SGU 158.Commuter Train】的更多相关文章

一道简单题. 火车停的位置不是在整点就是在二分之一点,坐标*2,然后枚举火车停的位置,计算总距离即可. code: #include <iostream> #include <cmath> #include <cstring> #include <algorithm> #include <iomanip> #define INF 309 using namespace std; int L, m, n, x, ans, p; int pg[INF…
Last night an explosion on a commuter train carriage in Taipei Songshan railway station wounded at least 21 people. The Criminal Investigation Bureau conducted an investigation on the crime scene and found the explosives. Some witness on that train s…
http://acm.sgu.ru/problemset.php?contest=0&volume=1 101 Domino 欧拉路 102 Coprime 枚举/数学方法 103 Traffic Lights 最短路 104 Little Shop of Flowers 动态规划 105 Div 3 找规律 106 The Equation 扩展欧几里德 107 987654321 Problem 找规律 108 Self-numbers II 枚举+筛法递推 109 Magic of Dav…
In late September, on a crowded commuter train in San Francisco, a man shot and killed 20-year-old student Justin Valdez. As security footage shows, before the gunman fired, he waved around his .45 caliber pistol and at one point even pointed it acro…
this blog from: http://blog.demofox.org/2017/03/09/how-to-train-neural-networks-with-backpropagation/ How to Train Neural Networks With Backpropagation Posted on March 9 2017 by Demofox This post is an attempt to demystify backpropagation, which is t…
题意: 给你一个数n,表示有n辆火车,编号从1到n,入站,问你有多少种出站的可能.    (题于文末) 知识点: ps:百度百科的卡特兰数讲的不错,注意看其参考的博客. 卡特兰数(Catalan):前几项为 : 1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670- 令h(0)=1,h(1)=1,catalan数满足递推式:      h(n)= h(0…
列车调度(Train) Description Figure 1 shows the structure of a station for train dispatching. Figure 1 In this station, A is the entrance for each train and B is the exit. S is the transfer end. All single tracks are one-way, which means that the train ca…
Organize Your Train part II Time Limit: 1000MS Memory Limit: 65536K Description RJ Freight, a Japanese railroad company for freight operations has recently constructed exchange lines at Hazawa, Yokohama. The layout of the lines is shown in Figure 1.…
Caused by: com.mysql.jdbc.exceptions.jdbc4.CommunicationsException: The last packet successfully received from the server was 2,926,157 milliseconds ago. The last packet sent successfully to the server was 2,926,158 milliseconds ago. is longer than t…
How to Train a GAN? Tips and tricks to make GANs work 转自:https://github.com/soumith/ganhacks While research in Generative Adversarial Networks (GANs) continues to improve the fundamental stability of these models, we use a bunch of tricks to train th…
A - Train Problem I Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 1022 Description As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to…
传送门 Description As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a pr…
水概率....SGU里难得的水题.... 495. Kids and Prizes Time limit per test: 0.5 second(s)Memory limit: 262144 kilobytes input: standardoutput: standard ICPC (International Cardboard Producing Company) is in the business of producing cardboard boxes. Recently the…
本篇用双向链表和模拟栈混洗过程两种解答方式具体解答“栈混洗”的应用问题 有关栈混洗的定义和解释在此篇:手记-栈与队列相关 列车调度(Train) 描述 某列车调度站的铁道联接结构如Figure 1所示. 其中,A为入口,B为出口,S为中转盲端.所有铁道均为单轨单向式:列车行驶的方向只能是从A到S,再从S到B:另外,不允许超车.因为车厢可在S中驻留,所以它们从B端驶出的次序,可能与从A端驶入的次序不同.不过S的容量有限,同时驻留的车厢不得超过m节. 设某列车由编号依次为{1, 2, ..., n}…
sgu 101 - Domino Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u Description Dominoes – game played with small, rectangular blocks of wood or other material, each identified by a number of dots, or pips, on its face. The bl…
