It is a characteristic of wisdom not to do desperate things. 不做孤注一掷的事情是智慧的表现. We are told that we had better not to put all our eggs into one basket, that means, we must learn how to spread the risks so as to reduce the potential risks to the minimum…

题目: Problem D. Great AgainInput file: standard inputOutput file: standard outputTime limit: 2 secondsMemory limit: 512 megabytesThe election in Berland is coming. The party United Berland is going to use its influence to win themagain. The crucial co…

题目:Problem K. PiecemakingInput file: standard inputOutput file: standard outputTime limit: 1 secondMemory limit: 512 mebibytesThe civil war in Berland continues for five years already. The United Nation decided to end the bloodshed.Berland consists o…

1000w的数组,一开始都是2^31-1,然后经过5*10^7次随机位置的随机修改,问你每次的全局最小值. 有效的随机修改的期望次数很少,只有当修改到的位置恰好是当前最小值的位置时才需要扫一下更新最小值. 分个块或者直接暴力都可以. #include<cstdio> #include<iostream> #include<cmath> #include<algorithm> using namespace std; int sz,l[10005],r[100…

Every man is a poet when he is in love. 每个恋爱中的人都是诗人. It is said this saying was from Plato, the famous ancient philosopher. Notice that when you are dating with a girl, the romantic process may involve flowers, chocolate, sweet words and surprising g…

Much effort, much prosperity. 越努力,越幸运. I have ever seen this sentence in many people's signature of their social media applications, like WeChat, Tecent QQ, and so on. I ever thought that may be true, if you make much more efforts, you would be able…

You only live once, but if you do it right, once is enough. 人生只有一次,但如果活对了,一次也就够了. Maybe I am going to have to do a lot more work on the project before it is presentable. A lot of people can finish their work effortlessly, but it is so stiff for me to…

Our lives stretched out ahead of us, like a perpetual sunrise. 生命如永恒的日出,生生不息. Please respect yourself, respect your efforts, self-respect leads to self-displine. And self-displine can lead to great achievements. Don't believe in pessimism, if somethi…

The only thing more painful than learning from experience is not learning from experience. 比从经验中学习更为痛苦的事情就是没能从经验中学习. Learning from experience is painful, but the alternative of not learning is worse. Yes, it it easy to understand why not learning fro…

Had I not seen the sun, I could have borne the shade. 我本可以忍受黑暗,如果我不曾见过阳光. A poem by Emily Dickinson, and the full text follows: Had I not seen the sun I could have borne the shade But light a newer wilderness My wilderness had made Aha, if just readi…

We read the world wrong and say that it deceives us. 我们把世界看错了,反而说它欺骗了我们. It is not a cakewalk to see the world clearly and read it in the right way. Sometimes we didn't attain the things we expected, we might complain that the world deceived us. Why…

Have you ever known the theory of chocie? There are a bunch of axiems, but there are only two thing we really need to concern with. One, the only person's behaviors that we can control are our own. And two, the only thing that we can truly give other…

Happiness is a way station between too much and too little. 幸福就是刚刚好. I don't want to talk about it any more. Maybe I am the one who was cursed by the Heaven or by the God, because there was no happiness or luckiness in return for my positive and usef…

In the evening one may praise the day. 入夜方能赞美白昼. I think that could be understand in different ways, at least two. One is that we praise the day in the evening, because the day brings us the light, in the evening we know the value of light. The other…

预计分数:100+60+60=220 实际分数:100+60+40=200 除了暴力什么都不会的我..... T1 2017.9.17巧克力棒(chocolate) 巧克力棒(chocolate)Time Limit:1000ms Memory Limit:64MB题目描述LYK 找到了一根巧克力棒,但是这根巧克力棒太长了,LYK 无法一口吞进去.具体地,这根巧克力棒长为 n,它想将这根巧克力棒折成 n 段长为 1 的巧克力棒,然后慢慢享用.它打算每次将一根长为 k 的巧克力棒折成两段长为 a…

win10 如何配置 java jdk 环境变量 2017.8.17 本篇还适用于 windows server 2012. windows server 2014+ 一.安装 下载 jdk 64位 windows 版本安装(默认安装) 默认安装的路径: C:\Program Files\Java\.0_144 C:\Program Files\Java\.0_144 二.配置环境变量 安装好后增加系统的环境变量,如图: 新建变量: JAVA_HOME C:\Program Files\Java…

题目: Problem F. Matrix GameInput file: standard inputOutput file: standard inputTime limit: 1 secondMemory limit: 256 mebibytesAlice and Bob are playing the next game. Both have same matrix N × M filled with digits from 0 to 9.Alice cuts the matrix ve…

题目:Problem J. TerminalInput file: standard inputOutput file: standard inputTime limit: 2 secondsMemory limit: 256 mebibytesN programmers from M teams are waiting at the terminal of airport. There are two shuttles at the exitof terminal, each shuttle…

题目:Problem L. Canonical duelInput file: standard inputOutput file: standard outputTime limit: 2 secondsMemory limit: 256 megabytesIn the game «Canonical duel» board N × M is used. Some of the cells of the board contain turrets. Aturret is the unit wi…

题目:Problem A. Arithmetic DerivativeInput file: standard inputOutput file: standard inputTime limit: 1 secondMemory limit: 256 mebibytesLets define an arithmetic derivative:• if p = 1 then p0 = 0;• if p is prime then p0 = 1;• if p is not prime then n0…

