Description

Modified from 10389 - text editor

In this problem we simulate a simple text editor. Given a series of keyboard input, output the final text content. Initially, the text content is empty and the cursor is at the beginning of the line. The user can:

1. Insert a lower-case alphabet ('a-z') before the cursor, denoted as a lower-case alphabet.

2. Erase an alphabet before the cursor, denoted as B.

3. Move the cursor to the right one character at a time, denoted as R.

4. Move the cursor to the left one word at a time, denoted as L.

Hints

In this problem, you are asked to use doubly linked list to reduce the time complexity. Three typical implementations as follows are possible. You can choose your favorite one, e.g., Impl 1 or Impl 2.

Remember to free memory when a linked list node is deleted. If you get Runtime Error on test case, probably there is something wrong with the pointers or there is memory leak in your program.

Make sure the time complexity is O(1) for each operation, otherwise you may get Time limit exceed

Input

The first line is an integer T (1 <= T <= 50), meaning that there are T subtasks in the input.

In each subtask, there are two lines. The first line is an integer N(1<= N <= 10⁶), being the length of the series of keyboard input. The next line is a string of length N, being the series of keyboard input.

It is guaranteed that L and B will not occur when the cursor is in the left-most position, R will not occur when the cursor is in the right-most position.

Output

For each subtask, output the final text content in a single line.

Sample Input  Download

Sample Output  Download

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Description

Receiving lucky money is the favorite activity of children during Chinese New Year.

Denny's family has N children, everyone wants to get the lucky money as soon as possible.

The elders in the family have prepared a game to decide the order of receiving the lucky money:

• At the beginning, all children sit in a circle, with seats numbered from 1 to N.
• Randomly decide two numbers A and K and the direction R/L, move the A_th youngest child K positions clockwise/counterclockwise.

For example:

N = 6

The given initial age sequence: 1 1 3 4 5 5

 

 

A = 3, K = 2, R

After processing:

 

 

A = 1, K = 3, L

After processing:

 

 

A = 1, K = 6, R

After processing:

 

 

A = 2, K = 2, R

After processing:

 

Output:
The age sequence in clockwise from the youngest: 1 3 5 4 5 1

Note: if two children have same age, consider the children with less order yonger.

 

Input

The first line contains two integers N M, the number of children and the number of position changes.

The second line contains N integers, giving the children's age sequence.

Each of the following M lines contains two integers A and K and the direction 'R' or 'L'.

 

test cases:

Output

The output contains one line.

Output the age sequence of the children in clockwise from the youngest.

Sample Input  Download

Sample Output  Download

Partial Judge Code

13442.c

Partial Judge Header

13442.h

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Description

After catching Suicune, Spongebob and Patrick thought they were happy enough so they went home. And then they found that Squidward had come home already (:

Squidward told them he need to drum up a marching band fast(for some reason). So he needed their help. Hence Spongebob and Patrick joined Squidward's band.

Today is their practicing day. Squidward is teaching all the members how to move at a performance. At a performance there are N rows. And in the i-th row, there are sz_i people. Note row may be empty(i.e. sz_i = 0). Everyone plays exactly one musical instrument. The musical instrument played by the j-th people in the i-th row is represented by a lower case letter c_ij.

Squidward ask them to move Q times. In each move, there are four types of moving:

1. all the people in the a_i-th row move to the front of the b_i-th row (without messing up the original order)
2. all the people in the a_i-th row move to the back of the b_i-th row (without messing up the original order)
3. exchange all the people in the a_i-th row and the b_i-th row (without messing up the original order)
4. reverse the order of all the people in the a_i-th row

Now it’s your turn! You need to help them to move and tell Squidward the final formation in the end.

 

Hint for reverse:

1. Easy way , but you can only get 2 more correct answers.

But, you should think another way to accelerate your performance to get AC.

 

Hint:

If you have no idea how to implement, please take it as your reference.

