【codechef】Nikhil and Commands（字符串删减，分类）

Nikhil learnt two new commands pwd and cd on the first day of Operating Systems lab.

pwd - command displays the current working directory and,
cd - changes the location of working directory.

If the cd parameter contains ".."(without quotes), that means to step one directory back.

The absolute path of directory is separated by slashes "/"(without quotes).

The default root directory is "/".

Your task is to print the current working directory.

 

Input

Input description.

• The first line of input contains T, denoting the number of test cases.
• The second line of input contains N, denoting the number of commands.
• Then N lines follow, each containing either a cd command or pwd command.

Output

Output description.

• For each pwd command, output the absolute path of current directory.

 

Constraints

Should contain all the constraints on the input data that you may have. Format it like:

• 1 ≤ T ≤ 100
• 1 ≤ N ≤ 100
• Maximum length of cd parameter ≤ 200

Example

Input:19pwdcd /home/csedpwdcd /lab/root/../dirpwdcd /homepwdcd labpwdOutput://home/csed//lab/dir//home//home/lab/

点击打开链接

碰到pwd就输出当前路径。出现/..就删它和前面的一个，如果开头没有‘/’则与上面的连接，否则重新开始记录。

功能：函数返回字符串str1中紧接“标记”的部分的指针, 字符串str2是作为标记的分隔符。如果分隔标记没有找到，函数返回NULL。为了将字符串转换成标记，第一次调用str1 指向作为标记的分隔符。之后所以的调用str1 都应为NULL。
例如:
char str[] = "now # is the time for all # good men to come to the # aid of their country";
char delims[] = "#";
char *result = NULL;
result = strtok( str, delims );
 while( result != NULL ) {
  printf( "result is \"%s\"\n", result );
   result = strtok( NULL, delims );
 }
以上代码的运行结果是：
    result is "now "
    result is " is the time for all "
    result is " good men to come to the "
    result is " aid of their country"

#include<bits/stdc++.h>using namespace std; int main(){    int tc; cin>>tc;    while(tc--) {      int cs;      char s[500];      vector<string> u;      scanf("%d",&cs);      while(cs--){          scanf("%s",s);          if(s==string("pwd")){              for(auto& x:u) printf("/%s",x.c_str()); //琢磨一下              puts("/");          }else{              scanf("%s",s);              if(*s=='/') u.clear();              for(char* i=strtok(s,"/");i;i=strtok(0,"/")){  //strtok的用法                  if(i==string("..") && u.size()) u.pop_back();                  <span style="font-size: 12.380952835083px; font-family: Arial, Helvetica, sans-serif;">if(i==string(".."))    </span>u.push_back(i);              }          }      }    }}

#include <bits/stdc++.h> using namespace std; int main(){int t;cin>>t;while(t--){int n;cin>>n;string x;getchar(); int q=0;string s="/";while(n--){getline(cin,x);if(x=="pwd"){cout<<s<<endl;}else{x.erase(0,3);if(x[0]=='/') //分类s=x;elses+=x;int p;for(int i=0;i<s.size();++i){if(s[i]=='/'&&i+1<s.size()&&s[i+1]!='.')p=i;if(i>0&&s[i]=='.'&&s[i-1]=='.'){s=s.substr(0,p)+s.substr(i+1,s.size()-i-1);i=-1;}}s+='/';}} }     return 0;} 

#include<bits/stdc++.h>#define fup(i,n) for(i=1;i<=n;i++)using namespace std;int main(){    int t;cin>>t;    while(t--){        int c,pos,i;        cin>>c;        string s = "/",com;        fup(i,c) {            cin>>com;            if(com == "pwd") {                if(s=="/") cout<<"/\n";                else cout<<s<<"/"<<endl;            }            else if( com == "cd") {                cin>>com;                if(com[0] != '/') {                    if(s=="/") s+=com;                    else s += "/" + com;                }                else s = com;                while( true)                {                    int k=s.find(".");                    if(k==-1)                        break;                    //for(int j=0;j<=)                    string s1=s.substr(0,k-1);                    string s2=s.substr(k+2,s.size()-k-2); //删除中间一段只要找出两边的子串，再拼接起来就行了                    k=s1.rfind("/");                    s1=s1.substr(0,k);                    s=s1+s2;                }            }        }    }}