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ACM_训练赛
早晨
训练赛
第一场 C题
早晨
训练赛
第一场 C题 C - Searching for Graph Time Limit:1000MS Memory Limit
·
2015-11-08 16:22
c
早晨
训练赛
第一场 A题
早晨
训练赛
第一场 A题 A - Nuts Time Limit:1000MS Memory Limit:262144KB&
·
2015-11-08 16:21
a
ACdream群
训练赛
(46)
A.Seinfeld http://acm.hdu.edu.cn/showproblem.php?pid=3351 用num来标记次数,用len来表示 { 的个数 从前开始遍历字符串,如果是 { 就不配对 ,len++,如果是 } 且如果len不为0就len--,在此情况下如果len为零了,就只能将此字符变成 { ,那么num++;遍历完了,如果len不为0就说明剩下了 { 没有配对 此时只需
·
2015-11-07 15:54
cd
2013-8-14大一大二暑期组队
训练赛
1001 Time Limit : 5000/2000ms (Java/Other) Memory Limit : 65535/65535K (Java/Other) Total Submission(s) : 11 Accepted Submission(s) : 2 Font: Times New Roman&nbs
·
2015-11-07 12:58
【2013南京区域赛】部分题解 hdu4802—hdu4812
上周末打了一场
训练赛
,题目是13年南京区域赛的 这场题目有好几个dp本来应该是我擅长的,但是可能是太久没做比赛了各种小错误代码写的也丑各种warusn trush搞得人很不爽 全场题之一的
·
2015-11-07 12:53
HDU
ACM_
高次同余方程
/*poj3243 *解决高次同余方程的应用,已知X^Y=KmodZ,及X,Z,K的值,求Y的值 */ #include #include #include usingnamespacestd; #definelint__int64 #defineMAXN131071 structHashNode{lintdata,id,next;}; HashNodehash[MAXN=1) { if(b&1
xiaotan1314
·
2015-11-07 10:00
高次同余方程
ACM_
扩展欧几里德算法
/* 扩展欧几里德算法 基本算法:对于不完全为0的非负整数a,b,gcd(a,b)表示a,b的最大公约数,必然存在整数对x,y,使得gcd(a,b)=ax+by。 证明:设a>b。 1,显然当b=0,gcd(a,b)=a。此时x=1,y=0; 2,ab!=0时 设ax1+by1=gcd(a,b); bx2+(amodb)y2=gcd(b,amodb); 根据朴素的欧几里德原理有gcd
xiaotan1314
·
2015-11-06 14:00
算法
ACM
gcd
【总结】10月
训练赛
第5场(8中)
模板题+送分题+数据结构不会搞题感觉好厉害的样子2015-10-712:19:13国庆第五场
训练赛
初总结1.一道数论:等比数列求和取模,矩阵乘法之类的做法,草稿了半小时无果,数据留情水50分;2.一道最短路
Formiko
·
2015-11-05 10:26
总结
网曝Dandy和Mata被回韩国 组战队狙击SKT
EDG一回国就传出Deft疑似要走的新闻,Spirit近期也被曝出WE
训练赛
不专心,而就在此时北美著名电竞记者莫凯西接连发推,曝出新闻:VG战队的Dandy和Mata双双打包回国,而MVP貌似要打造一支包揽
小苍mm
·
2015-11-04 00:00
hdu 4751 Divide Groups bfs (2013 ACM/ICPC Asia Regional Nanjing Online 1004)
SDUST的
训练赛
当时死磕这个水题3个小时,也无心去搞其他的 按照题意,转换成无向图,预处理去掉单向的边,然后判断剩下的图能否构成两个无向完全图(ps一个完全图也行或是一个完全图+一个孤点) 代码是赛后看的网上大神
·
2015-11-02 15:10
online
几道DP题
联合
训练赛
中遇到的(以后留给新生做): 1.排列的逆序数 题目描述 {1,2...n}的所有排列中逆序数为k的排列个数。
·
2015-11-02 11:53
dp
最后一周第二天
训练赛
之第二题
试题: B - B Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Practice SPOJ IC
·
2015-11-01 13:49
最后一周
训练赛
第一题
A - Problem A Time Limit:2000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Practice SPOJ Q
·
2015-11-01 13:49
[
ACM_
模拟] UVA 12504 Updating a Dictionary [字符串处理 字典增加、减少、改变问题]
Updating a Dictionary In this problem, a dictionary is collection of key-value pairs, where keys are lower-case letters, and values are non-negative integers. Given
·
2015-11-01 10:30
字符串处理
[
ACM_
水题] UVA 12502 Three Families [2人干3人的活后分钱,水]
Three Families Three families share a garden. They usually clean the garden together at the end of each week, but last week, family C was on holiday, so family A sp
·
2015-11-01 10:29
ACM
[
ACM_
模拟] UVA 12503 Robot Instructions [指令控制坐标轴上机器人移动 水]
Robot Instructions You have a robot standing on the origin of x axis. The robot will be given some instructions. Your task is to predict its position after executin
·
2015-11-01 10:29
struct
北邮新生
训练赛
3解题报告
