Who am I? Coming October 18, 2016! 我是谁?2016.10.18 拭目以待! Don't worry. You will be a wow. Don't worry. I will be a wow and I will walk on my new road towards a splendid expert in industrial control. How many roads must a man walk down, before you call…

I. Illegal or Not? time limit per test 1 second memory limit per test 512 megabytes input standard input output standard output The Schengen Agreement was signed by a number of countries to uniform many visa-related questions and to allow tourists fr…

Problem G. Garden Gathering Input file: standard input Output file: standard output Time limit: 3 seconds Memory limit: 512 megabytes Many of you may have been to St. Petersburg, but have you visited Peterhof Palace? It is a collection of splendid pa…

Problem D. Delay Time Input file: standard input Output file: standard output Time limit: 1 second Memory limit: 512 megabytes Petr and Egor are measuring the gravitational acceleration g on their physics lessons using a special device. An electromag…

Matt Pietrek October 1996 MSJ Matt Pietrek is the author of Windows 95 System Programming Secrets (IDG Books, 1995). He works at NuMega Technologies Inc., and can be reached at 71774.362@compuserve.com. QWay back in your July 1994 column, you wrote a…

October 23, 2013 - Fires and smoke in eastern China Satellite: Aqua Date Acquired: 10/12/2013 Resolutions: 1km (556.1 KB)500m (2.1 MB)250m (5.1 MB) Bands Used: 1,4,3 Credit: Jeff SchmaltzMODIS Land Rapid Response Team,NASA GSFC On October 12, 2013 th…

题目链接 思路 这个题和上个题类似,仔细推一下就知道这个题是判断是否是4的倍数 代码 #include<cstdio> #include<iostream> #define fi(s) freopen(s,"r",stdin); #define fo(s) freopen(s,"w",stdout); using namespace std; typedef long long ll; ll read() { ll x = 0,f = 1;c…

题目链接 思路 这个题思路挺巧妙的. 情况一: 首先如果这堆石子的数量是1~5,那么肯定是先手赢.因为先手可以直接拿走这些石子.如果石子数量恰好是6,那么肯定是后手赢.因为先手无论怎样拿也无法直接拿走六个石子. 情况二: 考虑继续推广,如果石子数是7~11,那么先手也能赢.因为先手可以先拿成6,然后就变成了情况1.如果石子数是12,那么一定是后手赢.因为根据上面讨论,当石子数量为6的时候,此时的先手一定输.如果石子数量为12,那么现在的人无论如何也无法拿成6,所以肯定会输. 结论 如果石子数是6…

Codechef October Challenge 2018 游记 CHSERVE - Chef and Serves 题目大意: 乒乓球比赛中,双方每累计得两分就会交换一次发球权. 不过,大厨和小厨用了另外一种规则:双方每累计得 K 分才会交换发球权.比赛开始时,由大厨发球. 给定大厨和小厨的当前得分(分别记为 P1 和 P2),请求出接下来由谁发球. 思路: \((P1+P2)\%K\)判断奇偶性即可. 代码链接 BITOBYT - Byte to Bit 题目大意: 在字节国里有三类居民…

October是一个免费,开源,自托管的基于laravel PHP框架CMS平台.在github平台上laravel应用排名第二,可以拿来研究一下.官方介绍:October是一个内容管理系统(CMS)和Web平台,其唯一目的是使您的开发工作流程简单.它诞生于对现有系统的失望.我们觉得建设网站已经成为一个令人费解和混乱的过程,让开发人员不满意.我们想把你转到更简单的一边,回到基础.下面我们随ytkah来安装测试一下 1.环境需求 PHP version 7.0 或更高 PDO PHP Extens…

洛谷 P4018 Roy&October之取石子 题目背景 Roy和October两人在玩一个取石子的游戏. 题目描述 游戏规则是这样的:共有n个石子,两人每次都只能取 p^kpk 个(p为质数,k为自然数,且 p^kpk 小于等于当前剩余石子数),谁取走最后一个石子,谁就赢了. 现在October先取,问她有没有必胜策略. 若她有必胜策略,输出一行"October wins!":否则输出一行"Roy wins!". 输入输出格式 输入格式: 第一行一个正整…

题目背景 Roy和October两人在玩一个取石子的游戏. 题目描述 游戏规则是这样的:共有n个石子,两人每次都只能取pk 个(p为质数,k为自然数,且pk小于等于当前剩余石子数),谁取走最后一个石子,谁就赢了. 现在October先取,问她有没有必胜策略. 若她有必胜策略,输出一行"October wins!":否则输出一行"Roy wins!". 输入输出格式 输入格式: 第一行一个正整数T,表示测试点组数. 第2行~第(T+1)行,一行一个正整数n,表示石子个…

P4018 Roy&October之取石子 题目背景 Roy和October两人在玩一个取石子的游戏. 题目描述 游戏规则是这样的:共有n个石子,两人每次都只能取p^kpk个(p为质数,k为自然数,且p^kpk小于等于当前剩余石子数),谁取走最后一个石子,谁就赢了. 现在October先取,问她有没有必胜策略. 若她有必胜策略,输出一行"October wins!":否则输出一行"Roy wins!". 输入输出格式 输入格式: 第一行一个正整数T,表示测试…

POJ 上的一套水题,哈哈~~~,最后一题很恶心,不想写了~~~ Rope Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7410   Accepted: 2603 Description Plotters have barberically hammered N nails into an innocent plane shape, so that one can see now only heads. Moreove…

