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 程式師世界 >> 編程語言 >> C語言 >> C++ >> C++入門知識 >> HDU-3328-Flipper(棧模擬)

HDU-3328-Flipper(棧模擬)

編輯:C++入門知識

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 . . . qm, where m is a positive integer and 1 ≤qin. Each qi 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 qi = 3, then the answer would be
Card 3 is a face up 4.

Sample Input
5
UUUDD
RRLL
5 1 2 3 4 5
10
UUDDUUDDUU
LLLRRRLRL
4 3 7 6 1
0

Sample Output
Pile 1
Card 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 2
Card 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.


思路:R操作就是把最右邊的一堆牌放到次右邊並改變每一張牌的方向,L操作就是把最左邊的一堆牌放到次左邊並改變每一張牌的方向。注意:每一對牌的最上面一個先取出,因此,直接用棧來模擬比較方便。

#include 
#include 
using namespace std;

struct S{
char s;
int num;
}que[101];

stacka[101];//用棧模擬

int main()
{
    int n,m,i,top,bottom,casenum=1;
    char c[101];
    struct S temp;

    while(~scanf("%d",&n) && n)
    {
        top=1;
        bottom=n;

        scanf("%s",c);

        for(i=1;i<=n;i++)
        {
            while(!a[i].empty())//讀入數據前先清棧
            {
                a[i].pop();
            }

            temp.s=c[i-1];
            temp.num=i;

            a[i].push(temp);//把數據壓棧
        }


        scanf("%s",c);

        for(i=1;i

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