精品亚洲午夜久久久久,国产成人精品一区二区三区,亚洲精品国产suv一区,欧美无遮挡国产欧美另类,久久永久免费视频,99er精品视频,在线免费观看韩国a视频,日韩天堂网
          全國

          熱門城市 | 全國 北京 上海 廣東

          華北地區 | 北京 天津 河北 山西 內蒙古

          東北地區 | 遼寧 吉林 黑龍江

          華東地區 | 上海 江蘇 浙江 安徽 福建 江西 山東

          華中地區 | 河南 湖北 湖南

          西南地區 | 重慶 四川 貴州 云南 西藏

          西北地區 | 陜西 甘肅 青海 寧夏 新疆

          華南地區 | 廣東 廣西 海南

          • 微 信
            高考

            關注高考網公眾號

            (www_gaokao_com)
            了解更多高考資訊

          首頁 > 高中頻道 > 信息學聯賽知識 > 信息學聯賽知識:Complete Search

          信息學聯賽知識:Complete Search

          2009-11-12 23:09:07網絡

            Complete Search

            The Idea

            Solving a problem using complete search is based on the ``Keep It Simple, Stupid'' principle. The goal of solving contest problems is to write programs that work in the time allowed, whether or not there is a faster algorithm.

            Complete search exploits the brute force, straight-forward, try-them-all method of finding the answer. This method should almost always be the first algorithm/solution you consider. If this works within time and space constraints, then do it: it's easy to code and usually easy to debug. This means you'll have more time to work on all the hard problems, where brute force doesn't work quickly enough.

            In the case of a problem with only fewer than a couple million possibilities, iterate through each one of them, and see if the answer works.

            Careful, Careful

            Sometimes, it's not obvious that you use this methodology.

            Problem: Party Lamps [IOI 98]

            You are given N lamps and four switches. The first switch toggles all lamps, the second the even lamps, the third the odd lamps, and last switch toggles lamps 1, 4, 7, 10, ... .

            Given the number of lamps, N, the number of button presses made (up to 10,000), and the state of some of the lamps (e.g., lamp 7 is off), output all the possible states the lamps could be in.

            Naively, for each button press, you have to try 4 possibilities, for a total of 410000 (about 106020 ), which means there's no way you could do complete search (this particular algorithm would exploit recursion).

            Noticing that the order of the button presses does not matter gets this number down to about 100004 (about 1016 ), still too big to completely search (but certainly closer by a factor of over 106000 ).

            However, pressing a button twice is the same as pressing the button no times, so all you really have to check is pressing each button either 0 or 1 times. That's only 24 = 16 possibilities, surely a number of iterations solvable within the time limit.

            Problem 3: The Clocks [IOI 94]

            A group of nine clocks inhabits a 3 x 3 grid; each is set to 12:00, 3:00, 6:00, or 9:00. Your goal is to manipulate them all to read 12:00. Unfortunately, the only way you can manipulate the clocks is by one of nine different types of move, each one of which rotates a certain subset of the clocks 90 degrees clockwise.

            Find the shortest sequence of moves which returns all the clocks to 12:00.

            The ``obvious'' thing to do is a recursive solution, which checks to see if there is a solution of 1 move, 2 moves, etc. until it finds a solution. This would take 9k time, where k is the number of moves. Since k might be fairly large, this is not going to run with reasonable time constraints.

            Note that the order of the moves does not matter. This reduces the time down to k9 , which isn't enough of an improvement.

            However, since doing each move 4 times is the same as doing it no times, you know that no move will be done more than 3 times. Thus, there are only 49 possibilities, which is only 262,072, which, given the rule of thumb for run-time of more than 10,000,000 operations in a second, should work in time. The brute-force solution, given this insight, is perfectly adequate.

            Sample Problems

            Milking Cows [USACO 1996 Competition Round]

            Given a cow milking schedule (Farmer A milks from time 300 to time 1000, Farmer B from 700 to 1200, etc.), calculate

            " The longest time interval in which at least one cow was being milked

            " The longest time interval in which no cow is being milked

            Perfect Cows & Perfect Cow Cousins [USACO 1995 Final Round]

            A perfect number is one in which the sum of the proper divisors add up to the number. For example, 28 = 1 + 2 + 4 + 7 + 14. A perfect pair is a pair of numbers such that the sum of the proper divisor of each one adds up to the other. There are, of course, longer perfect sets, such that the sum of the divisors of the first add up to the second, the second's divisors to the third, etc., until the sum of the last's proper divisors add up to the first number.

            Each cow in Farmer John's ranch is assigned a serial number. from 1 to 32000. A perfect cow is one which has a perfect number as its serial. A group of cows is a set of perfect cow cousins if their serial numbers form a perfect set. Find all perfect cows and perfect cow cousins.

            retrieved from http://ace.delos.com/usacogate

           

          [標簽:競賽聯賽 數學聯賽]

          分享:

          高考院校庫(挑大學·選專業,一步到位!)

          高考院校庫(挑大學·選專業,一步到位!)

          高校分數線

          專業分數線

          • 歡迎掃描二維碼
            關注高考網微信
            ID:gaokao_com

          • 👇掃描免費領
            近十年高考真題匯總
            備考、選科和專業解讀
            關注高考網官方服務號


          主站蜘蛛池模板: 东京热男人的av天堂| 两个人免费视频大全毛片| 马尔康县| 中文字幕午夜AV福利片| 在线无码精品秘 在线观看| 国产啪精品视频网站免| 开远市| 一区二区三区自拍偷拍亚洲| 精品的一区二区三区| 尤物yw午夜国产精品视频| 四虎国产精品免费久久麻豆| а的天堂网最新版在线| 临潭县| 精品在线观看视频二区| 国产无遮挡又黄又湿又爽毛片| 亚洲欧洲AV综合色无码| 亚洲AV成人无码天堂| 亚洲五月婷婷久久综合| 定襄县| 另类人妖在线观看一区二区| 桦甸市| 中国女人a毛片免费全部播放| 泾源县| 团风县| 淮北市| 国产一区二区欧美丝袜| 国产亚洲精品综合91| 国产成人综合亚洲av| 亚洲av午夜成人片精品| 国产麻豆精品久久一二三| 国产h视频在线观看网站免费| 亚洲一区二区三区无吗| 久久精品黄色免费热线| 亚洲乱在线播放| 亚洲中文字幕无码不卡电影| 久久亚洲av成人无码软件| japanese色国产在线看视频| 亚洲色婷婷综合开心网| 一本一本久久a久久精品综合| 国内色精品视频在线网址| 小13箩利洗澡无码免费视频|