原題地址：http://acm.uestc.edu.cn/#/problem/show/327

The new operating system BerOS has a nice feature. It is possible to use any number of characters?/?as a delimiter in path instead of one

?traditional?/. For example, strings?//usr///local//nginx/sbin//?and?/usr/local/nginx///sbin?are equivalent. The character?/?(or some?

sequence of such characters) at the end of the path is required only in case of the path to the root directory, which can be represented as?

single character?/.

A path called normalized if it contains the smallest possible number of characters?/.

Your task is to transform a given path to the normalized form.

Input

There are multi-cases. The first line of each case contains only lowercase Latin letters and character?/?— the path to some directory. All paths?

start with at least one character?/. The length of the given line is no more than?100?characters, it is not empty.

Output

The path in normalized form.

Sample input and output

Sample Input Sample Output

//usr///local//nginx/sbin	/usr/local/nginx/sbin

題目大意是將路徑化為linux下的最簡格式，即目錄間由一個斜杠隔開，根目錄前有一個斜杠。說白了就是將多個斜杠變為一個。
此題可以使用常規思路，利用開關變量，不斷判斷是否為字母，然后整個單詞輸出。但是，巧妙利用C++的流處理，能非常簡單的處理這道題。 首先 ，需要聲明庫<sstream>，這是處理字符串流的庫。然后創建一個輸入流isstream is(s)（注意，這里的輸入并非指從鍵盤敲入，而是從字符串中 輸入），其中s是題中的字符串。接下來，就將is當作cin用，就可以啦。當然，需要處理一下剛開始的字符串，將所有/替換為空格。且需要注意的是 ，有一種特例是全為/，這種情況只需判斷一下是否輸出即可。 獻上代碼： #include<iostream> #include<string> #include<sstream> using namespace std;int main() {string s, temp;while (cin >> s){for (int i = 0; i < s.length(); i++)if (s[i] == '/')s.replace(i, 1, 1, ' ');//將斜杠代換為空格istringstream is(s);//創建輸入流bool flag = 0;//判斷是否有輸入while (is >> temp){cout << '/' << temp; flag = 1;}if (!flag)cout << '/';cout << endl;}return 0; }

轉載于:https://www.cnblogs.com/HarryGuo2012/p/4524051.html

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