ACM_JAVA(ACM模板)

《ACM算法与程序设计》解题报告模板

电子科技大学期末解题报告课程：《ACM算法与程序设计》学院：学号：姓名：报告成绩：教师签名：

讨厌的青蛙 1、链接地址 https://www.360docs.net/doc/3112119385.html,/problem?id=2812 2、问题描述在韩国，有一种小的青蛙。每到晚上，这种青蛙会跳越稻田，从而踩踏稻子。农民在早上看到被踩踏的稻子，希望找到造成最大损害的那只青蛙经过的路径。每只青蛙总是沿着一条直线跳越稻田，而且每次跳跃的距离都相同，如图1所示。稻田里的稻子组成一个栅格，每棵稻子位于一个格点上，如图2所示。而青蛙总是从稻田的一侧跳进稻田，然后沿着某条直线穿越稻田，从另一侧跳出去，如图3所示。问题描述

青蛙的每一跳都恰好踩在一棵水稻上，将这棵水稻拍倒。可能会有多只青蛙从稻田穿越，有些水稻被多只青蛙踩踏，如图4所示。当然，农民所见到的是图5中的情形，看不到图4中的直线。根据图5，农民能够构造出青蛙穿越稻田时的行走路径，并且只关心那些在穿越稻田时至少踩踏了3 棵水稻的青蛙。因此，每条青蛙行走路径上至少包括3 棵被踩踏的水稻。而在一条青蛙行走路径的直线上，也可能会有些被踩踏的水稻不属于该行走路径。在图5中，格点(2, 1)、(6, 1)上的水稻可能是同一只青蛙踩踏的，但这条线上只有两棵被踩踏的水稻，因此不能作为一条青蛙行走路径；格点(2, 3)、(3, 4)、(6, 6)在同一条直线上，但它们的间距不等，因此不能作为一条青蛙行走路径；格点(2, 1)、(2, 3)、(2, 5)、(2, 7)是一条青蛙行走路径，该路径不包括格点(2, 6)。请你写一个程序，确定在所有的青蛙行路径中，踩踏水稻棵数最多的路径上有多少棵水稻被踩踏。例如，图5的答案是7，因为第6 行上全部水稻恰好构成一条青蛙行走路径。

ACM题目整理

题目来源：福州大学acm网站代码：fpcdq 一、入门熟悉ACM竞赛规则以及程序提交注意事项例题： Problem 1000 A+B Problem Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description Calculate a + b. Input The input will consist of a series of pairs of integers a and b,separated by a space, one pair of integers per line. Output For each pair of input integers a and b you should output the sum of a and b in one line,and with one line of output for each line in input. Sample Input 1 5 2 3 Sample Output 6 5

My answer： #include main() { long a,b; while((scanf("%ld%ld",&a,&b))!=EOF) { printf("%ld\n",a+b); } } 详情参考https://www.360docs.net/doc/3112119385.html,/faq.php 二、ACM分类主流算法： 1.搜索//回溯 Problem 1019 猫捉老鼠 Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description 一只猫和一只老鼠在10*10的迷宫中。迷宫中的每个方格可以是空的，或者含有障碍。猫和老鼠可以进入任意一个空的方格中。当他们相遇时，猫和老鼠在同一个方格中。但是，无论猫或老鼠都不能进入有障碍的方格。我们可以用字符组成的二维数组表示迷宫，如下图所示。

acm论文模板范文

acm论文模板范文 ACM是全世界领域影响力最大的专业学术组织。而acm模板，你们知道吗?这是 ___为大家了两篇acm论文，这样你们对模板会有直观的印象! [摘要] 鉴于ACM大学生程序设计竞赛(ACM/ICPC)在人才选拔和培养方面的显著作用，如何将ACM/ICPC竞赛活动嵌入常规教学，创新教学模式，结合专业教学，加强训练管理，提高培训效益，已成为人们关注的问题。针对这一应用需求，本文设计并开发了基于ACM/ICPC机制的大学生程序设计培训管理系统。系统采用B/S架构，以SQL Server xx作为后台管理数据库，Visual Studio 和https://www.360docs.net/doc/3112119385.html,为前端开发工具。在分析系统功能的基础上，着重阐述了该系统设计与实现的关键技术。该系统实际运行稳定、可靠，为开展ACM/ICPC竞赛培训和教学提供了一种有效管理途径。 [关键词] ACM/ICPC;培训管理系统;Web开发;https://www.360docs.net/doc/3112119385.html,;数据库技术 doi ： 10 . 3969 / j . issn . 1673 - 0194 . xx . 03. 015 [] TP311 [] A [] 1673 - 0194(xx)03- 0028- 03