.........A: 429. 学姐的数码管 时间限制 1000 ms 内存限制 65536 KB 题目描述 学姐的七段数码管玩的出神入化。 现在给你一个浮点数,你需要把它以七段数码管的形式输出出来。 一个 (2∗n+1)∗n 的矩阵来表示七段数码管,若下标均从0开始,则以第0列的两个,第 n−1 列的两个,第0行的一个,第
·
2015-11-01 10:16
[
ACM_
数据结构] POJ2352 [树状数组稍微变形]
Description Astronomers often examine star maps where stars are represented by points on a plane and each star has Cartesian coordinates. Let the level of a star be an amount of the stars tha
·
2015-10-31 11:37
数据结构
[
ACM_
图论] ZOJ 3708 [Density of Power Network 线路密度,a->b=b->a去重]
The vast power system is the most complicated man-made system and the greatest engineering innovation in the 20th century. The following diagram shows a typical 14 bus power system. In
·
2015-10-31 11:36
NetWork
[
ACM_
水题] ZOJ 3712 [Hard to Play 300 100 50 最大最小]
MightyHorse is playing a music game called osu!. After playing for several months, MightyHorse discovered the way of calculating score in osu!: 1. While p
·
2015-10-31 11:36
play
[
ACM_
暴力][
ACM_
几何] ZOJ 1426 Counting Rectangles (水平竖直线段组成的矩形个数,暴力)
Description We are given a figure consisting of only horizontal and vertical line segments. Our goal is to count the number of all different rectangles formed by these segments. As an example, the n
·
2015-10-31 11:36
count
[
ACM_
动态规划] UVA 12511 Virus [最长公共递增子序列 LCIS 动态规划]
Virus We have a log file, which is a sequence of recorded events. Naturally, the timestamps are strictly increasing. However, it is infected by a virus, so random
·
2015-10-31 11:36
动态规划
[
ACM_
动态规划] hdu 1176 免费馅饼 [变形数塔问题]
Problem Description 都说天上不会掉馅饼,但有一天gameboy正走在回家的小径上,忽然天上掉下大把大把的馅饼。说来gameboy的人品实在是太好了,这馅饼别处都不掉,就掉落在他身旁的10米范围内。馅饼如果掉在了地上当然就不能吃了,所以gameboy马上卸下身上的背包去接。但由于小径两侧都不能站人,所以他只能在小径上接。由于gameboy平时老
·
2015-10-31 11:36
动态规划
[
ACM_
模拟] ACM - Draw Something Cheat [n个长12的大写字母串,找出交集,按字母序输出]
Description Have you played Draw Something? It's currently one of the hottest social drawing games on Apple iOS and Android Devices! In this game, you and your friend play in turn. You
·
2015-10-31 11:36
ACM
[
ACM_
数学] LA 3708 Graveyard [墓地雕塑 圈上新加点 找规律]
Description Programming contests became so popular in the year 2397 that the governor of New Earck -- the largest human-inhabited planet of the galaxy -- opened a special Alley of
·
2015-10-31 11:36
ACM
[
ACM_
水题] UVA 11292 Dragon of Loowater [勇士斗恶龙 双数组排序 贪心]
Once upon a time, in the Kingdom of Loowater, a minor nuisance turned into a major problem. The shores of Rellau Creek in central Loowater had always been a prime breeding ground for g
·
2015-10-31 11:36
water
[
ACM_
模拟][
ACM_
数学] LA 2995 Image Is Everything [由6个视图计算立方体最大体积]
Description Your new company is building a robot that can hold small lightweight objects. The robot will have the intelligence to determine if an object is light enough to hold. It
·
2015-10-31 11:36
image
[
ACM_
图论] The Perfect Stall 完美的牛栏(匈牙利算法、最大二分匹配)
描述 农夫约翰上个星期刚刚建好了他的新牛棚,他使用了最新的挤奶技术。不幸的是,由于工程问题,每个牛栏都不一样。第一个星期,农夫约翰随便地让奶牛们进入牛栏,但是问题很快地显露出来:每头奶牛都只愿意在她们喜欢的那些牛栏中产奶。上个星期,农夫约翰刚刚收集到了奶牛们的爱好的信息(每头奶牛喜欢在哪些牛栏产奶)。一个牛栏只能容纳一头奶牛,当然,一头奶牛只能在一个牛栏中产奶。 给出奶牛们的爱好的信息,计算最
·
2015-10-31 11:35
ACM
[
ACM_
搜索] ZOJ 1103 || POJ 2415 Hike on a Graph (带条件移动3盘子到同一位置的最少步数 广搜)