题目:Problem D. Clones and TreasuresInput file: standard inputOutput file: standard outputTime limit: 1 secondMemory limit: 256 mebibytesThe magical treasury consists of n sequential rooms. Due to construction of treasury its impossible togo from room…

给你一个网格(n<=2000,m<=2000),有一些炸弹,你可以选择一个空的位置,再放一个炸弹并将其引爆,一个炸弹爆炸后,其所在行和列的所有炸弹都会爆炸,连锁反应. 问你所能引爆的最多炸弹数. 转化成: 将行列当成点,炸弹当成边,然后你可以给这个二分图加1条边,问你最大的连通块的边的数量. 可以通过枚举所有可以建的边,通过并查集来尝试更新答案.由于一条边必然会让总度数+2,所以一个连通块的边数是所有点的度数之和/2. 并查集不必要动态维护集合的大小,一开始就建好并查集,提前统计好即可. 最后…

有两辆车,容量都为K,有n(10w)个人被划分成m(2k)组,依次上车,每个人上车花一秒.每一组的人都要上同一辆车,一辆车的等待时间是其停留时间*其载的人数,问最小的两辆车的总等待时间. 是f(i,j)表示前i组,j个人是否可行.w(i)表示第i组的人数. if f(i,j)==1 then f(i+1,j+w(i+1))=1. 这是个bitset可以做的事情,每次左移以后或上f(i-1)的bitset即可.其实可以滚动数组. 然后每更新一次bitset,求一下其最左侧的1的位置,就是对于第一辆…

给你n个字符串,问你最小的长度的前缀,使得每个字符串任意循环滑动之后,这些前缀都两两不同. 二分答案mid之后,将每个字符串长度为mid的循环子串都哈希出来,相当于对每个字符串,找一个与其他字符串所选定的子串不同的子串,是个二分图最大匹配的模型,可以匈牙利或者Dinic跑最大流看是否满流. 一个小优化是对于某个字符串,如果其所有不同的子串数量超过n,那么一定满足,可以直接删去. 卡常数,不能用set,map啥的,采取了用数组记录哈希值,排序后二分的手段进行去重和离散化. #include<cst…

给你一个n*m的字符矩阵,将横向(或纵向)全部裂开,然后以任意顺序首尾相接,然后再从中间任意位置切开,问你能构成的字典序最大的字符串. 以横向切开为例,纵向类似. 将所有横排从大到小排序,枚举最后切开的位置在哪一横排,将这一排提到排序后的字符串数组最前面,求个“最大表示法”,如果最大表示法的位置恰好在第一排的位置,那么可以用来更新答案. 如果不在第一排的位置,那么其所构成的仍然是合法的串,而且一定不会影响答案. 这是一个最小表示法的板子. #include<cstdio> #include&l…

给你n,K,问你要选出最少几个长度为2的K进制数,才能让所有的n位K进制数删除n-2个元素后,所剩余的长度为2的子序列至少有一个是你所选定的. 如果n>K,那么根据抽屉原理,对于所有n位K进制数,必然会至少有1个数字出现2次或以上,所以00,11,...,K-1 K-1这样的数对是必选的. 对于其他的情况下,我们需要让他构造不出来n位不含重复数字的K进制数. 于是可以把K个数尽可能平均地分成n-1组,每一组内部让他们选出任意两个数都不合法,于是只能组间互相拼,这样他只能构造出最多n-1位的K进制…

给你一行房间,有的是隐身药水,有的是守卫,有的是金币. 你可以任选起点,向右走,每经过一个药水或金币就拿走,每经过一个守卫必须消耗1个药水,问你最多得几个金币. 药水看成左括号,守卫看成右括号, 就从每个位置贪心地向右走,如果在 r 遇到不匹配,则把下一次的左端点置成r+1,接着走. O(n)即可. 因为如果把左端点放在上次的l和r之间,要么会发生不匹配,要么答案无法比上次走的更优. 队友代码: #include <iostream> #include <cstdio> #incl…

假设一个数有n个质因子a1,a2,..,an,那么n'=Σ(a1*a2*...*an)/ai. 打个表出来,发现一个数x,如果x'=Kx,那么x一定由K个“基础因子”组成. 这些基础因子是2^2,3^3,5^5,7^7,11^11,13^13.只有6个,K不超过30,于是可以dfs. 要注意搜索顺序(每次枚举的时候,都从大于等于前项的开始搜)和可行性剪枝(如果超过r则剪枝,虽说有可能爆long long,但其实整除就可以判,而且没有精度误差). #include<cstdio> //#incl…

A smooth sea never made a skillful mariner. 平静的海洋练不出熟练的水手. A smooth sea never made a skillful mariner, but a violent sea may kill a greenhand. When we are engaged in a relatively new vacation, the best way may be to follow the principle from easy to…

环境: win7_x64旗舰版.VS2017企业版 一.安装VS2017 1.1 安装VS2017时,必须要勾选“使用C++的modules开发”选项 1.2 安装成功后,会在VS安装目录Microsoft Visual Studio\2017\Enterprise\VC\Tools\MSVC\14.10.25017生成ifc文件夹 二.创建和配置项目 2.1 创建Win32控制台项目 2.2 添加module.cpp文件内容 // module.cpp : 定义控制台应用程序的入口点. //…