1 引言 ACM国际大学生程序设计竞赛(ACM International Collegiate Programming Contest， ACM ICPC) 由美国计算机协会(ACM)主办，始于1970年，至今已经有40多年的，是世界公认的规模最大、水平最高、影响广泛的国际大学生程序设计竞赛，竞赛优胜者是各大IT企业和科研院所青睐和优先选拔的人才[1]。近些年来，伴随着ACM/ICPC大学生程序设计竞赛在国内如火如荼地开展，计算机界更加关注在人才培养方面，如何科学合理地引入、借鉴ACM/ICPC竞赛训练，将ACM/ICPC竞赛活动与常规专业课程教学有机结合起来，突破传统教学内容和，以有效培养学生的学习能力、创新意识和综合素质。这其中，如何有效组织开展ACM/ICPC竞赛训练，加强培训管理，提高培训效益，亦是人们关注的热点问题。但就目前情况来看，组织开展此项竞赛活动的训练指导或教学培训还没有一个成熟通用的、基于ACM/ICPC竞赛机制的ACM/ICPC 训练和活动的教学管理平台。具体表现在：(1)尽管一些知名院校搭建了自己的在线测试平台[2-3]，但由于大多采用英文表述问题，对于水平不高的低年级本科生和专科学生来说，在翻译题目和理解内容方面会出现偏差，导致在这些平台上进行在线模拟测验的效果并不理想;(2)很多网站虽然提供了ACM/ICPC竞赛的相关资料，比如网上题库、相关赛题的题解等，但这些资料在网上分布得比较分散，使

我的ACM算法模板

ACM模板 [ 王克纯 2020年9月21日

最大子串 int maxSum(int * a,int n) { int sum = a[0],b = 0; for(int i=0;i0) b += a[i]; else b = a[i]; if(b > sum) sum = b; } return sum; } int Kadane(const int array[], size_t length, unsigned int& left, unsigned int& right) { unsigned int i, cur_left, cur_right; int cur_max, max; cur_max = max = left = right = cur_left = cur_right = 0; for(i = 0; i < length; ++i){ cur_max += array[i]; if(cur_max > 0){ cur_right = i; if(max < cur_max){ max = cur_max; left = cur_left; right = cur_right; } } else{ cur_max = 0; cur_left = cur_right = i + 1; } } return max; } 快速幂 void js(int &a,int &b,int num) { b=1; while(num) { if(num&1) b*=a; num>>=1; a*=a; } } 矩阵乘法 struct mat{ int n,m;//n行m列 int data[MAX][MAX]; }; void mul(const mat& a,const mat& b,mat& c) //c=a*b { int i,j,k; if (a.m!=b.n); //报错 c.n=a.n,c.m=b.m; for (i=0;i

ACM动态规划问题简易模板(C++可编译)

1、0-1背包 #include #include //背包问题 /* 测试数据：输入： 8 23 8 4 5 1 6 6 7 3 7 8 3 3 4 9 6 2 输出： 1 0 1 0 1 0 1 1 */ int num,c; int v[10]; int w[10]; int m[10][30];//设m[i][j],则表示在前i个物品中，背包大小是j的情况下，背包所装东西的最大价值 void knapsack() { int n=num-1; int jmax,i,j; if(w[n]0;i--) { if(w[i]

} } m[0][c]=m[1][c]; if(c>=w[0]) { if(m[0][c]0) x[n]=1; else x[n]=0; } int main() { int i,x[10],j; scanf("%d %d",&num,&c); for(i=0;i