http://acm.sgu.ru/problem.php?contest=0&problem=495 题意:N个箱子M个人,初始N个箱子都有一个礼物,M个人依次等概率取一个箱子,如果有礼物则拿出礼物放回盒子,如果没有礼物则不操作.问M个人拿出礼物个数的期望.(N,M<=100000) #include <cstdio> using namespace std; double mpow(double a, int n) { double r=1; while(n) { if(n&…
追求“舒适”和“快感”的VR游戏设计方法   http://game.watch.impress.co.jp/docs/news/20160318_749016.html     [Bullet Train]   演讲的状况       在游戏的创造历史上,有那种决定性的创新,以及高完成度的作品,对于FPS风格来说,[DOOM]就是这样.1993年第一次登场依赖,给了后面的游戏不可估量的影响.       Epic Games为Oculus Rift和Oculus Touch开发的游戏[Bull…
题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=455 Due to the slow 'mod' and 'div' operations with int64 type, all Delphi solutions for the problem 455 (Sequence analysis) run much slower than the same code written in C++ or Java. We do not gua…
2015-02-14 Created By BaoXinjian…
题目大意 某人在打字机上打一个字符串,给出了他打每个字符出错的概率 q[i]. 打一个字符需要单位1的时间,删除一个字符也需要单位1的时间.在任意时刻,他可以花 t 的时间检查整个打出来的字符串,并且从当前光标删除到第一个出错的位置(留下的字符串要么为空,要么每个字符都是对的).问你,他正确的打完该字符串最少需要花费的时间的期望值是多少 字符串的长度小于等于3000 题意有点绕,具体可以看看原题是怎么描述的 做法分析 首先确定这是一个概率DP的问题. 定义状态 f[i] 表示正确的打完前 i 个…
104. Little shop of flowers time limit per test: 0.25 sec. memory limit per test: 4096 KB 问题: 你想要将你的花窗安排得最具美感.有F束花,每一束花都不一样,至少有F个按顺序排成一行的花瓶.花瓶从左到右,依次编号1-V.而花放置的位置是可以改变的,花依次编号1到F.花的序号有一个特征,即是序号决定了花束出现在花瓶里的顺序.例如,有两束花编号i和j,满足i<j,那么i所在的花瓶一定要在j所在花瓶的左边,即是i…
令一个点的属性值为:去除这个点以及与这个点相连的所有边后得到的连通分量的节点数的最大值. 则树的重心定义为:一个点,这个点的属性值在所有点中是最小的. SGU 134 即要找出所有的重心,并且找出重心的属性值. 考虑用树形DP. dp[u]表示割去u点,得到的连通分支的节点数的最大值. tot[u]记录以u为根的这棵子树的节点数总和(包括根). 则用一次dfs即可预处理出这两个数组.再枚举每个点,每个点的属性值其实为max(dp[u],n-tot[u]),因为有可能最大的连通分支在u的父亲及以上…
题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=170 解题报告:输入两个由'+'和'-'组成的字符串,让你判断第二个串能不能由第一个串中的字符交换位置得到,不能的话输出-1,能的话输出最少的需要交换的次数. 一开始以为是DP,后来仔细一看发现,+号之间相对的位置是固定的,就是说如果可以,+号在调换位置的过程中是不会跟+号调换的,只跟-号调换,只有这样才能保证交换的次数是最少的. #include<cstdio> #includ…
题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=179 解题报告:输入一个合法的括号串,求出这个括号串的字典序的下一个串.(认为'(' < ')') 我的做法主要是用了生成字典序的下一个序列的思想: 1.从序列的尾部开始往前找,找到第一个升序的位置例如 2 5 4 3 对于这个序列来说就是2 5这个位置 2.然后在后面这个降序的序列中找到一个比这个2稍大一点的数跟2进行交换得到3 5 4 2 3.然后把这个位置的后面的串转过来得到3…
题目链接:http://acm.sgu.ru/problem.php?contest=0&problem=180 解题报告:一个裸的求逆序对的题,离散化+线段树,也可以用离散化+树状数组.因为这题中的数列有重复的而且范围特别大,所以要进行离散化,离散化的方法是, 首先按照输入的数字排个序,然后把整个数列扫一遍,得出每个数字离散化的结果,然后再按照输入的顺序把顺序调整回来就OK 了.线段树部分就不说了. #include<cstdio> #include<cstring> #…
H - The equation Time Limit:250MS     Memory Limit:4096KB     64bit IO Format:%I64d & %I64u Submit Status Practice SGU 106 Description There is an equation ax + by + c = 0. Given a,b,c,x1,x2,y1,y2 you must determine, how many integer roots of this eq…
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 23574    Accepted Submission(s): 8907 Problem Description As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of studen…
题目传送门 /* 题意:n*n的矩阵,放置k个king,要求king互相不能攻击,即一个king的8个方向都没有另外的king,求方案个数 状态压缩DP:dp[i][num[j]][s] 代表在第i行,放置num[j]个king,其状态为s时的方案数 首先对state进行筛选,即一行king左右相邻没有另外的king,确定tot 接着对第一行进行处理,对2~n行DP递推,同样要满足各种条件限制,num个数和不超过k,s1,s2状态也要ok 最后累加放置k个king时的方案数 详细解释:http:…
题目传送门 /* 题意:几个判断,车上的人不能 <0 或 > C:车上初始和结束都不能有人在 (为0):车上满员时才有等候的人 水题:难点在于读懂题目意思,状态不佳,一直没搞懂意思,在这题卡住了 主要要理解have to wait 的意思 */ #include <cstdio> #include <iostream> #include <algorithm> #include <cstring> #include <string>…
题目地址:http://acm.sgu.ru/problem.php?contest=0&problem=107 /* 题意:n位数的平方的后面几位为987654321的个数 尼玛,我看描述这一句话都看了半天,其实只要先暴力程序测试一边就知道规律 详细解释:http://www.cnblogs.com/Rinyo/archive/2012/12/04/2802089.html */ #include <cstdio> #include <iostream> #include…