Challenges of csdn programming-book Transfer

Question details

Xiaoqiang has three boxes, A, B, and C, which are used to hold books. All the books (N in total) are stacked on a in the sequence from small to large, now he wants to put all the books in C. Every time he takes a book (not enough) from booka to B, a box is flipped, and B book is taken from B (not enough) to C and B box is flipped. Repeat the operation until all the books arrive at C and find the order of the books in C. For details, see the example.

Input description:

The input consists of multiple groups of data. Each group occupies one row, with three numbers. N (1 <= n <= 10 ^ 5 ), A (1 <= A <= 10 ^ 9), B (1 <= B <= 10 ^ 9). For the meaning, see the topic description. End with a file.

Output description:

Each group of data outputs a row. First, the number of data groups is output, then the number of N is output, and the arrangement of C books is made.

Answer description

Input example:

4 2 1

Output example

Case 1: 1 4 2 3

Hint

Initial status A: 4321 B: NULL C: NULL

First time: A-> B a: 21 B: 34 C: NULL

A flipped a: 12 B: 34 C: NULL

B-> c a: 12 B: 4 C: 3

B flip a: 12 B: 4 C: 3

Second time: A-> B a: null B: 214 C: 3

A flip a: null B: 214 C: 3

B-> c a: null B: 14 C: 23

B flip a: null B: 41 c: 23

Third time: B-> c a: null B: 1 C: 423

B flip a: null B: 1 C: 423

Fourth time: B-> c a: null B: NULL C: 1423

Idea: This question can be done with a two-way linked list, because a and B both have a flip operation, while a two-way linked list can traverse the elements of a table in two directions, so after the flip, it is equivalent to taking the linked list element from another direction. In the end, two-way linked lists are used to represent a and B, and a stack is used to represent C. We have been simulating the question operation. When A and B are empty, we can output the value of C in the stack.

Code:

# Include <iostream> # include <list> # include <stack> using namespace STD; int main () {int n, a, B, n_case = 0; while (CIN> N) {n_case ++; CIN> A> B; List <int> a_list; // list of elements storing a <int> B _list; // store B's element stack <int> S; // store c's element for (INT I = 1; I <= N; I ++) a_list.push_front (I ); // use 1-N to fill the aint COUNT = 1; while (! (A_list.empty () & B _list.empty () {int a_tmp = A; int B _tmp = B; If (count % 2 = 1) // obtain the element {While ((! A_list.empty () & a_tmp --) {B _list.push_front (a_list.front (); // obtain the first a element a_list.pop_front ();} while ((! B _list.empty () & B _tmp --) {S. push (B _list.front (); // obtain the first B element B _list.pop_front ();} else // obtain the element {While ((! A_list.empty () & a_tmp --) {B _list.push_back (a_list.back (); a_list.pop_back () ;}while ((! B _list.empty () & B _tmp --) {S. push (B _list.back (); B _list.pop_back () ;}count ++ ;}cout <"case" <n_case <":"; while (! S. Empty () {cout <''<S. Top (); S. Pop () ;}cout <Endl ;}return 0 ;}

Challenges of csdn programming-book Transfer