整理出ACM所有题目及答案

1111111杭电： 1000 A + B Problem (4) 1001 Sum Problem (5) 1002 A + B Problem II (6) 1005 Number Sequence (8) 1008 Elevator (9) 1009 FatMouse' Trade (11) 1021 Fibonacci Again (13) 1089 A+B for Input-Output Practice (I) (14) 1090 A+B for Input-Output Practice (II) (15) 1091 A+B for Input-Output Practice (III) (16) 1092 A+B for Input-Output Practice (IV) (17) 1093 A+B for Input-Output Practice (V) (18) 1094 A+B for Input-Output Practice (VI) (20) 1095 A+B for Input-Output Practice (VII) (21) 1096 A+B for Input-Output Practice (VIII) (22) 1176 免费馅饼 (23) 1204 糖果大战 (25) 1213 How Many Tables (26) 2000 ASCII码排序 (32) 2001 计算两点间的距离 (34) 2002 计算球体积 (35) 2003 求绝对值 (36) 2004 成绩转换 (37) 2005 第几天？ (38) 2006 求奇数的乘积 (40) 2007 平方和与立方和 (41) 2008 数值统计 (42) 2009 求数列的和 (43) 2010 水仙花数 (44) 2011 多项式求和 (46) 2012 素数判定 (47) 2014 青年歌手大奖赛_评委会打分 (49) 2015 偶数求和 (50) 2016 数据的交换输出 (52) 2017 字符串统计 (54) 2019 数列有序! (55) 2020 绝对值排序 (56) 2021 发工资咯：） (58) 2033 人见人爱A+B (59) 2037 今年暑假不AC (61) 2039 三角形 (63) 2040 亲和数 (64)

整理acm模板

1、KMP 算法 /* * next[]的含义：x[i-next[i]...i-1]=x[0...next[i]-1] * next[i]为满足x[i-z...i-1]=x[0...z-1]的最大z值（就是x的自身匹配） */ void kmp_pre(char x[],int m,int next[]) { int i,j; j=next[0]=-1; i=0; while(i=m) { ans++; j=next[j]; } } return ans; } 经典题目：POJ 3167 /* * POJ 3167 Cow Patterns * 模式串可以浮动的模式匹配问题 * 给出模式串的相对大小，需要找出模式串匹配次数和位置 * 比如说模式串：1，4，4，2，3，1 而主串：5,6,2,10,10,7,3,2,9 * 那么2,10,10,7,3,2就是匹配的 * * 统计比当前数小，和于当前数相等的，然后进行kmp */ #include #include #include #include #include using namespace std; const int MAXN=100010; const int MAXM=25010; int a[MAXN]; int b[MAXN];

ACM的论文写作格式标准

ACM Word Template for SIG Site 1st Author 1st author's affiliation 1st line of address 2nd line of address Telephone number, incl. country code 1st author's E-mail address 2nd Author 2nd author's affiliation 1st line of address 2nd line of address Telephone number, incl. country code 2nd E-mail 3rd Author 3rd author's affiliation 1st line of address 2nd line of address Telephone number, incl. country code 3rd E-mail ABSTRACT A s network speed continues to grow, new challenges of network processing is emerging. In this paper we first studied the progress of network processing from a hardware perspective and showed that I/O and memory systems become the main bottlenecks of performance promotion. Basing on the analysis, we get the conclusion that conventional solutions for reducing I/O and memory accessing latencies are insufficient for addressing the problems. Motivated by the studies, we proposed an improved DCA combined with INIC solution which has creations in optimized architectures, innovative I/O data transferring schemes and improved cache policies. Experimental results show that our solution reduces 52.3% and 14.3% cycles on average for receiving and transmitting respectively. Also I/O and memory traffics are significantly decreased. Moreover, an investigation to the behaviors of I/O and cache systems for network processing is performed. And some conclusions about the DCA method are also presented. Keywords Keywords are your own designated keywords. 1.INTRODUCTION Recently, many researchers found that I/O system becomes the bottleneck of network performance promotion in modern computer systems [1][2][3]. Aim to support computing intensive applications, conventional I/O system has obvious disadvantages for fast network processing in which bulk data transfer is performed. The lack of locality support and high latency are the two main problems for conventional I/O system, which have been wildly discussed before [2][4]. To overcome the limitations, an effective solution called Direct Cache Access (DCA) is suggested by INTEL [1]. It delivers network packages from Network Interface Card (NIC) into cache instead of memory, to reduce the data accessing latency. Although the solution is promising, it is proved that DCA is insufficient to reduce the accessing latency and memory traffic due to many limitations [3][5]. Another effective solution to solve the problem is Integrated Network Interface Card (INIC), which is used in many academic and industrial processor designs [6][7]. INIC is introduced to reduce the heavy burden for I/O registers access in Network Drivers and interruption handling. But recent report [8] shows that the benefit of INIC is insignificant for the state of the art 10GbE network system. In this paper, we focus on the high efficient I/O system design for network processing in general-purpose-processor (GPP). Basing on the analysis of existing methods, we proposed an improved DCA combined with INIC solution to reduce the I/O related data transfer latency. The key contributions of this paper are as follows: ?Review the network processing progress from a hardware perspective and point out that I/O and related last level memory systems have became the obstacle for performance promotion. ?Propose an improved DCA combined with INIC solution for I/O subsystem design to address the inefficient problem of a conventional I/O system. ?Give a framework of the improved I/O system architecture and evaluate the proposed solution with micro-benchmarks. ?Investigate I/O and Cache behaviors in the network processing progress basing on the proposed I/O system. The paper is organized as follows. In Section 2, we present the background and motivation. In Section 3, we describe the improved DCA combined INIC solution and give a framework of the proposed I/O system implementation. In Section 4, firstly we give the experiment environment and methods, and then analyze the experiment results. In Section 5, we show some related works. Finally, in Section 6, we carefully discuss our solutions with many existing technologies, and then draw some conclusions. 2.Background and Motivation In this section, firstly we revise the progress of network processing and the main network performance improvement bottlenecks nowadays. Then from the perspective of computer architecture, a deep analysis of network system is given. Also the motivation of this paper is presented. 2.1Network processing review Figure 1 illustrates the progress of network processing. Packages from physical line are sampled by Network Interface Card (NIC). NIC performs the address filtering and stream control operations, then send the frames to the socket buffer and notifies

