hdu3328（Flipper）经典栈类

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Problem Description

Little Bobby Roberts (son of Big Bob, of Problem G) plays this solitaire memory game called Flipper. He starts withn cards, numbered 1 through n, and lays them out in a row with the cards in order left-to-right. (Card 1 is on the far left; cardn is on the far right.) Some cards are face up and some are face down. Bobby then performsn - 1 flips — either right flips or left flips. In a right flip he takes the pile to the far right and flips it over onto the card to its immediate left. For example, if the rightmost pile has cards A, B, C (from top to bottom) and card D is to the immediate left, then flipping the pile over onto card D would result in a pile of 4 cards: C, B, A, D (from top to bottom). A left flip is analogous.

The very last flip performed will result in one pile of cards — some face up, some face down. For example, suppose Bobby deals out 5 cards (numbered 1 through 5) with cards 1 through 3 initially face up and cards 4 and 5 initially face down. If Bobby performs 2 right flips, then 2 left flips, the pile will be (from top to bottom) a face down 2, a face up 1, a face up 4, a face down 5, and a face up 3.

Now Bobby is very sharp and you can ask him what card is in any position and he can tell you!!! You will write a program that matches Bobby’s amazing feat.

 

Input

Each test case will consist of 4 lines. The first line will be a positive integern (2 ≤ n ≤ 100) which is the number of cards laid out. The second line will be a string ofn characters. A character U indicates the corresponding card is dealt face up and a character D indicates the card is face down. The third line is a string ofn - 1 characters indicating the order of the flips Bobby performs. Each character is either R, indicating a right flip, or L, indicating a left flip. The fourth line is of the formm q₁ q₂ . . . q_m, where m is a positive integer and 1 ≤q_i ≤ n. Each q_i is a query on a position of a card in the pile (1 being the top card,n being the bottom card). A line containing 0 indicates end of input.

 

Output

Each test case should generate m + 1 lines of output. The first line is of the form

Pile t

where t is the number of the test case (starting at 1). Each of the nextm lines should be of the form

Card qi is a face up k.

or

Card qi is a face down k.

accordingly, for i = 1, ..,m, where k is the number of the card.
For instance, in the above example with 5 cards, if q_i = 3, then the answer would be

Card 3 is a face up 4.

 

Sample Input

5UUUDDRRLL5 1 2 3 4 510UUDDUUDDUULLLRRRLRL4 3 7 6 10

 

Sample Output

Pile 1Card 1 is a face down 2.Card 2 is a face up 1.Card 3 is a face up 4.Card 4 is a face down 5.Card 5 is a face up 3.Pile 2Card 3 is a face down 1.Card 7 is a face down 9.Card 6 is a face up 7.Card 1 is a face down 5.

题意：给你几张牌的朝向（上下），然后在给你一个操作方案：将最左或者最右的牌翻到其后或前的牌堆上，翻的牌的朝向改变。最后给你一组牌在一堆的位置，叫你输出其朝向和牌的值

思路：可以把初始化的每张牌都放在一个栈中，然后通过栈来翻动，很形象的体现了栈的特点。直到都合为一个栈位置（实际上 ，合成一个栈还是可以操作的，只是这里不必了）

package stack;import java.util.Scanner;import java.util.Stack;public class P3328 {public static void main(String[] args) {Scanner sc=new Scanner(System.in);int t=1;while(sc.hasNext()){int n=sc.nextInt();if(n==0){break;}String dirc=sc.next();//输入初始朝向String operate=sc.next();//输入操作步骤int m=sc.nextInt();int[] a=new int[m];for(int i=0;i<m;i++){a[i]=sc.nextInt();}Stack[] stack=new Stack[n];//首先每张牌放入一个栈类中（由于前面的坑，有阴影了，所以这里就直接用java包中的栈类了，但是尽量还是自己写好点）char[] ch=dirc.toCharArray();Card[] card=new Card[n];//将每个牌放入牌类中，牌有属性：牌的值，和朝向for(int i=0;i<n;i++){card[i]=new Card(i+1,ch[i]);stack[i]=new Stack();stack[i].push(card[i]);}int low=0,high=n-1,i=0;Card c;//R、L表示每次都从最左和最右向中间翻牌，直到都叠在一起（也就是low=high）while(true){if(low==high){break;}if(operate.charAt(i++)=='R'){while(stack[high].isEmpty()!=true){//将当前最右边栈中的所有牌都翻到其前一个栈中c=(Card)stack[high].pop();c.changeDir();stack[high-1].push(c);}high--;}else{while(stack[low].isEmpty()!=true){//将当前最左边栈中的所有牌都翻到其后一个栈中c=(Card)stack[low].pop();c.changeDir();stack[low+1].push(c);}low++;}//System.out.println(low+"((((((("+high);}i=0;while(stack[low].isEmpty()!=true){//将最后一个栈中的所有牌都放入一个牌数组中，因为结果要求按所给的数a[i]（也就是在栈中的位置），输出第a[i]张牌的情况card[i++]=(Card) stack[low].pop();//System.out.println("value="+c.value+" dir="+c.dir);}System.out.println("Pile "+t);t++;for(int j=0;j<m;j++){System.out.println("Card "+a[j]+" is a face "+card[a[j]-1].dir+" "+card[a[j]-1].value+".");}}}}class Card{int value;String dir;public Card(int value,char dir){this.value=value;if(dir=='U'){this.dir="up";}else{this.dir="down";}  }public void changeDir(){if(this.dir=="up"){this.dir="down";}else{this.dir="up";}}}