Description "Hike on a Graph" is a game that is played on a board on which an undirected graph is drawn. The graph is complete and has all loops, i.e. for any two locations there is exactly
·
2015-10-31 11:35
Graph
[
ACM_
图论] Fire Net (ZOJ 1002 带障碍棋盘布炮,互不攻击最大数量)
Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall. A blockhouse is a small cas
·
2015-10-31 11:35
ACM
[
ACM_
图论] Sorting Slides(挑选幻灯片,二分匹配,中等)
Description Professor Clumsey is going to give an important talk this afternoon. Unfortunately, he is not a very tidy person and has put all his transparencies on one big heap. Before giving the talk
·
2015-10-31 11:35
sort
[
ACM_
搜索] Triangles(POJ1471,简单搜索,注意细节)
Description It is always very nice to have little brothers or sisters. You can tease them, lock them in the bathroom or put red hot chili in their sandwiches. But there is also a time when all meanne
·
2015-10-31 11:35
ACM
[
ACM_
动态规划] ZOJ 1425 Crossed Matchings(交叉最大匹配 动态规划)
Description There are two rows of positive integer numbers. We can draw one line segment between any two equal numbers, with values r, if one of them is located in the first row and the other one is
·
2015-10-31 11:35
match
[
ACM_
搜索] POJ 1096 Space Station Shielding (搜索 + 洪泛算法Flood_Fill)
Description Roger Wilco is in charge of the design of a low orbiting space station for the planet Mars. To simplify construction, the station is made up of a series of Airtight Cubical Modules (ACM's
·
2015-10-31 11:35
ACM
[
ACM_
几何] Transmitters (zoj 1041 ,可旋转半圆内的最多点)
Description In a wireless network with multiple transmitters sending on the same frequencies, it is often a requirement that signals don't overlap, or at least that they don't conflict. One way of a
·
2015-10-31 11:35
ACM
[
ACM_
模拟] The Willy Memorial Program (poj 1073 ,联通水管注水模拟)
Description Willy the spider used to live in the chemistry laboratory of Dr. Petro. He used to wander about the lab pipes and sometimes inside empty ones. One night while he was in a pipe, he fell as
·
2015-10-31 11:35
ACM
[
ACM_
其他] Square Ice (poj1099 规律)
Description Square Ice is a two-dimensional arrangement of water molecules H2O, with oxygen at the vertices of a square lattice and one hydrogen atom between each pair of adjacent oxygen atoms. The
·
2015-10-31 11:35
ACM
[
ACM_
图论] Highways (变形说法的最小生成树)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28972#problem/C 题目给出T种情况,每种情况有n个城镇,接下来每一行是第i个城镇到所有城镇的距离(其实就是个可达矩阵)。 求建设一条公路联通所有城镇并且要求最长的一段最小(其实就是最小生成树)!代码如下: #include<
·
2015-10-31 11:34
最小生成树
[
ACM_
数学] Counting Solutions to an Integral Equation (x+2y+2z=n 组合种类)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=27938#problem/E 题目大意:Given, n, count the number of solutions to the equation x+2y+2z=n, where x,y,z,n are non negative inte
·
2015-10-31 11:34
count
[
ACM_
几何] The Deadly Olympic Returns!!! (空间相对运动之最短距离)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28235#problem/B 题目大意: 有两个同时再空间中匀速运动的导弹,告诉一个时间以及各自的初始坐标和该时间时的坐标,求运动过程中的最短距离 解题思路: 求出相对初位置、相对速度,则答案就是原点到射线型轨迹的距离,注意是射线!!!