acm-计算几何模板

#include using namespace std; const double eps =1e-8; const double INF =1e20; const double pi = acos ; int dcmp (double x) { if (fabs (x) < eps) return0; return (x <0-1:1); 》 } inline double sqr (double x) {return x*x;} f %.2f\n", x, y);} bool operator== (const Point &b) const { return (dcmp ==0&& dcmp ==0); } bool operator< (const Point &b) const { return (dcmp ==0 dcmp <0: x < ; } ； Point operator+ (const Point &b) const { return Point (x+, y+; } Point operator- (const Point &b) const { return Point , ; } Point operator* (double a) { return Point (x*a, y*a); } Point operator/ (double a) { ` return Point (x/a, y/a); } double len2 () {f\n", r); } bool operator== (const Circle &a) const { return p ==&& (dcmp ==0); } double area () {otate_left ()); Line v = Line ((b+c)/2, ((b+c)/2) + (c-b).rotate_left ()); Point p = line_intersection (u, v); ]

ACM比赛模板

目录 1.最小生成树 (2) 2.最短路算法 (7) 3.素数打表 (11) 4.最大匹配 (12) 5.线段树（敌兵布阵） (14) 6．线段树（逆序树） (16) 7.树形dp (18) 8.树状数组（段跟新） (20) 9.Kmp模板 (22) 10.线段树（点跟新） (26) 11.强连通 (28) 12.最小割 (31) 13.单源最短路（spfa） (34) 14.三分查找 (36) 15.字典树（统计难题） (38) 16.最大流入门题1273 (40) 17.状态压缩 (43) 18.匈牙利（HDU 2063）（最大匹配） (45) 19.凸包（HDU1348） (47) 20.树状数组(HDU1166) (50) 21.强连通 (52) 22.前向星 (55) 23.矩阵 (58) 24.并查集 (60) 25. SORT (61) 26. STL (63) 27. LCA (HDU 2874) (67) 28. 01背包 (70) 29. 状态压缩代码： (72) 30. 快速幂 (74) 31.矩阵快速幂 (75) 32.GCD & LCM (77) 33.ACM小技巧： (78) 34. /** 大数（高精度）求幂**/ (80) 35. /** 大数除法与求余**/ (82) 36. /** 大数阶乘**/ (84) 37. /** 大数乘法**/ (85) 38. /** 大数累加**/ (86)