题目背景 Roy和October两人在玩一个取石子的游戏. 题目描述 游戏规则是这样的:共有n个石子,两人每次都只能取 p^kpk 个(p为质数,k为自然数,且 p^kpk 小于等于当前剩余石子数),谁取走最后一个石子,谁就赢了. 现在October先取,问她有没有必胜策略. 若她有必胜策略,输出一行"October wins!":否则输出一行"Roy wins!". 输入输出格式 输入格式: 第一行一个正整数T,表示测试点组数. 第2行~第(T+1)行,一行一个正…

题目背景 \(Roy\)和\(October\)两人在玩一个取石子的游戏. 题目描述 游戏规则是这样的:共有\(n\)个石子,两人每次都只能取\(p^k\)个(\(p\)为质数,\(k\)为自然数,且\(p^k\)小于等于当前剩余石子数),谁取走最后一个石子,谁就赢了. 现在\(October\)先取,问她有没有必胜策略. 若她有必胜策略,输出一行"\(October wins!\)":否则输出一行"\(Roy wins!\)". 输入输出格式 输入格式: 第一行一…

下载代码 composer create-project october/october myoctober 准备好数据库, create database october; 配置环境于安装 php artisan october:install `=========================== INSTALLATION ===========================' Database type: [0] MySQL [1] Postgres [2] SQLite [3] SQ…

题意 题解 如果n是6的倍数,先手必败,否则先手必胜. 因为6*x一定不是pk 所以取得话会变成6*y+a的形式a=1,2,3,4,5: 然后a一定为质数.我们把a取完就又成为了6*x的形式. 又因为总数不断减少,所以6*x的局面是必败局面. 做完这题让我想起了初中时就被这种问题被人坑过. 然后博弈论可以先把SG的表打出来. #include<iostream> #include<cstring> #include<cstdio> #include<cstring…

All LeetCode Questions List 题目汇总 Sorted by frequency of problems that appear in real interviews. Last updated: October 2, 2017Google (214)534 Design TinyURL388 Longest Absolute File Path683 K Empty Slots340 Longest Substring with At Most K Distinct C…

4的倍数不行,之间的数都可以到4的倍数,而6的倍数不能到4的倍数 #include <iostream> #include <cstdio> #include <queue> #include <algorithm> #include <cmath> #include <cstring> #define inf 2147483647 #define N 1000010 #define p(a) putchar(a) #define F…

拖了近一个月的总结.(可能源于最近不太想做事:() A题 给出n个长度都为n的字符串,你只可以对每个字符串分别排序,问当每个字符串按升序排序之后,每一列是否也是升序的. #include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { ios::sync_wi…

题目链接:https://www.luogu.org/problem/P4860 和<P4018 Roy&October之取石子>一样的推导思路,去找循环节. 可以发现:只要不能被4整除就是必胜态,只要能被4整除就是必败态. 实现代码如下: #include <bits/stdc++.h> using namespace std; int T, n; int main() { cin >> T; while (T --) { cin >> n; pu…

题目链接:https://www.luogu.org/problem/P4018 首先碰到这道题目还是没有思路,于是寻思还是枚举找一找规律. 然后写了一下代码: #include <bits/stdc++.h> using namespace std; const int maxn = 101; bool win[maxn]; bool isp(int a) { if (a < 2) return false; for (int i = 2; i * i <= a; i ++) i…

The music is full of  (a, an, the, /) life. It can release your pressure to a large extent, so I enjoy this music very much. 题目解析 考查冠词用法.此句意思是:这首曲子充满活力,很大程度上能帮助减压,所以我非常喜欢它.life这个名词表示“活力.生机”,是不可数名词,所以此处不填任何冠词.   National Day falls on the  (10月1日) when…

TIOBE Index for October 2018 from:https://www.tiobe.com/tiobe-index// October Headline: Swift is knocking at the door of the TIOBE index top 10 TIOBE 指数发布了 10 月份的编程语言排行榜,排名前三的依旧是Java.C.C++. The top 9 of programming languages in the TIOBE index is qui…

While there is life there is hope. 一息若存,希望不灭. Go on living even if there is no hope. Knowing is not enough, we must apply; Willing is not enough, we must do. 获得知识并不够,一切看应用: 立下志向也不够,一切看行动. Knowledge is a treasure, and practice is the key to abtain it.…

Genius is nothing but labor and diligence. 天才不过是勤奋而已. Labor and diligence are the requirements for success, especially for those ordinary. But sometimes, they can't guarantee that you are sure to succeed, the direction is also very important. Maybe t…

I am a slow walker, but I never walk backwards. 我走得慢,但我从不后退. I walked very slow, sometimes I even slided backwards. Success is failing nine times and getting up ten. 成功就是跌倒九次,第十次仍会站起来. I have failed many times before, I try to get up once again. I ho…

Life is not a problem to be solved, but a reality to be experienced. 人生不是待解决的难题,而是等着我们去体验的现实. Press forward. Don't stop, do not linger in your journey, but strive for the mark set before you. 继续前行!不要停下脚步!不要在路上流连! 向着设定的目标奋进!…

The art of being wise is the art of knowing what to overlook. 智慧之道在于懂得该忽略什么. Always do your best. What you plant now, you will harvest later. 总是去竭尽全力,你现在播种什么,将来就会收获什么.…