Cerinta completa

Implement a simple text editor. The editor initially contains an empty string, . Perform operations of the following types:

  1. append – Append string to the end of .
  2. delete – Delete the last characters of .
  3. print – Print the character of .
  4. undo – Undo the last (not previously undone) operation of type or , reverting to the state it was in prior to that operation.

Example 


operation
index   S       ops[index]  explanation
-----   ------  ----------  -----------
0       abcde   1 fg        append fg
1       abcdefg 3 6         print the 6th letter - f
2       abcdefg 2 5         delete the last 5 letters
3       ab      4           undo the last operation, index 2
4       abcdefg 3 7         print the 7th characgter - g
5       abcdefg 4           undo the last operation, index 0
6       abcde   3 4         print the 4th character - d

The results should be printed as: 

f
g
d

Input Format

The first line contains an integer, , denoting the number of operations.
Each line of the subsequent lines (where ) defines an operation to be performed. Each operation starts with a single integer, (where ), denoting a type of operation as defined in the Problem Statement above. If the operation requires an argument, is followed by its space-separated argument. For example, if and , line will be 1 abcd.

Constraints

  • The sum of the lengths of all in the input .
  • The sum of over all delete operations .
  • All input characters are lowercase English letters.
  • It is guaranteed that the sequence of operations given as input is possible to perform.

Output Format

Each operation of type must print the character on a new line.

Sample Input

STDIN   Function
-----   --------
8       Q = 8
1 abc   ops[0] = '1 abc'
3 3     ops[1] = '3 3'
2 3     ...
1 xy
3 2
4 
4 
3 1

Sample Output

c
y
a

Explanation

Initially, is empty. The following sequence of operations are described below:

  1. . We append to , so .
  2. Print the character on a new line. Currently, the character is c.
  3. Delete the last characters in (), so .
  4. Append to , so .
  5. Print the character on a new line. Currently, the character is y.
  6. Undo the last update to , making empty again (i.e., ).
  7. Undo the next to last update to (the deletion of the last characters), making .
  8. Print the character on a new line. Currently, the character is a.

Limbajul de programare folosit: cpp14

Cod:

#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <limits>
#include <tuple>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cassert>

using namespace std;

typedef struct query
{
    int type, remove;
    string add;
}query;

char stak[1000003];

int topp = 0;

void push(string s)
{
    for(int i = 0 ; i < s.length(); i++)
    {
        stak[++topp] = s[i];
    }
    assert(topp <= 1000000);
}

string pop(int remove)
{
    assert(remove >= 0 && remove <= topp);
    string popped = "";
    int del = remove;
    for (int  i = topp; del > 0; i--, del--) {
        popped = stak[i] + popped;
    }
    topp -= remove;
    return popped;
}

int main()
{
    int t, type, remove, k;
    stack<query> q_stack;
    cin>>t;
    while(t--)
    {
        cin>>type;
        if(type == 1)
        {
            string add;
            cin>>add;
            push(add);
            query last;
            last.type = type;
            last.add = add;
            q_stack.push(last);
        }
        else if(type == 2)
        {
            cin>>remove;
            string popped = pop(remove);
            query last;
            last.type = type;
            last.remove = remove;
            last.add = popped;
            q_stack.push(last);
        }
        else if(type == 3)
        {
            cin>>k;
            assert(k >= 1 && k <= topp);
            cout<<stak[k]<<endl;
        }
        else
        {
            query last = q_stack.top();
            q_stack.pop();
            if(last.type == 1)
            {
                int remove = (last.add).length();
                string popped = pop(remove);
            }
            else
            {
                push(last.add);
                assert(topp >= 1 && topp <= 1000000);
            }
        }
    }
    return 0;
}

Scor obtinut: 1.0

Submission ID: 464653243

Link challenge: https://www.hackerrank.com/challenges/simple-text-editor/problem

Simple Text Editor