·
2015-10-31 11:34
return
[
ACM_
几何] F. 3D Triangles (三维三角行相交)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28235#problem/A 题目大意:给出三维空间两个三角形三个顶点,判断二者是否有公共点,三角形顶点、边、内部算三角形的一部分。 解题思路:见模板 //**********************************************
·
2015-10-31 11:34
ACM
[
ACM_
动态规划] 轮廓线动态规划——铺放骨牌(状态压缩1)
Description Squares and rectangles fascinated the famous Dutch painter Piet Mondriaan. One night, after producing the drawings in his 'toilet series' (where he had to use his toilet paper to draw on
·
2015-10-31 11:34
动态规划
[
ACM_
图论] 棋盘问题 (棋盘上放棋子的方案数)
不能同行同列,给定形状和大小的棋盘,求摆放k个棋子的可行方案 Input 2表示是2X2的棋盘,1表示k,#表示可放,点不可放(-1 -1 结束) Output 输出摆放的方案数目C Sample Input 2 1 #. .# 4 4 ...# ..#. .#.. #... -1 -1 Sample Output 2 1
·
2015-10-31 11:34
ACM
[
ACM_
其他] 总和不小于S的连续子序列的长度的最小值——尺缩法
Description: 给定长度为n的整数数列,A[0],A[1],A[2]….A[n-1]以及整数S,求出总和不小于S的连续子序列的长度的最小值。如果解不存在,则输出0。 Input: 输入数据有多组,每组数据第一行输入n,S, (10<n<10^5,S<10^8)第二行输入A[0],A[1],A[2]….A[n-1] ( 0<A[i]≤10000)
·
2015-10-31 11:34
ACM
[
ACM_
几何] Metal Cutting(POJ1514)半平面割与全排暴力切割方案
Description In order to build a ship to travel to Eindhoven, The Netherlands, various sheet metal parts have to be cut from rectangular pieces of sheet metal. Each part is a convex polygon with at mo
·
2015-10-31 11:34
meta
[
ACM_
动态规划] Palindrome
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28415#problem/D 题目大意:给一个长为n的字符串,问最少插入几个字符成回文串 解题思路:总长-最长公共(原来的和其倒过来的串)子序列(LCS) 知识详解——LCS:给出两个子序列A,B,求长度最大的公共子序列(如152687和2356984——
·
2015-10-31 11:33
动态规划
[
ACM_
动态规划] Alignment (将军排队)
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28415#problem/F 题目大意:有n个士兵排成一列,将军想从中抽出最少人数使队伍中任何士兵都能够看到左边最远处或右边最远处 解题思路:①此题是最长上升子序列的升级版。 &
·
2015-10-31 11:33
动态规划
[
ACM_
贪心] Radar Installation
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28415#problem/A 题目大意:X轴为海岸线可放雷达监测目标点,告诉n个目标点和雷达监测半径,求最少多少个雷达可全覆盖,如果不能输出-1; 解题思路:赤裸裸的区间选点问题(数轴上有n个闭区间,去尽量少的点,使每个区间至少有一个点)。核心思想就是贪心算法:把所有区间按照b从小
·
2015-10-31 11:33
Install
[
ACM_
几何] Pipe
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28417#problem/B 本题大意: 给定一个管道上边界的拐点,管道宽为1,求一束光最远能照到的地方的X坐标,如果能照到终点,则输出...  
·
2015-10-31 11:33
ACM
[
ACM_
几何] Wall
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=28417#problem/E 题目大意:依次给n个点围成的一个城堡,在周围建围墙,要求围墙离城墙的距离大于一定的值,求围墙最短长度(结果四舍五入 解题思路:求围住所有点的凸包周长+一个圆的周长 #include<iostream> #include<cma
·
2015-10-31 11:33
ACM
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