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小强有3 一个盒子 A,B,C 用来装书，所有的书(干脆n这项)无论是在按序号堆升序 A在，现在，所有他想要把所有的书 C 往里走。
每次他 A 以书架 a 本书(够不上全拿完)至 B，A 翻转箱，然后，从 B 搭 b 本书()到 C，B 箱子翻转。
然后反复操作，直到全部书都到了 C，求最后的C 里面书的顺序，具体见例子。
输入描写叙述：
输入由多组数据构成。每组数据占一行。3 个数。n(1<=n<=10^5),a(1<=a<=10^9) ,b(1<=b<=10^9)，含义见题目表述。以文件结尾。
输出描写叙述：
每组数据输出一行，首先输出数据组数，然后输出 n 个数，C 中书的排列。
答题说明:
输入例子：
4 2 1
输出例子
Case 1: 1 4 2 3
Hint
初始状态A:4321 B: 空 C:空
第一次: A->B A:21 B:34 C: 空
A 翻转 A:12 B:34 C: 空
B->C A:12 B: 4 C: 3
B 翻转 A:12 B:4 C: 3
第二次: A->B A:空 B:214 C:3
A 翻转 A:空 B:214 C:3
B->C A:空 B:14 C:23
B 翻转 A:空 B:41 C:23
第三次: B->C A:空 B:1 C:423
B 翻转 A:空 B:1 C:423
第四次: B->C A:空 B:空 C:1423
#include <iostream>
using namespace std;
class CBook
{
public:
int *Abox, *Bbox, *Cbox;
int num, aMoveBook, bMoveBook;
CBook(int bookNum, int a, int b);
~CBook();
void setup();
void init();
void getABox();
void getBBox();
void reverse(int *box, int num);
void outPutCase();
bool isEmpty();
void operator=(CBook bookSource)
{
memcpy(Abox, bookSource.Abox, (num+1) * sizeof(int));
memcpy(Bbox, bookSource.Bbox, (num+1) * sizeof(int));
memcpy(Cbox, bookSource.Cbox, (num+1) * sizeof(int));
}
};
void setup(CBook book);
int main()
{
CBook bookMove(8, 5, 3);
bookMove.init();
setup(bookMove);
return 1;
}
CBook::CBook(int bookNum, int a, int b)
{
Abox = new int[bookNum + 1];
Bbox = new int[bookNum + 1];
Cbox = new int[bookNum + 1];
num = bookNum;
aMoveBook = a;
bMoveBook = b;
}
CBook::~CBook()
{
}
void CBook::init()
{
int temp = num;
Abox[0] = num;
Bbox[0] = 0;
Cbox[0] = 0;
int i = 1;
while (temp)
{
Abox[temp--] = i++;
}
}
void CBook::getABox()
{
//Abox is empty
if (Abox[0] == 0)
{
return;
}
if (Abox[0] < aMoveBook)
{
int tail = Bbox[0];
//bbox's book move back
while (tail)
{
Bbox[tail + Abox[0]] = Bbox[tail--];
}
tail = Abox[0];
for (int i=0; i<Abox[0]; ++i)
{
Bbox[tail--] = Abox[i+1];
}
Bbox[0] += Abox[0];
Abox[0] = 0;
}
else
{
int tail = Bbox[0];
//bbox's book move back
while (tail)
{
Bbox[tail + aMoveBook] = Bbox[tail--];
}
tail = aMoveBook;
for (int i=0; i<aMoveBook; ++i)
{
Bbox[tail--] = Abox[i+1];
}
Bbox[0] += aMoveBook;
reverse(Abox, aMoveBook);
}
}
void CBook::getBBox()
{
//Bbox is empty
if (Bbox[0] == 0)
{
return;
}
if (Bbox[0] < bMoveBook)
{
int tail = Cbox[0];
//Cbox's book move back
while (tail)
{
Cbox[tail + Bbox[0]] = Cbox[tail--];
}
tail = Bbox[0];
for (int i=0; i<Bbox[0]; ++i)
{
Cbox[tail--] = Bbox[i+1];
}
Cbox[0] += Bbox[0];
Bbox[0] = 0;
}
else
{
//Cbox's book move back
int tail = Cbox[0];
while (tail)
{
Cbox[tail + bMoveBook] = Cbox[tail--];
}
tail = bMoveBook;
for (int i=0; i<bMoveBook; ++i)
{
Cbox[tail--] = Bbox[i+1];
}
Cbox[0] += bMoveBook;
reverse(Bbox, bMoveBook);
}
}
void CBook::reverse(int *book, int num)
{
int tail = book[0];
int changeNum = -1;
for (int i=1; i<=num; ++i)
{
book[i] = book[tail--];
}
int head = num + 1;
while (head < tail)
{
changeNum = book[head];
book[head++] = book[tail];
book[tail--] = changeNum;
}
book[0] -= num;
}
void CBook::outPutCase()
{
if (Cbox[0] == num)
{
for (int i=1; i<=num; ++i)
{
cout<<Cbox[i]<<" ";
}
cout<<endl;
}
}
bool CBook::isEmpty()
{
return Cbox[0] == num;
}
void setup(CBook book)
{
if (book.isEmpty())
{
return;
}
//case1
//get book from A
//output
CBook temp(book.num, book.aMoveBook, book.bMoveBook);
temp = book;
if (book.Abox[0] > 0)
{
book.getABox();
book.outPutCase();
setup(book);
}
//case2
//get book from B
//output
book = temp;
if (book.Bbox[0] > 0)
{
book.getBBox();
book.outPutCase();
setup(book);
}
}
版权声明：本文博主原创文章，博客，未经同意不得转载。
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原文地址：http://www.cnblogs.com/mengfanrong/p/4803506.html