1.最小生成树要连通n个城市需要n－1条边线路。可以把边上的权值解释为线路的造价。则最小生成树表示使其造价最小的生成树。 prim算法（矩阵形式）： #define inf 0x3f3f3f3f int prim(int n,int sta)//n表示有n个顶点，sta表从sta这个顶点出发生成最小生成树{ int mark[M],dis[M]; int i,sum = 0; //sum是总的最小生成树边权值 for (i = 0;i < n;i ++) //初始化dis[i] 表从顶点sta到点i的权值 { dis[i] = mat[sta][i]; mark[i] = 0; } mark[sta] = 1; //sta 这个顶点加入最小生成树中 for (i = 1;i < n;i ++) //循环n-1次，每次找出一条最小权值的边 n个点的图 { //只有n-1条边 int min = inf; //inf 表无穷大 for (j = 0;j < n;j ++)//找出当前未在最小生成树中边权最小的顶点 if (!mark[j] && dis[j] < min) min = dis[j],flag = j; mark[flag] = 1; //把该顶点加入最小生成树中 sum += dis[flag]; //sum加上其边权值 for (j = 0;j < n;j ++) //以falg为起点更新到各点是最小权值 if (dis[j] > mat[flag][j]) dis[j] = mat[flag][j]; } return sum; //返回边权总和 } prim算法（边表形式）： struct Edge//frm为起点,to为终点，w为边权，nxt指向下一个顶点 { // int frm; int to,w,nxt; }edge[M]; int vis[M],head[M],dis[M]; void addedge (int cu,int cv,int cw)//生成边的函数

ACM Word Template for SIGCOMM Submissions

ACM Word Template for SIGCOMM Submissions 1st Author 1st author's affiliation 1st line of address 2nd line of address Telephone number, incl. country code 1st author's email address 2nd Author 2nd author's affiliation 1st line of address 2nd line of address Telephone number, incl. country code 2nd E-mail 3rd Author 3rd author's affiliation 1st line of address 2nd line of address Telephone number, incl. country code 3rd E-mail ABSTRACT In this paper, we describe the formatting guidelines for ACM SIG Proceedings. Categories and Subject Descriptors D.3.3 [Programming Languages]: Language Contructs and Features –abstract data types, polymorphism, control structures. This is just an example, please use the correct category and subject descriptors for your submission.The ACM Computing Classification Scheme: https://www.360docs.net/doc/3112119385.html,/class/1998/ General Terms Your general terms must be any of the following 16 designated terms: Algorithms, Management, Measurement, Documentation, Performance, Design, Economics, Reliability, Experimentation, Security, Human Factors, Standardization, Languages, Theory, Legal Aspects, Verification. Keywords Keywords are your own designated keywords. 1.INTRODUCTION The proceedings are the records of the conference. ACM hopes to give these conference by-products a single, high-quality appearance. To do this, we ask that authors follow some simple guidelines. In essence, we ask you to make your paper look exactly like this document. The easiest way to do this is simply to down-load a template from [2], and replace the content with your own material. 2.PAGE SIZE All material on each page should fit within a rectangle of 18 x 23.5 cm (7" x 9.25"), centered on the page, beginning 1.9 cm (0.75") from the top of the page and ending with 2.54 cm (1") from the bottom. The right and left margins should be 1.9 cm (.75”). The text should be in two 8.45 cm (3.33") columns with a .83 cm (.33") gutter. 3.TYPESET TEXT 3.1Normal or Body Text Please use a 10-point Times Roman font, or other Roman font with serifs, as close as possible in appearance to Times Roman in which these guidelines have been set. The goal is to have a 10-point text on 12-point leading, as you see here. Please use sans-serif or non-proportional fonts only for special purposes, such as distinguishing source code text. If Times Roman is not available, try the font named Computer Modern Roman. On a Macintosh, use the font named Times. Right margins should be justified, not ragged. 3.2Title and Authors The title (Helvetica 18-point bold), authors' names (Helvetica 12-point) and affiliations (Helvetica 10-point) run across the full width of the page –one column wide. We also recommend phone number (Helvetica 10-point) and e-mail address (Helvetica 12-point). See the top of this page for three addresses. If only one address is needed, center all address text. For two addresses, use two centered tabs, and so on. For more than three authors, you may have to improvise.1 3.3First Page Copyright Notice Please leave 3.81 cm (1.5") of blank text box at the bottom of the left column of the first page for the copyright notice. 3.4Subsequent Pages For pages other than the first page, start at the top of the page, and continue in double-column format. The two columns on the last page should be as close to equal length as possible. 1If necessary, you may place some address information in a footnote, or in a named section at the end of your paper. Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are not made or distributed for profit or commercial advantage and that copies bear this notice and the full citation on the first page. To copy otherwise, or republish, to post on servers or to redistribute to lists, requires prior specific permission and/or a fee. Conference’04, Month 1–2, 2004, City, State, Country. Copyright 2004 ACM 1-58113-000-0/00/0004…$5.00.