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用两个map实现一一对应的判定。 Crypt Kicker IITime limit:1sec.Submitted:122Memory limit:32MAccepted:33 Source: Waterloo ACM Programming Contest Oct 4, 1998A common but insecure method of encrypting text is to permute the letters of the alphabet. That is, in the text, each letter of the alphabet is consistently re 阅读全文
posted @ 2011-06-09 22:29
归雾
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stl set实现字符串搜索。。效率一般。(附二分搜索。) Compound WordsTime limit:1sec.Submitted:233Memory limit:32MAccepted:81 Source: Waterloo ACM Programming Contest Sep 28, 1996You are to find all the two-word compound words in a dictionary. A two-word compound word is a word in the dictionary that is the concatenation of 阅读全文
posted @ 2011-06-09 20:41
归雾
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简单题。map简单应用。 排序后输出。 Etaoin ShrdluTime limit:1sec.Submitted:84Memory limit:32MAccepted:54 Source: University of Ulm Internal Contest 2001The relative frequency of characters in natural language texts is very important for cryptography. However, the statistics vary for different languages. Here are th 阅读全文
posted @ 2011-06-09 19:48
归雾
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字符串的排序。匹配。用stl vector实现。 Anagram GroupsTime limit:1sec.Submitted:114Memory limit:32MAccepted:35 Source: University of Ulm Internal Contest 2000World-renowned Prof. A. N. Agram's current research deals with large anagram groups. He has just found a new application for his theory on the distributi 阅读全文
posted @ 2011-06-09 18:37
归雾
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将中缀表达式转换为后缀表达式确定运算顺序。运算通过stl stack实现。结果通过stl map存储。 FriendsTime limit:1sec.Submitted:41Memory limit:32MAccepted:26 Source: University of Ulm Internal Contest 1999You want to plan a big birthday party with your friends. On planning you notice that you have to do a lot of operations with sets of frien 阅读全文
posted @ 2011-06-09 14:14
归雾
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简单的模拟。stl map实现各变量值的变化与存储。 Evaluating Simple C ExpressionsTime limit:1sec.Submitted:66Memory limit:32MAccepted:34 Source: UVA - V3 The task in this problem is to evaluate a sequence of simple Cexpressions, buy you need not know C to solve the problem! Each ofthe expressions will appear on a line by 阅读全文
posted @ 2011-06-09 13:58
归雾
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next_permutation()是一个求一个排序的下一个排列的函数。如果要走遍所有的排列，你必须先排序。这是这两个函数使用需要注意的地方。其函数原形为：template<class BidIt> bool next_permutation(BidIt first, BidIt last);template<class BidIt, class Pred>bool next_permutation(BidIt first, BidIt last, Pred pr);prev_permutation()与之相反，是求一个排列的前一个排序。 ID CodesTime li 阅读全文
posted @ 2011-06-09 13:33
归雾
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先将含有多余括号的中缀表达式转换为无括号的后缀表达式。再转换为含有最少括号的中缀表达式。stl stack实现。Complicated ExpressionsTime limit:5sec.Submitted:86Memory limit:32MAccepted:40 Source: ACM ICPC Central European Regional 2000The most important activity of ACM is the GSM network. As the mobile phoneoperator, ACM must build its own transmittin 阅读全文
posted @ 2011-06-09 13:13
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