立志用最少的代碼做最高效的表達

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PAT甲級最優題解——>傳送門

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People often have a preference among synonyms of the same word. For example, some may prefer “the police”, while others may prefer “the cops”. Analyzing such patterns can help to narrow down a speaker’s identity, which is useful when validating, for example, whether it’s still the same person behind an online avatar.

Now given a paragraph of text sampled from someone’s speech, can you find the person’s most commonly used word?

Input Specification:
Each input file contains one test case. For each case, there is one line of text no more than 1048576 characters in length, terminated by a carriage return \n. The input contains at least one alphanumerical character, i.e., one character from the set [0-9 A-Z a-z].

Output Specification:
For each test case, print in one line the most commonly occurring word in the input text, followed by a space and the number of times it has occurred in the input. If there are more than one such words, print the lexicographically smallest one. The word should be printed in all lower case. Here a “word” is defined as a continuous sequence of alphanumerical characters separated by non-alphanumerical characters or the line beginning/end.

Note that words are case insensitive.

Sample Input:
Can1: “Can a can can a can? It can!”

Sample Output:
can 5

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題意：輸入一行句子。 每個單詞由數字或字母組成，不區分大小寫， 求出現次數最多的單詞

分析：按題意，單詞一定被非數字和字母分割。因此逐個字符讀入，一旦遇到非空格或字母就將單詞存入散列表中， 最后輸出max_word即可。

注意：可以對字符串的末尾手動添加空格和回車，否則最后一個單詞無法被讀入(因為其后沒有特殊字符)。

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#include<bits/stdc++.h> using namespace std; int main() {string s; getline(cin, s);s += ' ', s += '\n'; int len = s.length();unordered_map<string, int>um; int i = 0, max_times = -1;string temp, max_word;while(s[i] != '\n') {s[i] = tolower(s[i]);if(isalpha(s[i]) || isdigit(s[i])) temp += s[i];else if(!temp.empty()) {um[temp]++;if(um[temp] > max_times) {max_times = um[temp];max_word = temp;} else if(um[temp] == max_times) if(max_word > temp) max_word = temp;temp.clear();}i++;}cout << max_word << ' ' << max_times << '\n';return 0; }

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耗時：

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求贊哦~ (?ω?)

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