#include <stdlib.h>
#include <stdio.h>
typedef struct _Node {
    char val;
    struct _Node* next;
} Node;
 
//list[i]'s head node
Node *head[100005] = {};
 
//list[i]'s tail node
Node *tail[100005] = {};
 
//reverse(list[i])'s head node
Node *rev_head[100005] = {};
 
//reverse(list[i])'s tail node
Node *rev_tail[100005] = {};
 
/*
list[a] = 1 -> 2 -> 3 -> 4
rev_list[a] = 4 -> 3 -> 2 -> 1
head[a]->val = 1, tail[a]->val = 4
rev_head[a]->val = 4, rev_tail[a]->val = 1
you can use rev_head and rev_tail to get O(1) reverse
*/
 
void swap(int a, int b) {
    //swap list[a] and list[b]
    Node *tmp = (Node *)malloc(sizeof(Node));
 
    /* swap(head_node)
    tmp = head[a];
    head[a] = head[b];
    head[b] = tmp;
    */
 
    /* swap(tail_node)
 
    */

    /*swap(rev_head)
 
    */

    /*swap(rev_tail)
 
    */
}
void append(int a, int b) {
    //append list[a] to list[b]'s behind
    if(head[a] == NULL) return;
    if(head[b] == NULL) {
        swap(a, b);
        return;
    }
    /*
    tail[b]->next = head[a];
    tail[b] = ...
    ...
    */
}
void reverse(int a) {
    //reverse list[a]
    if(head[a] == NULL) return;
    /*
    use rev_head and rev_tail to reverse list[a] in O(1)
    hint: swap something
    */
}

Input

The first line contains one integer N (1 ≤ N ≤ 10⁵) – the number of rows.

Then N lines follow. First the i+1-th line contains one integer sz_i – the number of people in the i-th row. If there are some people in the row, a space character and sz_i characters c_ij are followed where c_ij is the musical instrument played by the j-th people in the i-th row. The sum of all the sz_i is less than or equal to 10⁶.

The N+2-th line contains one integer Q (1 ≤ Q ≤ 10⁵) – the times they need to move.

Then Q lines follow. The i+N+3-th line contains two integer type_i, a_i. If type_i isn't equal to 4, then a integer b_i follows(1 ≤ type_i ≤ 4, 1 ≤ a_i, b_i ≤ N, a_i ≠ b_i) – they are the corresponding information for each moving.

 

It's guaranteed that:

• The 1st testcase must be identical to the Sample #1 below
• For the first three testcases: 1 ≤ N, Q ≤ 10³, the sum of all the sz_i ≤ 10⁴, all the type_i ≠ 4
• For the first six testcases: all the type_i ≠ 4
• For the first eight testcases: the number of the fourth type of moving ≤ 100

Output

After all the moving, for each row output all the musical instrument played in that row and then print a newline('\n') character.

If there is no people in some row, just output an empty line.

 

Note: there are two sample below. "# Sample Input 1/2" and "# Sample Output 1/2" are not the part of input and output.

They are just for marking that the following content corresponds to which sample.

Sample Input  Download

Sample Output  Download

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Description

Given a linked list of length N, where the last node points to the M-th node in the list. Note that when M equals 0, it indicates that the last node points to NULL.

 

Your task is to get an input T and perform T insertion operations.

For each operation, you need to find the first node with value A (where V = A) in the list and insert a new node with value B after it. If no node with value A is found in the list, skip this operation.

After completing all operations, output each node's value in the linked list once, starting from the head.

Example Operation 1:

Insert a node with value 2 after the node with value 1.

Example Operation 2:

Insert a node with value 6 after the node with value 5.

Other Example Operation:

This question is a partial judge.
Please implement functions defined in the header file.

• insertNode: get an input T and perform T insertion operations
• printLinkedList: output each node's value in the linked list once, starting from the head

Input

The first line contains two integers N and M, where N represents the length of the linked list, and the last node in the linked list points to the M-th node.

The second line contains N integers V_i , representing the value of the i-th node.

The third line contains one integer T, representing T insertion operations to be performed.

The following T lines each contain two integers A and B, indicating to insert a node with value B after the first node with value V equal to A.

Constraints

• 1 ≤ N ≤ 10⁶
• 0 ≤ M ≤ N
• 1 ≤ T ≤ 100
• -10⁹ ≤ A, B, V ≤ 10⁹

^​Subtasks

• Testcases 1 ~ 3: 0 ≤ V ≤ 10⁶, V is unique, M = 0
• Testcases 4 ~ 6: 0 ≤ V ≤ 10⁶, V is unique
• Testcases 7 ~ 8: No additional restrictions

Output

 

After completing T operations, output each node's value in the linked list once, starting from the head.

 

Please remember to print "\n" at the end.

Sample Input  Download

Sample Output  Download

Partial Judge Code

14576.c

Partial Judge Header

14576.h

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