平方,如果溢出的话,计算器会显示结果的最高n位.如果一直这样做下去,能得到的最大数是多少? 思路: 这个肯定是会循环的. 比较普通的做法就是用set来判断是否出现过来终止循环. 另一个高效算法:Floyd判圈算法!! 想象一下,假设有两个小孩子在一个“可以无限向前跑”的跑道上赛跑,同时出发,但其中一个孩子的速度是另一个两倍.如果跑到是直的,跑得快的小孩永远…

这道题使用暴力解法O(n*n)会超时,那么用动态维护最大值可以优化到O(n).这种思想非常实用. #include<iostream> #include<cstdio> #include<cstdlib> #include<cstring> #include<string> #include<cmath> #include<map> #include<set> #include<vector> #i…

题意: 有一个\(base(2 \leq base \leq 6)\)进制系统,这里面的数都是整数,不含前导0,相邻两个数字不相同. 而且每个数字有一个得分\(score(1 \leq score \leq 10^9)\),得分为 相邻两个数字之差的平方之和. 给出\(base\)和\(score\),求满足条件的整数的个数 \(mod \, 2^{32}\). 分析: 首先考虑DP的做法: 设\(dp(i, j)\)表示满足当前分数为\(i\)最后一个数字是\(j\)的数字的个数. 递推就是枚…

[Link]: [Description] 你要构建一个供电系统; 给你n种灯泡来构建这么一个系统; 每种灯泡有4个参数 1.灯泡的工作电压 2.灯泡的所需的电源的花费(只要买一个电源就能供这种灯泡的所有灯泡使用); 3.灯泡的单个价格 4.灯泡的所需个数; 现在,你可以把某一些灯泡换成另外一种灯泡电压要严格更高; 然后所需的灯泡个数不变,其他的都变成另外一种电压的属性; 问你最少需要花费多少钱构建这么一个供电系统; (即买电源的钱+买灯泡的钱) [Solution] 把所有的灯泡,按照电压的大…

Fast Food Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 662 Appoint description:  System Crawler  (2015-08-27) Description   The fastfood chain McBurger owns several restaurants along a highway. Rec…

Scheduling Lectures Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 607 Appoint description:  System Crawler  (2015-08-26) Description   You are teaching a course and must cover n ( ) topics. The leng…

Matrix Chain Multiplication Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 442 Appoint description:  System Crawler  (2015-08-25) Description   Suppose you have to evaluate an expression like A*B*C*D…

 Always on the run Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 590 Appoint description:  System Crawler  (2015-08-26) Description   Screeching tires. Searching lights. Wailing sirens. Police cars…

The Tower of Babylon Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 437 Appoint description:  System Crawler  (2015-08-29) Description   Perhaps you have heard of the legend of the Tower of Babylon.…

十五 Twenty Questions Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1252 Appoint description:  System Crawler  (2015-08-25) Description   Consider a closed world and a set of features that are defined…

String Compression Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1351 Appoint description:  System Crawler  (2015-08-21) Description   Run Length Encoding(RLE) is a simple form of compression. RLE c…

控制台程序. 在这个版本的银行示例中,把借款和贷款事务创建为在不同线程中执行的任务,它们把事务提交给职员.创建事务的任务是Callable<>任务,因为它们需要返回已为每个账户创建的借款或贷款事务的总额. // Defines a customer account public class Account { // Constructor public Account(int accountNumber, int balance) { this.accountNumber = accountN…

K - 4 Values whose Sum is 0 Crawling in process... Crawling failed Time Limit:9000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1152 Appoint description: System Crawler (2015-03-12) Description   The SUM problem c…

F - Remember the Word Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1401 Appoint description:  System Crawler  (2015-03-07) Description   Neal is very curious about combinatorial problems, and now h…

In an open credit system, the students can choose any course they like, but there is a problem. Some of the students are more senior than other students. The professor of such a course has found quite a number of such students who came from senior cl…

简评:臭不要脸 AI 名单,another side of AI. 这是一个可怕的 AI 清单,上面的各种商用 AI 项目都用于一些很恶劣的目的.请大家保持警惕. 区别对待类 · HireVue - 扫描你的脸并告诉公司你是否值得雇用的应用程序[ 摘要 ] · 同性恋检测 AI - 根据一项新的研究表明,机器可以比人类具有更好的同性恋判断,人工智能可以根据他们的脸部照片准确地猜测人们是否是同性恋[ 摘要 ] · 种族主义者聊天机器人 - 微软聊天机器人叫做 Tay,它从 Twitter 学习了一…

KUANGBIN带你飞 全专题整理 https://www.cnblogs.com/slzk/articles/7402292.html 专题一 简单搜索 POJ 1321 棋盘问题    //2019.3.18 POJ 2251 Dungeon Master POJ 3278 Catch That Cow  //4.8 POJ 3279 Fliptile POJ 1426 Find The Multiple  //4.8 POJ 3126 Prime Path POJ 3087 Shuffle…