淺談設(shè)計(jì)獲勝策略-全文預(yù)覽
【正文】 effort, focus on the future: how can you get the most points in the next hour with what you have? Have a checklist to use before turning in your solutions: Code freeze five minutes before end of contest? Turn asserts off. Turn off debugging output. Turn on all optimizations. Make sure input and output are to correct filenames. Make sure the input and output formats are correct. Repile and test once more. Copy files to correct locations (floppy?) with correct names. Tips amp。 if you have to, put tolerances in everywhere (never test equality) Comments on ments: Not long prose, just brief notes Explain highlevel functionality: ++i。39。 of course, you should solve the easier of the two only. Alternatively, like induction, for some problems one can make a small change to the solution of a slightly smaller problem to find the full answer. Receipted from 返回9 / 9。s execution speed or memory usage will double when the problem size doubles. An algorithm of O(N 2) will run about four times slower (or use 4x more space) when the problem size doubles. Constanttime or space algorithms are denoted O(1). This concept applies to time and space both。 document functional sections As if to someone intelligent who knows the problem, but not the code Anything you had to think about Anything you looked at even once saying, now what does that do again? Always ment order of array indices Keep a log of your performance in each contest: successes, mistakes, and what you could have done better。t delete your extra debugging output, ment it out Optimize progressively, and only as much as needed Keep all working versions! Code to debug: whitespace is good, use meaningful variable names, don39。s going wrong? A short amount (20 mins) of debugging is better than switching to anything else。 sketch notes with algorithm, plexity, the numbers, data structs, tricky details, ... Brainstorm many possible algorithms then pick the stupidest that works! DO THE MATH! (space amp。另外，像歸納法一樣，你可以對一個(gè)較小的問題的解答作一些小小的改變以得到原問題的完整答案。 例如，對于4路對稱，你只需解決問題的四分之一，就可以寫下4個(gè)結(jié)果，這四個(gè)結(jié)果和你所解決的一個(gè)結(jié)果是對稱的（注意自對稱的解答，他當(dāng)然只應(yīng)該被輸出一次或兩次）。注意，有時(shí)你可以對你的數(shù)據(jù)的不同（獨(dú)立）部分重復(fù)使用相同的算法以有效地改進(jìn)程序的運(yùn)行時(shí)間。比如說，一個(gè)程序需要將大寫字母轉(zhuǎn)化為小寫字母，可以不需任何條件地利用一個(gè)表格進(jìn)行快速查找來實(shí)現(xiàn)。通過數(shù)學(xué)計(jì)算來判斷一個(gè)過濾器是否足夠好或者是否你需要嘗試制作一個(gè)產(chǎn)生器。（？？原文是Plug in the numbers.） 例子： 一個(gè)簡單的重復(fù)N次的循環(huán)復(fù)雜度為O(N)： 1 sum=0 2 fori=1ton 3 sum=sum+i 一個(gè)雙重嵌套循環(huán)的復(fù)雜度通常為O(N 2)： fill array a with N elements 1 fori=1ton1 2 forj=i+1ton 3 if(a[i]a[j]) swap(a[i],a[j]) 注意，雖然這個(gè)循環(huán)執(zhí)行了N(N+1)/2 次if語句，但他的復(fù)雜度仍然是O(N 2)，因?yàn)镹加倍后執(zhí)行時(shí)間增加了四倍。 如果你的程序有l(wèi)層遞歸，每層遞歸有b個(gè)遞歸調(diào)用，該程序復(fù)雜度為O(b l)。復(fù)雜的算法或許只能處理這個(gè)數(shù)目的一半。 當(dāng)然，遞歸也像循環(huán)一樣計(jì)算，并且遞歸程序可以有像O(b N)， O(N!)， 甚至O(N N)的階。這個(gè)概念同時(shí)適用于時(shí)間和空間；這里我們將集中討論時(shí)間。 document functional sections） 好像是寫給某個(gè)了解該問題但并不了解程序代碼
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