给出一段程序,求运行时间. 现在只考虑一层LOOP,不妨用数组a[i]来表示n的i次方的系数.如果输入OP m,那么就在a[0]上加m,遇到END,就说明循环结束了,需要在系数上乘以循环次数.如果次数为数字,那么每个系数都乘以它:如果为n,那么全部右移一位(是指把n^2的系数给n^3),记得a[0] = 0. 当有多层LOOP时,递归调用即可. 输出的时候需要注意几个错误:如 1*n^3 (应该为n^3), 2*n^1 (应该为n^2), 注意n^0的系数,只要不是0,就要输出. #includ…

Instant Complexity Time Limit: 1000MS Memory Limit: 10000K Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm that runs in linear time is usually much f…

Instant Complexity Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 1908   Accepted: 658 Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm…

Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 1535   Accepted: 529 Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm that runs in linear…

题目:http://poj.org/problem?id=1472 这个题目是分到“模拟题”一类的,我觉得模拟的成分比较少,主要考察编程能力.独立写完这个题特别兴奋...所以我必须好好说一说,独家哦... 题意是计算一段伪代码的时间复杂度,整体代码夹在“BEGIN”和“END”之间,中间有很多循环语句,夹在“LOOP X”和“END”之间,表示循环X次.第一眼一看很难下手,其实只要想到一点这个题就非常水了: “BEGIN”语句可以看做”LOOP 1“. 这样的话整个程序的递归结构就明显了,就是L…

Instant Complexity Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 20000/10000K (Java/Other) Total Submission(s) : 8   Accepted Submission(s) : 7 Problem Description Analyzing the run-time complexity of algorithms is an important tool for desi…

Instant Complexity Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 2017   Accepted: 698 Description Analyzing the run-time complexity of algorithms is an important tool for designing efficient programs that solve a problem. An algorithm…

acm之pku题目分类 对ACM有兴趣的同学们可以看看 DP:  1011   NTA                 简单题  1013   Great Equipment     简单题  1024   Calendar Game       简单题  1027   Human Gene Functions   简单题  1037   Gridland            简单题  1052   Algernon s Noxious Emissions 简单题  1409   Commun…

以下是poj百道水题,新手可以考虑从这里刷起 搜索1002 Fire Net1004 Anagrams by Stack1005 Jugs1008 Gnome Tetravex1091 Knight Moves1101 Gamblers1204 Additive equations 1221 Risk1230 Legendary Pokemon1249 Pushing Boxes 1364 Machine Schedule1368 BOAT1406 Jungle Roads1411 Annive…

Uva 11181 Probability|Given Problem's Link:   http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=18546 Mean: n个人去逛超市,第i个人会购买东西的概率是Pi.出超市以后发现有r个人买了东西,问你每个人购买东西的实际概率是多少. analyse: 转换模型: 有n个员工,每个员工被选出来的概率是Pi.最后选出了r个,问你第i个员工在这r个中的概率是多少. 设: 事件A---…